J. Duke | 319a3b9 | 2007-12-01 00:00:00 +0000 | [diff] [blame^] | 1 | /* |
| 2 | * Copyright 2003-2004 Sun Microsystems, Inc. All Rights Reserved. |
| 3 | * DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER. |
| 4 | * |
| 5 | * This code is free software; you can redistribute it and/or modify it |
| 6 | * under the terms of the GNU General Public License version 2 only, as |
| 7 | * published by the Free Software Foundation. Sun designates this |
| 8 | * particular file as subject to the "Classpath" exception as provided |
| 9 | * by Sun in the LICENSE file that accompanied this code. |
| 10 | * |
| 11 | * This code is distributed in the hope that it will be useful, but WITHOUT |
| 12 | * ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or |
| 13 | * FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License |
| 14 | * version 2 for more details (a copy is included in the LICENSE file that |
| 15 | * accompanied this code). |
| 16 | * |
| 17 | * You should have received a copy of the GNU General Public License version |
| 18 | * 2 along with this work; if not, write to the Free Software Foundation, |
| 19 | * Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA. |
| 20 | * |
| 21 | * Please contact Sun Microsystems, Inc., 4150 Network Circle, Santa Clara, |
| 22 | * CA 95054 USA or visit www.sun.com if you need additional information or |
| 23 | * have any questions. |
| 24 | */ |
| 25 | |
| 26 | package sun.misc; |
| 27 | |
| 28 | import sun.misc.FpUtils; |
| 29 | import sun.misc.DoubleConsts; |
| 30 | import sun.misc.FloatConsts; |
| 31 | import java.util.regex.*; |
| 32 | |
| 33 | public class FormattedFloatingDecimal{ |
| 34 | boolean isExceptional; |
| 35 | boolean isNegative; |
| 36 | int decExponent; // value set at construction, then immutable |
| 37 | int decExponentRounded; |
| 38 | char digits[]; |
| 39 | int nDigits; |
| 40 | int bigIntExp; |
| 41 | int bigIntNBits; |
| 42 | boolean mustSetRoundDir = false; |
| 43 | boolean fromHex = false; |
| 44 | int roundDir = 0; // set by doubleValue |
| 45 | int precision; // number of digits to the right of decimal |
| 46 | |
| 47 | public enum Form { SCIENTIFIC, COMPATIBLE, DECIMAL_FLOAT, GENERAL }; |
| 48 | |
| 49 | private Form form; |
| 50 | |
| 51 | private FormattedFloatingDecimal( boolean negSign, int decExponent, char []digits, int n, boolean e, int precision, Form form ) |
| 52 | { |
| 53 | isNegative = negSign; |
| 54 | isExceptional = e; |
| 55 | this.decExponent = decExponent; |
| 56 | this.digits = digits; |
| 57 | this.nDigits = n; |
| 58 | this.precision = precision; |
| 59 | this.form = form; |
| 60 | } |
| 61 | |
| 62 | /* |
| 63 | * Constants of the implementation |
| 64 | * Most are IEEE-754 related. |
| 65 | * (There are more really boring constants at the end.) |
| 66 | */ |
| 67 | static final long signMask = 0x8000000000000000L; |
| 68 | static final long expMask = 0x7ff0000000000000L; |
| 69 | static final long fractMask= ~(signMask|expMask); |
| 70 | static final int expShift = 52; |
| 71 | static final int expBias = 1023; |
| 72 | static final long fractHOB = ( 1L<<expShift ); // assumed High-Order bit |
| 73 | static final long expOne = ((long)expBias)<<expShift; // exponent of 1.0 |
| 74 | static final int maxSmallBinExp = 62; |
| 75 | static final int minSmallBinExp = -( 63 / 3 ); |
| 76 | static final int maxDecimalDigits = 15; |
| 77 | static final int maxDecimalExponent = 308; |
| 78 | static final int minDecimalExponent = -324; |
| 79 | static final int bigDecimalExponent = 324; // i.e. abs(minDecimalExponent) |
| 80 | |
| 81 | static final long highbyte = 0xff00000000000000L; |
| 82 | static final long highbit = 0x8000000000000000L; |
| 83 | static final long lowbytes = ~highbyte; |
| 84 | |
| 85 | static final int singleSignMask = 0x80000000; |
| 86 | static final int singleExpMask = 0x7f800000; |
| 87 | static final int singleFractMask = ~(singleSignMask|singleExpMask); |
| 88 | static final int singleExpShift = 23; |
| 89 | static final int singleFractHOB = 1<<singleExpShift; |
| 90 | static final int singleExpBias = 127; |
| 91 | static final int singleMaxDecimalDigits = 7; |
| 92 | static final int singleMaxDecimalExponent = 38; |
| 93 | static final int singleMinDecimalExponent = -45; |
| 94 | |
| 95 | static final int intDecimalDigits = 9; |
| 96 | |
| 97 | |
| 98 | /* |
| 99 | * count number of bits from high-order 1 bit to low-order 1 bit, |
| 100 | * inclusive. |
| 101 | */ |
| 102 | private static int |
| 103 | countBits( long v ){ |
| 104 | // |
| 105 | // the strategy is to shift until we get a non-zero sign bit |
| 106 | // then shift until we have no bits left, counting the difference. |
| 107 | // we do byte shifting as a hack. Hope it helps. |
| 108 | // |
| 109 | if ( v == 0L ) return 0; |
| 110 | |
| 111 | while ( ( v & highbyte ) == 0L ){ |
| 112 | v <<= 8; |
| 113 | } |
| 114 | while ( v > 0L ) { // i.e. while ((v&highbit) == 0L ) |
| 115 | v <<= 1; |
| 116 | } |
| 117 | |
| 118 | int n = 0; |
| 119 | while (( v & lowbytes ) != 0L ){ |
| 120 | v <<= 8; |
| 121 | n += 8; |
| 122 | } |
| 123 | while ( v != 0L ){ |
| 124 | v <<= 1; |
| 125 | n += 1; |
| 126 | } |
| 127 | return n; |
| 128 | } |
| 129 | |
| 130 | /* |
| 131 | * Keep big powers of 5 handy for future reference. |
| 132 | */ |
| 133 | private static FDBigInt b5p[]; |
| 134 | |
| 135 | private static synchronized FDBigInt |
| 136 | big5pow( int p ){ |
| 137 | assert p >= 0 : p; // negative power of 5 |
| 138 | if ( b5p == null ){ |
| 139 | b5p = new FDBigInt[ p+1 ]; |
| 140 | }else if (b5p.length <= p ){ |
| 141 | FDBigInt t[] = new FDBigInt[ p+1 ]; |
| 142 | System.arraycopy( b5p, 0, t, 0, b5p.length ); |
| 143 | b5p = t; |
| 144 | } |
| 145 | if ( b5p[p] != null ) |
| 146 | return b5p[p]; |
| 147 | else if ( p < small5pow.length ) |
| 148 | return b5p[p] = new FDBigInt( small5pow[p] ); |
| 149 | else if ( p < long5pow.length ) |
| 150 | return b5p[p] = new FDBigInt( long5pow[p] ); |
| 151 | else { |
| 152 | // construct the value. |
| 153 | // recursively. |
| 154 | int q, r; |
| 155 | // in order to compute 5^p, |
| 156 | // compute its square root, 5^(p/2) and square. |
| 157 | // or, let q = p / 2, r = p -q, then |
| 158 | // 5^p = 5^(q+r) = 5^q * 5^r |
| 159 | q = p >> 1; |
| 160 | r = p - q; |
| 161 | FDBigInt bigq = b5p[q]; |
| 162 | if ( bigq == null ) |
| 163 | bigq = big5pow ( q ); |
| 164 | if ( r < small5pow.length ){ |
| 165 | return (b5p[p] = bigq.mult( small5pow[r] ) ); |
| 166 | }else{ |
| 167 | FDBigInt bigr = b5p[ r ]; |
| 168 | if ( bigr == null ) |
| 169 | bigr = big5pow( r ); |
| 170 | return (b5p[p] = bigq.mult( bigr ) ); |
| 171 | } |
| 172 | } |
| 173 | } |
| 174 | |
| 175 | // |
| 176 | // a common operation |
| 177 | // |
| 178 | private static FDBigInt |
| 179 | multPow52( FDBigInt v, int p5, int p2 ){ |
| 180 | if ( p5 != 0 ){ |
| 181 | if ( p5 < small5pow.length ){ |
| 182 | v = v.mult( small5pow[p5] ); |
| 183 | } else { |
| 184 | v = v.mult( big5pow( p5 ) ); |
| 185 | } |
| 186 | } |
| 187 | if ( p2 != 0 ){ |
| 188 | v.lshiftMe( p2 ); |
| 189 | } |
| 190 | return v; |
| 191 | } |
| 192 | |
| 193 | // |
| 194 | // another common operation |
| 195 | // |
| 196 | private static FDBigInt |
| 197 | constructPow52( int p5, int p2 ){ |
| 198 | FDBigInt v = new FDBigInt( big5pow( p5 ) ); |
| 199 | if ( p2 != 0 ){ |
| 200 | v.lshiftMe( p2 ); |
| 201 | } |
| 202 | return v; |
| 203 | } |
| 204 | |
| 205 | /* |
| 206 | * Make a floating double into a FDBigInt. |
| 207 | * This could also be structured as a FDBigInt |
| 208 | * constructor, but we'd have to build a lot of knowledge |
| 209 | * about floating-point representation into it, and we don't want to. |
| 210 | * |
| 211 | * AS A SIDE EFFECT, THIS METHOD WILL SET THE INSTANCE VARIABLES |
| 212 | * bigIntExp and bigIntNBits |
| 213 | * |
| 214 | */ |
| 215 | private FDBigInt |
| 216 | doubleToBigInt( double dval ){ |
| 217 | long lbits = Double.doubleToLongBits( dval ) & ~signMask; |
| 218 | int binexp = (int)(lbits >>> expShift); |
| 219 | lbits &= fractMask; |
| 220 | if ( binexp > 0 ){ |
| 221 | lbits |= fractHOB; |
| 222 | } else { |
| 223 | assert lbits != 0L : lbits; // doubleToBigInt(0.0) |
| 224 | binexp +=1; |
| 225 | while ( (lbits & fractHOB ) == 0L){ |
| 226 | lbits <<= 1; |
| 227 | binexp -= 1; |
| 228 | } |
| 229 | } |
| 230 | binexp -= expBias; |
| 231 | int nbits = countBits( lbits ); |
| 232 | /* |
| 233 | * We now know where the high-order 1 bit is, |
| 234 | * and we know how many there are. |
| 235 | */ |
| 236 | int lowOrderZeros = expShift+1-nbits; |
| 237 | lbits >>>= lowOrderZeros; |
| 238 | |
| 239 | bigIntExp = binexp+1-nbits; |
| 240 | bigIntNBits = nbits; |
| 241 | return new FDBigInt( lbits ); |
| 242 | } |
| 243 | |
| 244 | /* |
| 245 | * Compute a number that is the ULP of the given value, |
| 246 | * for purposes of addition/subtraction. Generally easy. |
| 247 | * More difficult if subtracting and the argument |
| 248 | * is a normalized a power of 2, as the ULP changes at these points. |
| 249 | */ |
| 250 | private static double |
| 251 | ulp( double dval, boolean subtracting ){ |
| 252 | long lbits = Double.doubleToLongBits( dval ) & ~signMask; |
| 253 | int binexp = (int)(lbits >>> expShift); |
| 254 | double ulpval; |
| 255 | if ( subtracting && ( binexp >= expShift ) && ((lbits&fractMask) == 0L) ){ |
| 256 | // for subtraction from normalized, powers of 2, |
| 257 | // use next-smaller exponent |
| 258 | binexp -= 1; |
| 259 | } |
| 260 | if ( binexp > expShift ){ |
| 261 | ulpval = Double.longBitsToDouble( ((long)(binexp-expShift))<<expShift ); |
| 262 | } else if ( binexp == 0 ){ |
| 263 | ulpval = Double.MIN_VALUE; |
| 264 | } else { |
| 265 | ulpval = Double.longBitsToDouble( 1L<<(binexp-1) ); |
| 266 | } |
| 267 | if ( subtracting ) ulpval = - ulpval; |
| 268 | |
| 269 | return ulpval; |
| 270 | } |
| 271 | |
| 272 | /* |
| 273 | * Round a double to a float. |
| 274 | * In addition to the fraction bits of the double, |
| 275 | * look at the class instance variable roundDir, |
| 276 | * which should help us avoid double-rounding error. |
| 277 | * roundDir was set in hardValueOf if the estimate was |
| 278 | * close enough, but not exact. It tells us which direction |
| 279 | * of rounding is preferred. |
| 280 | */ |
| 281 | float |
| 282 | stickyRound( double dval ){ |
| 283 | long lbits = Double.doubleToLongBits( dval ); |
| 284 | long binexp = lbits & expMask; |
| 285 | if ( binexp == 0L || binexp == expMask ){ |
| 286 | // what we have here is special. |
| 287 | // don't worry, the right thing will happen. |
| 288 | return (float) dval; |
| 289 | } |
| 290 | lbits += (long)roundDir; // hack-o-matic. |
| 291 | return (float)Double.longBitsToDouble( lbits ); |
| 292 | } |
| 293 | |
| 294 | |
| 295 | /* |
| 296 | * This is the easy subcase -- |
| 297 | * all the significant bits, after scaling, are held in lvalue. |
| 298 | * negSign and decExponent tell us what processing and scaling |
| 299 | * has already been done. Exceptional cases have already been |
| 300 | * stripped out. |
| 301 | * In particular: |
| 302 | * lvalue is a finite number (not Inf, nor NaN) |
| 303 | * lvalue > 0L (not zero, nor negative). |
| 304 | * |
| 305 | * The only reason that we develop the digits here, rather than |
| 306 | * calling on Long.toString() is that we can do it a little faster, |
| 307 | * and besides want to treat trailing 0s specially. If Long.toString |
| 308 | * changes, we should re-evaluate this strategy! |
| 309 | */ |
| 310 | private void |
| 311 | developLongDigits( int decExponent, long lvalue, long insignificant ){ |
| 312 | char digits[]; |
| 313 | int ndigits; |
| 314 | int digitno; |
| 315 | int c; |
| 316 | // |
| 317 | // Discard non-significant low-order bits, while rounding, |
| 318 | // up to insignificant value. |
| 319 | int i; |
| 320 | for ( i = 0; insignificant >= 10L; i++ ) |
| 321 | insignificant /= 10L; |
| 322 | if ( i != 0 ){ |
| 323 | long pow10 = long5pow[i] << i; // 10^i == 5^i * 2^i; |
| 324 | long residue = lvalue % pow10; |
| 325 | lvalue /= pow10; |
| 326 | decExponent += i; |
| 327 | if ( residue >= (pow10>>1) ){ |
| 328 | // round up based on the low-order bits we're discarding |
| 329 | lvalue++; |
| 330 | } |
| 331 | } |
| 332 | if ( lvalue <= Integer.MAX_VALUE ){ |
| 333 | assert lvalue > 0L : lvalue; // lvalue <= 0 |
| 334 | // even easier subcase! |
| 335 | // can do int arithmetic rather than long! |
| 336 | int ivalue = (int)lvalue; |
| 337 | ndigits = 10; |
| 338 | digits = (char[])(perThreadBuffer.get()); |
| 339 | digitno = ndigits-1; |
| 340 | c = ivalue%10; |
| 341 | ivalue /= 10; |
| 342 | while ( c == 0 ){ |
| 343 | decExponent++; |
| 344 | c = ivalue%10; |
| 345 | ivalue /= 10; |
| 346 | } |
| 347 | while ( ivalue != 0){ |
| 348 | digits[digitno--] = (char)(c+'0'); |
| 349 | decExponent++; |
| 350 | c = ivalue%10; |
| 351 | ivalue /= 10; |
| 352 | } |
| 353 | digits[digitno] = (char)(c+'0'); |
| 354 | } else { |
| 355 | // same algorithm as above (same bugs, too ) |
| 356 | // but using long arithmetic. |
| 357 | ndigits = 20; |
| 358 | digits = (char[])(perThreadBuffer.get()); |
| 359 | digitno = ndigits-1; |
| 360 | c = (int)(lvalue%10L); |
| 361 | lvalue /= 10L; |
| 362 | while ( c == 0 ){ |
| 363 | decExponent++; |
| 364 | c = (int)(lvalue%10L); |
| 365 | lvalue /= 10L; |
| 366 | } |
| 367 | while ( lvalue != 0L ){ |
| 368 | digits[digitno--] = (char)(c+'0'); |
| 369 | decExponent++; |
| 370 | c = (int)(lvalue%10L); |
| 371 | lvalue /= 10; |
| 372 | } |
| 373 | digits[digitno] = (char)(c+'0'); |
| 374 | } |
| 375 | char result []; |
| 376 | ndigits -= digitno; |
| 377 | result = new char[ ndigits ]; |
| 378 | System.arraycopy( digits, digitno, result, 0, ndigits ); |
| 379 | this.digits = result; |
| 380 | this.decExponent = decExponent+1; |
| 381 | this.nDigits = ndigits; |
| 382 | } |
| 383 | |
| 384 | // |
| 385 | // add one to the least significant digit. |
| 386 | // in the unlikely event there is a carry out, |
| 387 | // deal with it. |
| 388 | // assert that this will only happen where there |
| 389 | // is only one digit, e.g. (float)1e-44 seems to do it. |
| 390 | // |
| 391 | private void |
| 392 | roundup(){ |
| 393 | int i; |
| 394 | int q = digits[ i = (nDigits-1)]; |
| 395 | if ( q == '9' ){ |
| 396 | while ( q == '9' && i > 0 ){ |
| 397 | digits[i] = '0'; |
| 398 | q = digits[--i]; |
| 399 | } |
| 400 | if ( q == '9' ){ |
| 401 | // carryout! High-order 1, rest 0s, larger exp. |
| 402 | decExponent += 1; |
| 403 | digits[0] = '1'; |
| 404 | return; |
| 405 | } |
| 406 | // else fall through. |
| 407 | } |
| 408 | digits[i] = (char)(q+1); |
| 409 | } |
| 410 | |
| 411 | // Given the desired number of digits predict the result's exponent. |
| 412 | private int checkExponent(int length) { |
| 413 | if (length >= nDigits || length < 0) |
| 414 | return decExponent; |
| 415 | |
| 416 | for (int i = 0; i < length; i++) |
| 417 | if (digits[i] != '9') |
| 418 | // a '9' anywhere in digits will absorb the round |
| 419 | return decExponent; |
| 420 | return decExponent + (digits[length] >= '5' ? 1 : 0); |
| 421 | } |
| 422 | |
| 423 | // Unlike roundup(), this method does not modify digits. It also |
| 424 | // rounds at a particular precision. |
| 425 | private char [] applyPrecision(int length) { |
| 426 | char [] result = new char[nDigits]; |
| 427 | for (int i = 0; i < result.length; i++) result[i] = '0'; |
| 428 | |
| 429 | if (length >= nDigits || length < 0) { |
| 430 | // no rounding necessary |
| 431 | System.arraycopy(digits, 0, result, 0, nDigits); |
| 432 | return result; |
| 433 | } |
| 434 | if (length == 0) { |
| 435 | // only one digit (0 or 1) is returned because the precision |
| 436 | // excludes all significant digits |
| 437 | if (digits[0] >= '5') { |
| 438 | result[0] = '1'; |
| 439 | } |
| 440 | return result; |
| 441 | } |
| 442 | |
| 443 | int i = length; |
| 444 | int q = digits[i]; |
| 445 | if (q >= '5' && i > 0) { |
| 446 | q = digits[--i]; |
| 447 | if ( q == '9' ) { |
| 448 | while ( q == '9' && i > 0 ){ |
| 449 | q = digits[--i]; |
| 450 | } |
| 451 | if ( q == '9' ){ |
| 452 | // carryout! High-order 1, rest 0s, larger exp. |
| 453 | result[0] = '1'; |
| 454 | return result; |
| 455 | } |
| 456 | } |
| 457 | result[i] = (char)(q + 1); |
| 458 | } |
| 459 | while (--i >= 0) { |
| 460 | result[i] = digits[i]; |
| 461 | } |
| 462 | return result; |
| 463 | } |
| 464 | |
| 465 | /* |
| 466 | * FIRST IMPORTANT CONSTRUCTOR: DOUBLE |
| 467 | */ |
| 468 | public FormattedFloatingDecimal( double d ) |
| 469 | { |
| 470 | this(d, Integer.MAX_VALUE, Form.COMPATIBLE); |
| 471 | } |
| 472 | |
| 473 | public FormattedFloatingDecimal( double d, int precision, Form form ) |
| 474 | { |
| 475 | long dBits = Double.doubleToLongBits( d ); |
| 476 | long fractBits; |
| 477 | int binExp; |
| 478 | int nSignificantBits; |
| 479 | |
| 480 | this.precision = precision; |
| 481 | this.form = form; |
| 482 | |
| 483 | // discover and delete sign |
| 484 | if ( (dBits&signMask) != 0 ){ |
| 485 | isNegative = true; |
| 486 | dBits ^= signMask; |
| 487 | } else { |
| 488 | isNegative = false; |
| 489 | } |
| 490 | // Begin to unpack |
| 491 | // Discover obvious special cases of NaN and Infinity. |
| 492 | binExp = (int)( (dBits&expMask) >> expShift ); |
| 493 | fractBits = dBits&fractMask; |
| 494 | if ( binExp == (int)(expMask>>expShift) ) { |
| 495 | isExceptional = true; |
| 496 | if ( fractBits == 0L ){ |
| 497 | digits = infinity; |
| 498 | } else { |
| 499 | digits = notANumber; |
| 500 | isNegative = false; // NaN has no sign! |
| 501 | } |
| 502 | nDigits = digits.length; |
| 503 | return; |
| 504 | } |
| 505 | isExceptional = false; |
| 506 | // Finish unpacking |
| 507 | // Normalize denormalized numbers. |
| 508 | // Insert assumed high-order bit for normalized numbers. |
| 509 | // Subtract exponent bias. |
| 510 | if ( binExp == 0 ){ |
| 511 | if ( fractBits == 0L ){ |
| 512 | // not a denorm, just a 0! |
| 513 | decExponent = 0; |
| 514 | digits = zero; |
| 515 | nDigits = 1; |
| 516 | return; |
| 517 | } |
| 518 | while ( (fractBits&fractHOB) == 0L ){ |
| 519 | fractBits <<= 1; |
| 520 | binExp -= 1; |
| 521 | } |
| 522 | nSignificantBits = expShift + binExp +1; // recall binExp is - shift count. |
| 523 | binExp += 1; |
| 524 | } else { |
| 525 | fractBits |= fractHOB; |
| 526 | nSignificantBits = expShift+1; |
| 527 | } |
| 528 | binExp -= expBias; |
| 529 | // call the routine that actually does all the hard work. |
| 530 | dtoa( binExp, fractBits, nSignificantBits ); |
| 531 | } |
| 532 | |
| 533 | /* |
| 534 | * SECOND IMPORTANT CONSTRUCTOR: SINGLE |
| 535 | */ |
| 536 | public FormattedFloatingDecimal( float f ) |
| 537 | { |
| 538 | this(f, Integer.MAX_VALUE, Form.COMPATIBLE); |
| 539 | } |
| 540 | public FormattedFloatingDecimal( float f, int precision, Form form ) |
| 541 | { |
| 542 | int fBits = Float.floatToIntBits( f ); |
| 543 | int fractBits; |
| 544 | int binExp; |
| 545 | int nSignificantBits; |
| 546 | |
| 547 | this.precision = precision; |
| 548 | this.form = form; |
| 549 | |
| 550 | // discover and delete sign |
| 551 | if ( (fBits&singleSignMask) != 0 ){ |
| 552 | isNegative = true; |
| 553 | fBits ^= singleSignMask; |
| 554 | } else { |
| 555 | isNegative = false; |
| 556 | } |
| 557 | // Begin to unpack |
| 558 | // Discover obvious special cases of NaN and Infinity. |
| 559 | binExp = (int)( (fBits&singleExpMask) >> singleExpShift ); |
| 560 | fractBits = fBits&singleFractMask; |
| 561 | if ( binExp == (int)(singleExpMask>>singleExpShift) ) { |
| 562 | isExceptional = true; |
| 563 | if ( fractBits == 0L ){ |
| 564 | digits = infinity; |
| 565 | } else { |
| 566 | digits = notANumber; |
| 567 | isNegative = false; // NaN has no sign! |
| 568 | } |
| 569 | nDigits = digits.length; |
| 570 | return; |
| 571 | } |
| 572 | isExceptional = false; |
| 573 | // Finish unpacking |
| 574 | // Normalize denormalized numbers. |
| 575 | // Insert assumed high-order bit for normalized numbers. |
| 576 | // Subtract exponent bias. |
| 577 | if ( binExp == 0 ){ |
| 578 | if ( fractBits == 0 ){ |
| 579 | // not a denorm, just a 0! |
| 580 | decExponent = 0; |
| 581 | digits = zero; |
| 582 | nDigits = 1; |
| 583 | return; |
| 584 | } |
| 585 | while ( (fractBits&singleFractHOB) == 0 ){ |
| 586 | fractBits <<= 1; |
| 587 | binExp -= 1; |
| 588 | } |
| 589 | nSignificantBits = singleExpShift + binExp +1; // recall binExp is - shift count. |
| 590 | binExp += 1; |
| 591 | } else { |
| 592 | fractBits |= singleFractHOB; |
| 593 | nSignificantBits = singleExpShift+1; |
| 594 | } |
| 595 | binExp -= singleExpBias; |
| 596 | // call the routine that actually does all the hard work. |
| 597 | dtoa( binExp, ((long)fractBits)<<(expShift-singleExpShift), nSignificantBits ); |
| 598 | } |
| 599 | |
| 600 | private void |
| 601 | dtoa( int binExp, long fractBits, int nSignificantBits ) |
| 602 | { |
| 603 | int nFractBits; // number of significant bits of fractBits; |
| 604 | int nTinyBits; // number of these to the right of the point. |
| 605 | int decExp; |
| 606 | |
| 607 | // Examine number. Determine if it is an easy case, |
| 608 | // which we can do pretty trivially using float/long conversion, |
| 609 | // or whether we must do real work. |
| 610 | nFractBits = countBits( fractBits ); |
| 611 | nTinyBits = Math.max( 0, nFractBits - binExp - 1 ); |
| 612 | if ( binExp <= maxSmallBinExp && binExp >= minSmallBinExp ){ |
| 613 | // Look more closely at the number to decide if, |
| 614 | // with scaling by 10^nTinyBits, the result will fit in |
| 615 | // a long. |
| 616 | if ( (nTinyBits < long5pow.length) && ((nFractBits + n5bits[nTinyBits]) < 64 ) ){ |
| 617 | /* |
| 618 | * We can do this: |
| 619 | * take the fraction bits, which are normalized. |
| 620 | * (a) nTinyBits == 0: Shift left or right appropriately |
| 621 | * to align the binary point at the extreme right, i.e. |
| 622 | * where a long int point is expected to be. The integer |
| 623 | * result is easily converted to a string. |
| 624 | * (b) nTinyBits > 0: Shift right by expShift-nFractBits, |
| 625 | * which effectively converts to long and scales by |
| 626 | * 2^nTinyBits. Then multiply by 5^nTinyBits to |
| 627 | * complete the scaling. We know this won't overflow |
| 628 | * because we just counted the number of bits necessary |
| 629 | * in the result. The integer you get from this can |
| 630 | * then be converted to a string pretty easily. |
| 631 | */ |
| 632 | long halfULP; |
| 633 | if ( nTinyBits == 0 ) { |
| 634 | if ( binExp > nSignificantBits ){ |
| 635 | halfULP = 1L << ( binExp-nSignificantBits-1); |
| 636 | } else { |
| 637 | halfULP = 0L; |
| 638 | } |
| 639 | if ( binExp >= expShift ){ |
| 640 | fractBits <<= (binExp-expShift); |
| 641 | } else { |
| 642 | fractBits >>>= (expShift-binExp) ; |
| 643 | } |
| 644 | developLongDigits( 0, fractBits, halfULP ); |
| 645 | return; |
| 646 | } |
| 647 | /* |
| 648 | * The following causes excess digits to be printed |
| 649 | * out in the single-float case. Our manipulation of |
| 650 | * halfULP here is apparently not correct. If we |
| 651 | * better understand how this works, perhaps we can |
| 652 | * use this special case again. But for the time being, |
| 653 | * we do not. |
| 654 | * else { |
| 655 | * fractBits >>>= expShift+1-nFractBits; |
| 656 | * fractBits *= long5pow[ nTinyBits ]; |
| 657 | * halfULP = long5pow[ nTinyBits ] >> (1+nSignificantBits-nFractBits); |
| 658 | * developLongDigits( -nTinyBits, fractBits, halfULP ); |
| 659 | * return; |
| 660 | * } |
| 661 | */ |
| 662 | } |
| 663 | } |
| 664 | /* |
| 665 | * This is the hard case. We are going to compute large positive |
| 666 | * integers B and S and integer decExp, s.t. |
| 667 | * d = ( B / S ) * 10^decExp |
| 668 | * 1 <= B / S < 10 |
| 669 | * Obvious choices are: |
| 670 | * decExp = floor( log10(d) ) |
| 671 | * B = d * 2^nTinyBits * 10^max( 0, -decExp ) |
| 672 | * S = 10^max( 0, decExp) * 2^nTinyBits |
| 673 | * (noting that nTinyBits has already been forced to non-negative) |
| 674 | * I am also going to compute a large positive integer |
| 675 | * M = (1/2^nSignificantBits) * 2^nTinyBits * 10^max( 0, -decExp ) |
| 676 | * i.e. M is (1/2) of the ULP of d, scaled like B. |
| 677 | * When we iterate through dividing B/S and picking off the |
| 678 | * quotient bits, we will know when to stop when the remainder |
| 679 | * is <= M. |
| 680 | * |
| 681 | * We keep track of powers of 2 and powers of 5. |
| 682 | */ |
| 683 | |
| 684 | /* |
| 685 | * Estimate decimal exponent. (If it is small-ish, |
| 686 | * we could double-check.) |
| 687 | * |
| 688 | * First, scale the mantissa bits such that 1 <= d2 < 2. |
| 689 | * We are then going to estimate |
| 690 | * log10(d2) ~=~ (d2-1.5)/1.5 + log(1.5) |
| 691 | * and so we can estimate |
| 692 | * log10(d) ~=~ log10(d2) + binExp * log10(2) |
| 693 | * take the floor and call it decExp. |
| 694 | * FIXME -- use more precise constants here. It costs no more. |
| 695 | */ |
| 696 | double d2 = Double.longBitsToDouble( |
| 697 | expOne | ( fractBits &~ fractHOB ) ); |
| 698 | decExp = (int)Math.floor( |
| 699 | (d2-1.5D)*0.289529654D + 0.176091259 + (double)binExp * 0.301029995663981 ); |
| 700 | int B2, B5; // powers of 2 and powers of 5, respectively, in B |
| 701 | int S2, S5; // powers of 2 and powers of 5, respectively, in S |
| 702 | int M2, M5; // powers of 2 and powers of 5, respectively, in M |
| 703 | int Bbits; // binary digits needed to represent B, approx. |
| 704 | int tenSbits; // binary digits needed to represent 10*S, approx. |
| 705 | FDBigInt Sval, Bval, Mval; |
| 706 | |
| 707 | B5 = Math.max( 0, -decExp ); |
| 708 | B2 = B5 + nTinyBits + binExp; |
| 709 | |
| 710 | S5 = Math.max( 0, decExp ); |
| 711 | S2 = S5 + nTinyBits; |
| 712 | |
| 713 | M5 = B5; |
| 714 | M2 = B2 - nSignificantBits; |
| 715 | |
| 716 | /* |
| 717 | * the long integer fractBits contains the (nFractBits) interesting |
| 718 | * bits from the mantissa of d ( hidden 1 added if necessary) followed |
| 719 | * by (expShift+1-nFractBits) zeros. In the interest of compactness, |
| 720 | * I will shift out those zeros before turning fractBits into a |
| 721 | * FDBigInt. The resulting whole number will be |
| 722 | * d * 2^(nFractBits-1-binExp). |
| 723 | */ |
| 724 | fractBits >>>= (expShift+1-nFractBits); |
| 725 | B2 -= nFractBits-1; |
| 726 | int common2factor = Math.min( B2, S2 ); |
| 727 | B2 -= common2factor; |
| 728 | S2 -= common2factor; |
| 729 | M2 -= common2factor; |
| 730 | |
| 731 | /* |
| 732 | * HACK!! For exact powers of two, the next smallest number |
| 733 | * is only half as far away as we think (because the meaning of |
| 734 | * ULP changes at power-of-two bounds) for this reason, we |
| 735 | * hack M2. Hope this works. |
| 736 | */ |
| 737 | if ( nFractBits == 1 ) |
| 738 | M2 -= 1; |
| 739 | |
| 740 | if ( M2 < 0 ){ |
| 741 | // oops. |
| 742 | // since we cannot scale M down far enough, |
| 743 | // we must scale the other values up. |
| 744 | B2 -= M2; |
| 745 | S2 -= M2; |
| 746 | M2 = 0; |
| 747 | } |
| 748 | /* |
| 749 | * Construct, Scale, iterate. |
| 750 | * Some day, we'll write a stopping test that takes |
| 751 | * account of the assymetry of the spacing of floating-point |
| 752 | * numbers below perfect powers of 2 |
| 753 | * 26 Sept 96 is not that day. |
| 754 | * So we use a symmetric test. |
| 755 | */ |
| 756 | char digits[] = this.digits = new char[18]; |
| 757 | int ndigit = 0; |
| 758 | boolean low, high; |
| 759 | long lowDigitDifference; |
| 760 | int q; |
| 761 | |
| 762 | /* |
| 763 | * Detect the special cases where all the numbers we are about |
| 764 | * to compute will fit in int or long integers. |
| 765 | * In these cases, we will avoid doing FDBigInt arithmetic. |
| 766 | * We use the same algorithms, except that we "normalize" |
| 767 | * our FDBigInts before iterating. This is to make division easier, |
| 768 | * as it makes our fist guess (quotient of high-order words) |
| 769 | * more accurate! |
| 770 | * |
| 771 | * Some day, we'll write a stopping test that takes |
| 772 | * account of the assymetry of the spacing of floating-point |
| 773 | * numbers below perfect powers of 2 |
| 774 | * 26 Sept 96 is not that day. |
| 775 | * So we use a symmetric test. |
| 776 | */ |
| 777 | Bbits = nFractBits + B2 + (( B5 < n5bits.length )? n5bits[B5] : ( B5*3 )); |
| 778 | tenSbits = S2+1 + (( (S5+1) < n5bits.length )? n5bits[(S5+1)] : ( (S5+1)*3 )); |
| 779 | if ( Bbits < 64 && tenSbits < 64){ |
| 780 | if ( Bbits < 32 && tenSbits < 32){ |
| 781 | // wa-hoo! They're all ints! |
| 782 | int b = ((int)fractBits * small5pow[B5] ) << B2; |
| 783 | int s = small5pow[S5] << S2; |
| 784 | int m = small5pow[M5] << M2; |
| 785 | int tens = s * 10; |
| 786 | /* |
| 787 | * Unroll the first iteration. If our decExp estimate |
| 788 | * was too high, our first quotient will be zero. In this |
| 789 | * case, we discard it and decrement decExp. |
| 790 | */ |
| 791 | ndigit = 0; |
| 792 | q = b / s; |
| 793 | b = 10 * ( b % s ); |
| 794 | m *= 10; |
| 795 | low = (b < m ); |
| 796 | high = (b+m > tens ); |
| 797 | assert q < 10 : q; // excessively large digit |
| 798 | if ( (q == 0) && ! high ){ |
| 799 | // oops. Usually ignore leading zero. |
| 800 | decExp--; |
| 801 | } else { |
| 802 | digits[ndigit++] = (char)('0' + q); |
| 803 | } |
| 804 | /* |
| 805 | * HACK! Java spec sez that we always have at least |
| 806 | * one digit after the . in either F- or E-form output. |
| 807 | * Thus we will need more than one digit if we're using |
| 808 | * E-form |
| 809 | */ |
| 810 | if (! (form == Form.COMPATIBLE && -3 < decExp && decExp < 8)) { |
| 811 | high = low = false; |
| 812 | } |
| 813 | while( ! low && ! high ){ |
| 814 | q = b / s; |
| 815 | b = 10 * ( b % s ); |
| 816 | m *= 10; |
| 817 | assert q < 10 : q; // excessively large digit |
| 818 | if ( m > 0L ){ |
| 819 | low = (b < m ); |
| 820 | high = (b+m > tens ); |
| 821 | } else { |
| 822 | // hack -- m might overflow! |
| 823 | // in this case, it is certainly > b, |
| 824 | // which won't |
| 825 | // and b+m > tens, too, since that has overflowed |
| 826 | // either! |
| 827 | low = true; |
| 828 | high = true; |
| 829 | } |
| 830 | digits[ndigit++] = (char)('0' + q); |
| 831 | } |
| 832 | lowDigitDifference = (b<<1) - tens; |
| 833 | } else { |
| 834 | // still good! they're all longs! |
| 835 | long b = (fractBits * long5pow[B5] ) << B2; |
| 836 | long s = long5pow[S5] << S2; |
| 837 | long m = long5pow[M5] << M2; |
| 838 | long tens = s * 10L; |
| 839 | /* |
| 840 | * Unroll the first iteration. If our decExp estimate |
| 841 | * was too high, our first quotient will be zero. In this |
| 842 | * case, we discard it and decrement decExp. |
| 843 | */ |
| 844 | ndigit = 0; |
| 845 | q = (int) ( b / s ); |
| 846 | b = 10L * ( b % s ); |
| 847 | m *= 10L; |
| 848 | low = (b < m ); |
| 849 | high = (b+m > tens ); |
| 850 | assert q < 10 : q; // excessively large digit |
| 851 | if ( (q == 0) && ! high ){ |
| 852 | // oops. Usually ignore leading zero. |
| 853 | decExp--; |
| 854 | } else { |
| 855 | digits[ndigit++] = (char)('0' + q); |
| 856 | } |
| 857 | /* |
| 858 | * HACK! Java spec sez that we always have at least |
| 859 | * one digit after the . in either F- or E-form output. |
| 860 | * Thus we will need more than one digit if we're using |
| 861 | * E-form |
| 862 | */ |
| 863 | if (! (form == Form.COMPATIBLE && -3 < decExp && decExp < 8)) { |
| 864 | high = low = false; |
| 865 | } |
| 866 | while( ! low && ! high ){ |
| 867 | q = (int) ( b / s ); |
| 868 | b = 10 * ( b % s ); |
| 869 | m *= 10; |
| 870 | assert q < 10 : q; // excessively large digit |
| 871 | if ( m > 0L ){ |
| 872 | low = (b < m ); |
| 873 | high = (b+m > tens ); |
| 874 | } else { |
| 875 | // hack -- m might overflow! |
| 876 | // in this case, it is certainly > b, |
| 877 | // which won't |
| 878 | // and b+m > tens, too, since that has overflowed |
| 879 | // either! |
| 880 | low = true; |
| 881 | high = true; |
| 882 | } |
| 883 | digits[ndigit++] = (char)('0' + q); |
| 884 | } |
| 885 | lowDigitDifference = (b<<1) - tens; |
| 886 | } |
| 887 | } else { |
| 888 | FDBigInt tenSval; |
| 889 | int shiftBias; |
| 890 | |
| 891 | /* |
| 892 | * We really must do FDBigInt arithmetic. |
| 893 | * Fist, construct our FDBigInt initial values. |
| 894 | */ |
| 895 | Bval = multPow52( new FDBigInt( fractBits ), B5, B2 ); |
| 896 | Sval = constructPow52( S5, S2 ); |
| 897 | Mval = constructPow52( M5, M2 ); |
| 898 | |
| 899 | |
| 900 | // normalize so that division works better |
| 901 | Bval.lshiftMe( shiftBias = Sval.normalizeMe() ); |
| 902 | Mval.lshiftMe( shiftBias ); |
| 903 | tenSval = Sval.mult( 10 ); |
| 904 | /* |
| 905 | * Unroll the first iteration. If our decExp estimate |
| 906 | * was too high, our first quotient will be zero. In this |
| 907 | * case, we discard it and decrement decExp. |
| 908 | */ |
| 909 | ndigit = 0; |
| 910 | q = Bval.quoRemIteration( Sval ); |
| 911 | Mval = Mval.mult( 10 ); |
| 912 | low = (Bval.cmp( Mval ) < 0); |
| 913 | high = (Bval.add( Mval ).cmp( tenSval ) > 0 ); |
| 914 | assert q < 10 : q; // excessively large digit |
| 915 | if ( (q == 0) && ! high ){ |
| 916 | // oops. Usually ignore leading zero. |
| 917 | decExp--; |
| 918 | } else { |
| 919 | digits[ndigit++] = (char)('0' + q); |
| 920 | } |
| 921 | /* |
| 922 | * HACK! Java spec sez that we always have at least |
| 923 | * one digit after the . in either F- or E-form output. |
| 924 | * Thus we will need more than one digit if we're using |
| 925 | * E-form |
| 926 | */ |
| 927 | if (! (form == Form.COMPATIBLE && -3 < decExp && decExp < 8)) { |
| 928 | high = low = false; |
| 929 | } |
| 930 | while( ! low && ! high ){ |
| 931 | q = Bval.quoRemIteration( Sval ); |
| 932 | Mval = Mval.mult( 10 ); |
| 933 | assert q < 10 : q; // excessively large digit |
| 934 | low = (Bval.cmp( Mval ) < 0); |
| 935 | high = (Bval.add( Mval ).cmp( tenSval ) > 0 ); |
| 936 | digits[ndigit++] = (char)('0' + q); |
| 937 | } |
| 938 | if ( high && low ){ |
| 939 | Bval.lshiftMe(1); |
| 940 | lowDigitDifference = Bval.cmp(tenSval); |
| 941 | } else |
| 942 | lowDigitDifference = 0L; // this here only for flow analysis! |
| 943 | } |
| 944 | this.decExponent = decExp+1; |
| 945 | this.digits = digits; |
| 946 | this.nDigits = ndigit; |
| 947 | /* |
| 948 | * Last digit gets rounded based on stopping condition. |
| 949 | */ |
| 950 | if ( high ){ |
| 951 | if ( low ){ |
| 952 | if ( lowDigitDifference == 0L ){ |
| 953 | // it's a tie! |
| 954 | // choose based on which digits we like. |
| 955 | if ( (digits[nDigits-1]&1) != 0 ) roundup(); |
| 956 | } else if ( lowDigitDifference > 0 ){ |
| 957 | roundup(); |
| 958 | } |
| 959 | } else { |
| 960 | roundup(); |
| 961 | } |
| 962 | } |
| 963 | } |
| 964 | |
| 965 | public String |
| 966 | toString(){ |
| 967 | // most brain-dead version |
| 968 | StringBuffer result = new StringBuffer( nDigits+8 ); |
| 969 | if ( isNegative ){ result.append( '-' ); } |
| 970 | if ( isExceptional ){ |
| 971 | result.append( digits, 0, nDigits ); |
| 972 | } else { |
| 973 | result.append( "0."); |
| 974 | result.append( digits, 0, nDigits ); |
| 975 | result.append('e'); |
| 976 | result.append( decExponent ); |
| 977 | } |
| 978 | return new String(result); |
| 979 | } |
| 980 | |
| 981 | // This method should only ever be called if this object is constructed |
| 982 | // without Form.DECIMAL_FLOAT because the perThreadBuffer is not large |
| 983 | // enough to handle floating-point numbers of large precision. |
| 984 | public String toJavaFormatString() { |
| 985 | char result[] = (char[])(perThreadBuffer.get()); |
| 986 | int i = getChars(result); |
| 987 | return new String(result, 0, i); |
| 988 | } |
| 989 | |
| 990 | // returns the exponent before rounding |
| 991 | public int getExponent() { |
| 992 | return decExponent - 1; |
| 993 | } |
| 994 | |
| 995 | // returns the exponent after rounding has been done by applyPrecision |
| 996 | public int getExponentRounded() { |
| 997 | return decExponentRounded - 1; |
| 998 | } |
| 999 | |
| 1000 | public int getChars(char[] result) { |
| 1001 | assert nDigits <= 19 : nDigits; // generous bound on size of nDigits |
| 1002 | int i = 0; |
| 1003 | if (isNegative) { result[0] = '-'; i = 1; } |
| 1004 | if (isExceptional) { |
| 1005 | System.arraycopy(digits, 0, result, i, nDigits); |
| 1006 | i += nDigits; |
| 1007 | } else { |
| 1008 | char digits [] = this.digits; |
| 1009 | int exp = decExponent; |
| 1010 | switch (form) { |
| 1011 | case COMPATIBLE: |
| 1012 | break; |
| 1013 | case DECIMAL_FLOAT: |
| 1014 | exp = checkExponent(decExponent + precision); |
| 1015 | digits = applyPrecision(decExponent + precision); |
| 1016 | break; |
| 1017 | case SCIENTIFIC: |
| 1018 | exp = checkExponent(precision + 1); |
| 1019 | digits = applyPrecision(precision + 1); |
| 1020 | break; |
| 1021 | case GENERAL: |
| 1022 | exp = checkExponent(precision); |
| 1023 | digits = applyPrecision(precision); |
| 1024 | // adjust precision to be the number of digits to right of decimal |
| 1025 | // the real exponent to be output is actually exp - 1, not exp |
| 1026 | if (exp - 1 < -4 || exp - 1 >= precision) { |
| 1027 | form = Form.SCIENTIFIC; |
| 1028 | precision--; |
| 1029 | } else { |
| 1030 | form = Form.DECIMAL_FLOAT; |
| 1031 | precision = precision - exp; |
| 1032 | } |
| 1033 | break; |
| 1034 | default: |
| 1035 | assert false; |
| 1036 | } |
| 1037 | decExponentRounded = exp; |
| 1038 | |
| 1039 | if (exp > 0 |
| 1040 | && ((form == Form.COMPATIBLE && (exp < 8)) |
| 1041 | || (form == Form.DECIMAL_FLOAT))) |
| 1042 | { |
| 1043 | // print digits.digits. |
| 1044 | int charLength = Math.min(nDigits, exp); |
| 1045 | System.arraycopy(digits, 0, result, i, charLength); |
| 1046 | i += charLength; |
| 1047 | if (charLength < exp) { |
| 1048 | charLength = exp-charLength; |
| 1049 | for (int nz = 0; nz < charLength; nz++) |
| 1050 | result[i++] = '0'; |
| 1051 | // Do not append ".0" for formatted floats since the user |
| 1052 | // may request that it be omitted. It is added as necessary |
| 1053 | // by the Formatter. |
| 1054 | if (form == Form.COMPATIBLE) { |
| 1055 | result[i++] = '.'; |
| 1056 | result[i++] = '0'; |
| 1057 | } |
| 1058 | } else { |
| 1059 | // Do not append ".0" for formatted floats since the user |
| 1060 | // may request that it be omitted. It is added as necessary |
| 1061 | // by the Formatter. |
| 1062 | if (form == Form.COMPATIBLE) { |
| 1063 | result[i++] = '.'; |
| 1064 | if (charLength < nDigits) { |
| 1065 | int t = Math.min(nDigits - charLength, precision); |
| 1066 | System.arraycopy(digits, charLength, result, i, t); |
| 1067 | i += t; |
| 1068 | } else { |
| 1069 | result[i++] = '0'; |
| 1070 | } |
| 1071 | } else { |
| 1072 | int t = Math.min(nDigits - charLength, precision); |
| 1073 | if (t > 0) { |
| 1074 | result[i++] = '.'; |
| 1075 | System.arraycopy(digits, charLength, result, i, t); |
| 1076 | i += t; |
| 1077 | } |
| 1078 | } |
| 1079 | } |
| 1080 | } else if (exp <= 0 |
| 1081 | && ((form == Form.COMPATIBLE && exp > -3) |
| 1082 | || (form == Form.DECIMAL_FLOAT))) |
| 1083 | { |
| 1084 | // print 0.0* digits |
| 1085 | result[i++] = '0'; |
| 1086 | if (exp != 0) { |
| 1087 | // write '0' s before the significant digits |
| 1088 | int t = Math.min(-exp, precision); |
| 1089 | if (t > 0) { |
| 1090 | result[i++] = '.'; |
| 1091 | for (int nz = 0; nz < t; nz++) |
| 1092 | result[i++] = '0'; |
| 1093 | } |
| 1094 | } |
| 1095 | int t = Math.min(digits.length, precision + exp); |
| 1096 | if (t > 0) { |
| 1097 | if (i == 1) |
| 1098 | result[i++] = '.'; |
| 1099 | // copy only when significant digits are within the precision |
| 1100 | System.arraycopy(digits, 0, result, i, t); |
| 1101 | i += t; |
| 1102 | } |
| 1103 | } else { |
| 1104 | result[i++] = digits[0]; |
| 1105 | if (form == Form.COMPATIBLE) { |
| 1106 | result[i++] = '.'; |
| 1107 | if (nDigits > 1) { |
| 1108 | System.arraycopy(digits, 1, result, i, nDigits-1); |
| 1109 | i += nDigits-1; |
| 1110 | } else { |
| 1111 | result[i++] = '0'; |
| 1112 | } |
| 1113 | result[i++] = 'E'; |
| 1114 | } else { |
| 1115 | if (nDigits > 1) { |
| 1116 | int t = Math.min(nDigits -1, precision); |
| 1117 | if (t > 0) { |
| 1118 | result[i++] = '.'; |
| 1119 | System.arraycopy(digits, 1, result, i, t); |
| 1120 | i += t; |
| 1121 | } |
| 1122 | } |
| 1123 | result[i++] = 'e'; |
| 1124 | } |
| 1125 | int e; |
| 1126 | if (exp <= 0) { |
| 1127 | result[i++] = '-'; |
| 1128 | e = -exp+1; |
| 1129 | } else { |
| 1130 | if (form != Form.COMPATIBLE) |
| 1131 | result[i++] = '+'; |
| 1132 | e = exp-1; |
| 1133 | } |
| 1134 | // decExponent has 1, 2, or 3, digits |
| 1135 | if (e <= 9) { |
| 1136 | if (form != Form.COMPATIBLE) |
| 1137 | result[i++] = '0'; |
| 1138 | result[i++] = (char)(e+'0'); |
| 1139 | } else if (e <= 99) { |
| 1140 | result[i++] = (char)(e/10 +'0'); |
| 1141 | result[i++] = (char)(e%10 + '0'); |
| 1142 | } else { |
| 1143 | result[i++] = (char)(e/100+'0'); |
| 1144 | e %= 100; |
| 1145 | result[i++] = (char)(e/10+'0'); |
| 1146 | result[i++] = (char)(e%10 + '0'); |
| 1147 | } |
| 1148 | } |
| 1149 | } |
| 1150 | return i; |
| 1151 | } |
| 1152 | |
| 1153 | // Per-thread buffer for string/stringbuffer conversion |
| 1154 | private static ThreadLocal perThreadBuffer = new ThreadLocal() { |
| 1155 | protected synchronized Object initialValue() { |
| 1156 | return new char[26]; |
| 1157 | } |
| 1158 | }; |
| 1159 | |
| 1160 | // This method should only ever be called if this object is constructed |
| 1161 | // without Form.DECIMAL_FLOAT because the perThreadBuffer is not large |
| 1162 | // enough to handle floating-point numbers of large precision. |
| 1163 | public void appendTo(Appendable buf) { |
| 1164 | char result[] = (char[])(perThreadBuffer.get()); |
| 1165 | int i = getChars(result); |
| 1166 | if (buf instanceof StringBuilder) |
| 1167 | ((StringBuilder) buf).append(result, 0, i); |
| 1168 | else if (buf instanceof StringBuffer) |
| 1169 | ((StringBuffer) buf).append(result, 0, i); |
| 1170 | else |
| 1171 | assert false; |
| 1172 | } |
| 1173 | |
| 1174 | public static FormattedFloatingDecimal |
| 1175 | readJavaFormatString( String in ) throws NumberFormatException { |
| 1176 | boolean isNegative = false; |
| 1177 | boolean signSeen = false; |
| 1178 | int decExp; |
| 1179 | char c; |
| 1180 | |
| 1181 | parseNumber: |
| 1182 | try{ |
| 1183 | in = in.trim(); // don't fool around with white space. |
| 1184 | // throws NullPointerException if null |
| 1185 | int l = in.length(); |
| 1186 | if ( l == 0 ) throw new NumberFormatException("empty String"); |
| 1187 | int i = 0; |
| 1188 | switch ( c = in.charAt( i ) ){ |
| 1189 | case '-': |
| 1190 | isNegative = true; |
| 1191 | //FALLTHROUGH |
| 1192 | case '+': |
| 1193 | i++; |
| 1194 | signSeen = true; |
| 1195 | } |
| 1196 | |
| 1197 | // Check for NaN and Infinity strings |
| 1198 | c = in.charAt(i); |
| 1199 | if(c == 'N' || c == 'I') { // possible NaN or infinity |
| 1200 | boolean potentialNaN = false; |
| 1201 | char targetChars[] = null; // char arrary of "NaN" or "Infinity" |
| 1202 | |
| 1203 | if(c == 'N') { |
| 1204 | targetChars = notANumber; |
| 1205 | potentialNaN = true; |
| 1206 | } else { |
| 1207 | targetChars = infinity; |
| 1208 | } |
| 1209 | |
| 1210 | // compare Input string to "NaN" or "Infinity" |
| 1211 | int j = 0; |
| 1212 | while(i < l && j < targetChars.length) { |
| 1213 | if(in.charAt(i) == targetChars[j]) { |
| 1214 | i++; j++; |
| 1215 | } |
| 1216 | else // something is amiss, throw exception |
| 1217 | break parseNumber; |
| 1218 | } |
| 1219 | |
| 1220 | // For the candidate string to be a NaN or infinity, |
| 1221 | // all characters in input string and target char[] |
| 1222 | // must be matched ==> j must equal targetChars.length |
| 1223 | // and i must equal l |
| 1224 | if( (j == targetChars.length) && (i == l) ) { // return NaN or infinity |
| 1225 | return (potentialNaN ? new FormattedFloatingDecimal(Double.NaN) // NaN has no sign |
| 1226 | : new FormattedFloatingDecimal(isNegative? |
| 1227 | Double.NEGATIVE_INFINITY: |
| 1228 | Double.POSITIVE_INFINITY)) ; |
| 1229 | } |
| 1230 | else { // something went wrong, throw exception |
| 1231 | break parseNumber; |
| 1232 | } |
| 1233 | |
| 1234 | } else if (c == '0') { // check for hexadecimal floating-point number |
| 1235 | if (l > i+1 ) { |
| 1236 | char ch = in.charAt(i+1); |
| 1237 | if (ch == 'x' || ch == 'X' ) // possible hex string |
| 1238 | return parseHexString(in); |
| 1239 | } |
| 1240 | } // look for and process decimal floating-point string |
| 1241 | |
| 1242 | char[] digits = new char[ l ]; |
| 1243 | int nDigits= 0; |
| 1244 | boolean decSeen = false; |
| 1245 | int decPt = 0; |
| 1246 | int nLeadZero = 0; |
| 1247 | int nTrailZero= 0; |
| 1248 | digitLoop: |
| 1249 | while ( i < l ){ |
| 1250 | switch ( c = in.charAt( i ) ){ |
| 1251 | case '0': |
| 1252 | if ( nDigits > 0 ){ |
| 1253 | nTrailZero += 1; |
| 1254 | } else { |
| 1255 | nLeadZero += 1; |
| 1256 | } |
| 1257 | break; // out of switch. |
| 1258 | case '1': |
| 1259 | case '2': |
| 1260 | case '3': |
| 1261 | case '4': |
| 1262 | case '5': |
| 1263 | case '6': |
| 1264 | case '7': |
| 1265 | case '8': |
| 1266 | case '9': |
| 1267 | while ( nTrailZero > 0 ){ |
| 1268 | digits[nDigits++] = '0'; |
| 1269 | nTrailZero -= 1; |
| 1270 | } |
| 1271 | digits[nDigits++] = c; |
| 1272 | break; // out of switch. |
| 1273 | case '.': |
| 1274 | if ( decSeen ){ |
| 1275 | // already saw one ., this is the 2nd. |
| 1276 | throw new NumberFormatException("multiple points"); |
| 1277 | } |
| 1278 | decPt = i; |
| 1279 | if ( signSeen ){ |
| 1280 | decPt -= 1; |
| 1281 | } |
| 1282 | decSeen = true; |
| 1283 | break; // out of switch. |
| 1284 | default: |
| 1285 | break digitLoop; |
| 1286 | } |
| 1287 | i++; |
| 1288 | } |
| 1289 | /* |
| 1290 | * At this point, we've scanned all the digits and decimal |
| 1291 | * point we're going to see. Trim off leading and trailing |
| 1292 | * zeros, which will just confuse us later, and adjust |
| 1293 | * our initial decimal exponent accordingly. |
| 1294 | * To review: |
| 1295 | * we have seen i total characters. |
| 1296 | * nLeadZero of them were zeros before any other digits. |
| 1297 | * nTrailZero of them were zeros after any other digits. |
| 1298 | * if ( decSeen ), then a . was seen after decPt characters |
| 1299 | * ( including leading zeros which have been discarded ) |
| 1300 | * nDigits characters were neither lead nor trailing |
| 1301 | * zeros, nor point |
| 1302 | */ |
| 1303 | /* |
| 1304 | * special hack: if we saw no non-zero digits, then the |
| 1305 | * answer is zero! |
| 1306 | * Unfortunately, we feel honor-bound to keep parsing! |
| 1307 | */ |
| 1308 | if ( nDigits == 0 ){ |
| 1309 | digits = zero; |
| 1310 | nDigits = 1; |
| 1311 | if ( nLeadZero == 0 ){ |
| 1312 | // we saw NO DIGITS AT ALL, |
| 1313 | // not even a crummy 0! |
| 1314 | // this is not allowed. |
| 1315 | break parseNumber; // go throw exception |
| 1316 | } |
| 1317 | |
| 1318 | } |
| 1319 | |
| 1320 | /* Our initial exponent is decPt, adjusted by the number of |
| 1321 | * discarded zeros. Or, if there was no decPt, |
| 1322 | * then its just nDigits adjusted by discarded trailing zeros. |
| 1323 | */ |
| 1324 | if ( decSeen ){ |
| 1325 | decExp = decPt - nLeadZero; |
| 1326 | } else { |
| 1327 | decExp = nDigits+nTrailZero; |
| 1328 | } |
| 1329 | |
| 1330 | /* |
| 1331 | * Look for 'e' or 'E' and an optionally signed integer. |
| 1332 | */ |
| 1333 | if ( (i < l) && (((c = in.charAt(i) )=='e') || (c == 'E') ) ){ |
| 1334 | int expSign = 1; |
| 1335 | int expVal = 0; |
| 1336 | int reallyBig = Integer.MAX_VALUE / 10; |
| 1337 | boolean expOverflow = false; |
| 1338 | switch( in.charAt(++i) ){ |
| 1339 | case '-': |
| 1340 | expSign = -1; |
| 1341 | //FALLTHROUGH |
| 1342 | case '+': |
| 1343 | i++; |
| 1344 | } |
| 1345 | int expAt = i; |
| 1346 | expLoop: |
| 1347 | while ( i < l ){ |
| 1348 | if ( expVal >= reallyBig ){ |
| 1349 | // the next character will cause integer |
| 1350 | // overflow. |
| 1351 | expOverflow = true; |
| 1352 | } |
| 1353 | switch ( c = in.charAt(i++) ){ |
| 1354 | case '0': |
| 1355 | case '1': |
| 1356 | case '2': |
| 1357 | case '3': |
| 1358 | case '4': |
| 1359 | case '5': |
| 1360 | case '6': |
| 1361 | case '7': |
| 1362 | case '8': |
| 1363 | case '9': |
| 1364 | expVal = expVal*10 + ( (int)c - (int)'0' ); |
| 1365 | continue; |
| 1366 | default: |
| 1367 | i--; // back up. |
| 1368 | break expLoop; // stop parsing exponent. |
| 1369 | } |
| 1370 | } |
| 1371 | int expLimit = bigDecimalExponent+nDigits+nTrailZero; |
| 1372 | if ( expOverflow || ( expVal > expLimit ) ){ |
| 1373 | // |
| 1374 | // The intent here is to end up with |
| 1375 | // infinity or zero, as appropriate. |
| 1376 | // The reason for yielding such a small decExponent, |
| 1377 | // rather than something intuitive such as |
| 1378 | // expSign*Integer.MAX_VALUE, is that this value |
| 1379 | // is subject to further manipulation in |
| 1380 | // doubleValue() and floatValue(), and I don't want |
| 1381 | // it to be able to cause overflow there! |
| 1382 | // (The only way we can get into trouble here is for |
| 1383 | // really outrageous nDigits+nTrailZero, such as 2 billion. ) |
| 1384 | // |
| 1385 | decExp = expSign*expLimit; |
| 1386 | } else { |
| 1387 | // this should not overflow, since we tested |
| 1388 | // for expVal > (MAX+N), where N >= abs(decExp) |
| 1389 | decExp = decExp + expSign*expVal; |
| 1390 | } |
| 1391 | |
| 1392 | // if we saw something not a digit ( or end of string ) |
| 1393 | // after the [Ee][+-], without seeing any digits at all |
| 1394 | // this is certainly an error. If we saw some digits, |
| 1395 | // but then some trailing garbage, that might be ok. |
| 1396 | // so we just fall through in that case. |
| 1397 | // HUMBUG |
| 1398 | if ( i == expAt ) |
| 1399 | break parseNumber; // certainly bad |
| 1400 | } |
| 1401 | /* |
| 1402 | * We parsed everything we could. |
| 1403 | * If there are leftovers, then this is not good input! |
| 1404 | */ |
| 1405 | if ( i < l && |
| 1406 | ((i != l - 1) || |
| 1407 | (in.charAt(i) != 'f' && |
| 1408 | in.charAt(i) != 'F' && |
| 1409 | in.charAt(i) != 'd' && |
| 1410 | in.charAt(i) != 'D'))) { |
| 1411 | break parseNumber; // go throw exception |
| 1412 | } |
| 1413 | |
| 1414 | return new FormattedFloatingDecimal( isNegative, decExp, digits, nDigits, false, Integer.MAX_VALUE, Form.COMPATIBLE ); |
| 1415 | } catch ( StringIndexOutOfBoundsException e ){ } |
| 1416 | throw new NumberFormatException("For input string: \"" + in + "\""); |
| 1417 | } |
| 1418 | |
| 1419 | /* |
| 1420 | * Take a FormattedFloatingDecimal, which we presumably just scanned in, |
| 1421 | * and find out what its value is, as a double. |
| 1422 | * |
| 1423 | * AS A SIDE EFFECT, SET roundDir TO INDICATE PREFERRED |
| 1424 | * ROUNDING DIRECTION in case the result is really destined |
| 1425 | * for a single-precision float. |
| 1426 | */ |
| 1427 | |
| 1428 | public double |
| 1429 | doubleValue(){ |
| 1430 | int kDigits = Math.min( nDigits, maxDecimalDigits+1 ); |
| 1431 | long lValue; |
| 1432 | double dValue; |
| 1433 | double rValue, tValue; |
| 1434 | |
| 1435 | // First, check for NaN and Infinity values |
| 1436 | if(digits == infinity || digits == notANumber) { |
| 1437 | if(digits == notANumber) |
| 1438 | return Double.NaN; |
| 1439 | else |
| 1440 | return (isNegative?Double.NEGATIVE_INFINITY:Double.POSITIVE_INFINITY); |
| 1441 | } |
| 1442 | else { |
| 1443 | if (mustSetRoundDir) { |
| 1444 | roundDir = 0; |
| 1445 | } |
| 1446 | /* |
| 1447 | * convert the lead kDigits to a long integer. |
| 1448 | */ |
| 1449 | // (special performance hack: start to do it using int) |
| 1450 | int iValue = (int)digits[0]-(int)'0'; |
| 1451 | int iDigits = Math.min( kDigits, intDecimalDigits ); |
| 1452 | for ( int i=1; i < iDigits; i++ ){ |
| 1453 | iValue = iValue*10 + (int)digits[i]-(int)'0'; |
| 1454 | } |
| 1455 | lValue = (long)iValue; |
| 1456 | for ( int i=iDigits; i < kDigits; i++ ){ |
| 1457 | lValue = lValue*10L + (long)((int)digits[i]-(int)'0'); |
| 1458 | } |
| 1459 | dValue = (double)lValue; |
| 1460 | int exp = decExponent-kDigits; |
| 1461 | /* |
| 1462 | * lValue now contains a long integer with the value of |
| 1463 | * the first kDigits digits of the number. |
| 1464 | * dValue contains the (double) of the same. |
| 1465 | */ |
| 1466 | |
| 1467 | if ( nDigits <= maxDecimalDigits ){ |
| 1468 | /* |
| 1469 | * possibly an easy case. |
| 1470 | * We know that the digits can be represented |
| 1471 | * exactly. And if the exponent isn't too outrageous, |
| 1472 | * the whole thing can be done with one operation, |
| 1473 | * thus one rounding error. |
| 1474 | * Note that all our constructors trim all leading and |
| 1475 | * trailing zeros, so simple values (including zero) |
| 1476 | * will always end up here |
| 1477 | */ |
| 1478 | if (exp == 0 || dValue == 0.0) |
| 1479 | return (isNegative)? -dValue : dValue; // small floating integer |
| 1480 | else if ( exp >= 0 ){ |
| 1481 | if ( exp <= maxSmallTen ){ |
| 1482 | /* |
| 1483 | * Can get the answer with one operation, |
| 1484 | * thus one roundoff. |
| 1485 | */ |
| 1486 | rValue = dValue * small10pow[exp]; |
| 1487 | if ( mustSetRoundDir ){ |
| 1488 | tValue = rValue / small10pow[exp]; |
| 1489 | roundDir = ( tValue == dValue ) ? 0 |
| 1490 | :( tValue < dValue ) ? 1 |
| 1491 | : -1; |
| 1492 | } |
| 1493 | return (isNegative)? -rValue : rValue; |
| 1494 | } |
| 1495 | int slop = maxDecimalDigits - kDigits; |
| 1496 | if ( exp <= maxSmallTen+slop ){ |
| 1497 | /* |
| 1498 | * We can multiply dValue by 10^(slop) |
| 1499 | * and it is still "small" and exact. |
| 1500 | * Then we can multiply by 10^(exp-slop) |
| 1501 | * with one rounding. |
| 1502 | */ |
| 1503 | dValue *= small10pow[slop]; |
| 1504 | rValue = dValue * small10pow[exp-slop]; |
| 1505 | |
| 1506 | if ( mustSetRoundDir ){ |
| 1507 | tValue = rValue / small10pow[exp-slop]; |
| 1508 | roundDir = ( tValue == dValue ) ? 0 |
| 1509 | :( tValue < dValue ) ? 1 |
| 1510 | : -1; |
| 1511 | } |
| 1512 | return (isNegative)? -rValue : rValue; |
| 1513 | } |
| 1514 | /* |
| 1515 | * Else we have a hard case with a positive exp. |
| 1516 | */ |
| 1517 | } else { |
| 1518 | if ( exp >= -maxSmallTen ){ |
| 1519 | /* |
| 1520 | * Can get the answer in one division. |
| 1521 | */ |
| 1522 | rValue = dValue / small10pow[-exp]; |
| 1523 | tValue = rValue * small10pow[-exp]; |
| 1524 | if ( mustSetRoundDir ){ |
| 1525 | roundDir = ( tValue == dValue ) ? 0 |
| 1526 | :( tValue < dValue ) ? 1 |
| 1527 | : -1; |
| 1528 | } |
| 1529 | return (isNegative)? -rValue : rValue; |
| 1530 | } |
| 1531 | /* |
| 1532 | * Else we have a hard case with a negative exp. |
| 1533 | */ |
| 1534 | } |
| 1535 | } |
| 1536 | |
| 1537 | /* |
| 1538 | * Harder cases: |
| 1539 | * The sum of digits plus exponent is greater than |
| 1540 | * what we think we can do with one error. |
| 1541 | * |
| 1542 | * Start by approximating the right answer by, |
| 1543 | * naively, scaling by powers of 10. |
| 1544 | */ |
| 1545 | if ( exp > 0 ){ |
| 1546 | if ( decExponent > maxDecimalExponent+1 ){ |
| 1547 | /* |
| 1548 | * Lets face it. This is going to be |
| 1549 | * Infinity. Cut to the chase. |
| 1550 | */ |
| 1551 | return (isNegative)? Double.NEGATIVE_INFINITY : Double.POSITIVE_INFINITY; |
| 1552 | } |
| 1553 | if ( (exp&15) != 0 ){ |
| 1554 | dValue *= small10pow[exp&15]; |
| 1555 | } |
| 1556 | if ( (exp>>=4) != 0 ){ |
| 1557 | int j; |
| 1558 | for( j = 0; exp > 1; j++, exp>>=1 ){ |
| 1559 | if ( (exp&1)!=0) |
| 1560 | dValue *= big10pow[j]; |
| 1561 | } |
| 1562 | /* |
| 1563 | * The reason for the weird exp > 1 condition |
| 1564 | * in the above loop was so that the last multiply |
| 1565 | * would get unrolled. We handle it here. |
| 1566 | * It could overflow. |
| 1567 | */ |
| 1568 | double t = dValue * big10pow[j]; |
| 1569 | if ( Double.isInfinite( t ) ){ |
| 1570 | /* |
| 1571 | * It did overflow. |
| 1572 | * Look more closely at the result. |
| 1573 | * If the exponent is just one too large, |
| 1574 | * then use the maximum finite as our estimate |
| 1575 | * value. Else call the result infinity |
| 1576 | * and punt it. |
| 1577 | * ( I presume this could happen because |
| 1578 | * rounding forces the result here to be |
| 1579 | * an ULP or two larger than |
| 1580 | * Double.MAX_VALUE ). |
| 1581 | */ |
| 1582 | t = dValue / 2.0; |
| 1583 | t *= big10pow[j]; |
| 1584 | if ( Double.isInfinite( t ) ){ |
| 1585 | return (isNegative)? Double.NEGATIVE_INFINITY : Double.POSITIVE_INFINITY; |
| 1586 | } |
| 1587 | t = Double.MAX_VALUE; |
| 1588 | } |
| 1589 | dValue = t; |
| 1590 | } |
| 1591 | } else if ( exp < 0 ){ |
| 1592 | exp = -exp; |
| 1593 | if ( decExponent < minDecimalExponent-1 ){ |
| 1594 | /* |
| 1595 | * Lets face it. This is going to be |
| 1596 | * zero. Cut to the chase. |
| 1597 | */ |
| 1598 | return (isNegative)? -0.0 : 0.0; |
| 1599 | } |
| 1600 | if ( (exp&15) != 0 ){ |
| 1601 | dValue /= small10pow[exp&15]; |
| 1602 | } |
| 1603 | if ( (exp>>=4) != 0 ){ |
| 1604 | int j; |
| 1605 | for( j = 0; exp > 1; j++, exp>>=1 ){ |
| 1606 | if ( (exp&1)!=0) |
| 1607 | dValue *= tiny10pow[j]; |
| 1608 | } |
| 1609 | /* |
| 1610 | * The reason for the weird exp > 1 condition |
| 1611 | * in the above loop was so that the last multiply |
| 1612 | * would get unrolled. We handle it here. |
| 1613 | * It could underflow. |
| 1614 | */ |
| 1615 | double t = dValue * tiny10pow[j]; |
| 1616 | if ( t == 0.0 ){ |
| 1617 | /* |
| 1618 | * It did underflow. |
| 1619 | * Look more closely at the result. |
| 1620 | * If the exponent is just one too small, |
| 1621 | * then use the minimum finite as our estimate |
| 1622 | * value. Else call the result 0.0 |
| 1623 | * and punt it. |
| 1624 | * ( I presume this could happen because |
| 1625 | * rounding forces the result here to be |
| 1626 | * an ULP or two less than |
| 1627 | * Double.MIN_VALUE ). |
| 1628 | */ |
| 1629 | t = dValue * 2.0; |
| 1630 | t *= tiny10pow[j]; |
| 1631 | if ( t == 0.0 ){ |
| 1632 | return (isNegative)? -0.0 : 0.0; |
| 1633 | } |
| 1634 | t = Double.MIN_VALUE; |
| 1635 | } |
| 1636 | dValue = t; |
| 1637 | } |
| 1638 | } |
| 1639 | |
| 1640 | /* |
| 1641 | * dValue is now approximately the result. |
| 1642 | * The hard part is adjusting it, by comparison |
| 1643 | * with FDBigInt arithmetic. |
| 1644 | * Formulate the EXACT big-number result as |
| 1645 | * bigD0 * 10^exp |
| 1646 | */ |
| 1647 | FDBigInt bigD0 = new FDBigInt( lValue, digits, kDigits, nDigits ); |
| 1648 | exp = decExponent - nDigits; |
| 1649 | |
| 1650 | correctionLoop: |
| 1651 | while(true){ |
| 1652 | /* AS A SIDE EFFECT, THIS METHOD WILL SET THE INSTANCE VARIABLES |
| 1653 | * bigIntExp and bigIntNBits |
| 1654 | */ |
| 1655 | FDBigInt bigB = doubleToBigInt( dValue ); |
| 1656 | |
| 1657 | /* |
| 1658 | * Scale bigD, bigB appropriately for |
| 1659 | * big-integer operations. |
| 1660 | * Naively, we multipy by powers of ten |
| 1661 | * and powers of two. What we actually do |
| 1662 | * is keep track of the powers of 5 and |
| 1663 | * powers of 2 we would use, then factor out |
| 1664 | * common divisors before doing the work. |
| 1665 | */ |
| 1666 | int B2, B5; // powers of 2, 5 in bigB |
| 1667 | int D2, D5; // powers of 2, 5 in bigD |
| 1668 | int Ulp2; // powers of 2 in halfUlp. |
| 1669 | if ( exp >= 0 ){ |
| 1670 | B2 = B5 = 0; |
| 1671 | D2 = D5 = exp; |
| 1672 | } else { |
| 1673 | B2 = B5 = -exp; |
| 1674 | D2 = D5 = 0; |
| 1675 | } |
| 1676 | if ( bigIntExp >= 0 ){ |
| 1677 | B2 += bigIntExp; |
| 1678 | } else { |
| 1679 | D2 -= bigIntExp; |
| 1680 | } |
| 1681 | Ulp2 = B2; |
| 1682 | // shift bigB and bigD left by a number s. t. |
| 1683 | // halfUlp is still an integer. |
| 1684 | int hulpbias; |
| 1685 | if ( bigIntExp+bigIntNBits <= -expBias+1 ){ |
| 1686 | // This is going to be a denormalized number |
| 1687 | // (if not actually zero). |
| 1688 | // half an ULP is at 2^-(expBias+expShift+1) |
| 1689 | hulpbias = bigIntExp+ expBias + expShift; |
| 1690 | } else { |
| 1691 | hulpbias = expShift + 2 - bigIntNBits; |
| 1692 | } |
| 1693 | B2 += hulpbias; |
| 1694 | D2 += hulpbias; |
| 1695 | // if there are common factors of 2, we might just as well |
| 1696 | // factor them out, as they add nothing useful. |
| 1697 | int common2 = Math.min( B2, Math.min( D2, Ulp2 ) ); |
| 1698 | B2 -= common2; |
| 1699 | D2 -= common2; |
| 1700 | Ulp2 -= common2; |
| 1701 | // do multiplications by powers of 5 and 2 |
| 1702 | bigB = multPow52( bigB, B5, B2 ); |
| 1703 | FDBigInt bigD = multPow52( new FDBigInt( bigD0 ), D5, D2 ); |
| 1704 | // |
| 1705 | // to recap: |
| 1706 | // bigB is the scaled-big-int version of our floating-point |
| 1707 | // candidate. |
| 1708 | // bigD is the scaled-big-int version of the exact value |
| 1709 | // as we understand it. |
| 1710 | // halfUlp is 1/2 an ulp of bigB, except for special cases |
| 1711 | // of exact powers of 2 |
| 1712 | // |
| 1713 | // the plan is to compare bigB with bigD, and if the difference |
| 1714 | // is less than halfUlp, then we're satisfied. Otherwise, |
| 1715 | // use the ratio of difference to halfUlp to calculate a fudge |
| 1716 | // factor to add to the floating value, then go 'round again. |
| 1717 | // |
| 1718 | FDBigInt diff; |
| 1719 | int cmpResult; |
| 1720 | boolean overvalue; |
| 1721 | if ( (cmpResult = bigB.cmp( bigD ) ) > 0 ){ |
| 1722 | overvalue = true; // our candidate is too big. |
| 1723 | diff = bigB.sub( bigD ); |
| 1724 | if ( (bigIntNBits == 1) && (bigIntExp > -expBias) ){ |
| 1725 | // candidate is a normalized exact power of 2 and |
| 1726 | // is too big. We will be subtracting. |
| 1727 | // For our purposes, ulp is the ulp of the |
| 1728 | // next smaller range. |
| 1729 | Ulp2 -= 1; |
| 1730 | if ( Ulp2 < 0 ){ |
| 1731 | // rats. Cannot de-scale ulp this far. |
| 1732 | // must scale diff in other direction. |
| 1733 | Ulp2 = 0; |
| 1734 | diff.lshiftMe( 1 ); |
| 1735 | } |
| 1736 | } |
| 1737 | } else if ( cmpResult < 0 ){ |
| 1738 | overvalue = false; // our candidate is too small. |
| 1739 | diff = bigD.sub( bigB ); |
| 1740 | } else { |
| 1741 | // the candidate is exactly right! |
| 1742 | // this happens with surprising fequency |
| 1743 | break correctionLoop; |
| 1744 | } |
| 1745 | FDBigInt halfUlp = constructPow52( B5, Ulp2 ); |
| 1746 | if ( (cmpResult = diff.cmp( halfUlp ) ) < 0 ){ |
| 1747 | // difference is small. |
| 1748 | // this is close enough |
| 1749 | if (mustSetRoundDir) { |
| 1750 | roundDir = overvalue ? -1 : 1; |
| 1751 | } |
| 1752 | break correctionLoop; |
| 1753 | } else if ( cmpResult == 0 ){ |
| 1754 | // difference is exactly half an ULP |
| 1755 | // round to some other value maybe, then finish |
| 1756 | dValue += 0.5*ulp( dValue, overvalue ); |
| 1757 | // should check for bigIntNBits == 1 here?? |
| 1758 | if (mustSetRoundDir) { |
| 1759 | roundDir = overvalue ? -1 : 1; |
| 1760 | } |
| 1761 | break correctionLoop; |
| 1762 | } else { |
| 1763 | // difference is non-trivial. |
| 1764 | // could scale addend by ratio of difference to |
| 1765 | // halfUlp here, if we bothered to compute that difference. |
| 1766 | // Most of the time ( I hope ) it is about 1 anyway. |
| 1767 | dValue += ulp( dValue, overvalue ); |
| 1768 | if ( dValue == 0.0 || dValue == Double.POSITIVE_INFINITY ) |
| 1769 | break correctionLoop; // oops. Fell off end of range. |
| 1770 | continue; // try again. |
| 1771 | } |
| 1772 | |
| 1773 | } |
| 1774 | return (isNegative)? -dValue : dValue; |
| 1775 | } |
| 1776 | } |
| 1777 | |
| 1778 | /* |
| 1779 | * Take a FormattedFloatingDecimal, which we presumably just scanned in, |
| 1780 | * and find out what its value is, as a float. |
| 1781 | * This is distinct from doubleValue() to avoid the extremely |
| 1782 | * unlikely case of a double rounding error, wherein the converstion |
| 1783 | * to double has one rounding error, and the conversion of that double |
| 1784 | * to a float has another rounding error, IN THE WRONG DIRECTION, |
| 1785 | * ( because of the preference to a zero low-order bit ). |
| 1786 | */ |
| 1787 | |
| 1788 | public float |
| 1789 | floatValue(){ |
| 1790 | int kDigits = Math.min( nDigits, singleMaxDecimalDigits+1 ); |
| 1791 | int iValue; |
| 1792 | float fValue; |
| 1793 | |
| 1794 | // First, check for NaN and Infinity values |
| 1795 | if(digits == infinity || digits == notANumber) { |
| 1796 | if(digits == notANumber) |
| 1797 | return Float.NaN; |
| 1798 | else |
| 1799 | return (isNegative?Float.NEGATIVE_INFINITY:Float.POSITIVE_INFINITY); |
| 1800 | } |
| 1801 | else { |
| 1802 | /* |
| 1803 | * convert the lead kDigits to an integer. |
| 1804 | */ |
| 1805 | iValue = (int)digits[0]-(int)'0'; |
| 1806 | for ( int i=1; i < kDigits; i++ ){ |
| 1807 | iValue = iValue*10 + (int)digits[i]-(int)'0'; |
| 1808 | } |
| 1809 | fValue = (float)iValue; |
| 1810 | int exp = decExponent-kDigits; |
| 1811 | /* |
| 1812 | * iValue now contains an integer with the value of |
| 1813 | * the first kDigits digits of the number. |
| 1814 | * fValue contains the (float) of the same. |
| 1815 | */ |
| 1816 | |
| 1817 | if ( nDigits <= singleMaxDecimalDigits ){ |
| 1818 | /* |
| 1819 | * possibly an easy case. |
| 1820 | * We know that the digits can be represented |
| 1821 | * exactly. And if the exponent isn't too outrageous, |
| 1822 | * the whole thing can be done with one operation, |
| 1823 | * thus one rounding error. |
| 1824 | * Note that all our constructors trim all leading and |
| 1825 | * trailing zeros, so simple values (including zero) |
| 1826 | * will always end up here. |
| 1827 | */ |
| 1828 | if (exp == 0 || fValue == 0.0f) |
| 1829 | return (isNegative)? -fValue : fValue; // small floating integer |
| 1830 | else if ( exp >= 0 ){ |
| 1831 | if ( exp <= singleMaxSmallTen ){ |
| 1832 | /* |
| 1833 | * Can get the answer with one operation, |
| 1834 | * thus one roundoff. |
| 1835 | */ |
| 1836 | fValue *= singleSmall10pow[exp]; |
| 1837 | return (isNegative)? -fValue : fValue; |
| 1838 | } |
| 1839 | int slop = singleMaxDecimalDigits - kDigits; |
| 1840 | if ( exp <= singleMaxSmallTen+slop ){ |
| 1841 | /* |
| 1842 | * We can multiply dValue by 10^(slop) |
| 1843 | * and it is still "small" and exact. |
| 1844 | * Then we can multiply by 10^(exp-slop) |
| 1845 | * with one rounding. |
| 1846 | */ |
| 1847 | fValue *= singleSmall10pow[slop]; |
| 1848 | fValue *= singleSmall10pow[exp-slop]; |
| 1849 | return (isNegative)? -fValue : fValue; |
| 1850 | } |
| 1851 | /* |
| 1852 | * Else we have a hard case with a positive exp. |
| 1853 | */ |
| 1854 | } else { |
| 1855 | if ( exp >= -singleMaxSmallTen ){ |
| 1856 | /* |
| 1857 | * Can get the answer in one division. |
| 1858 | */ |
| 1859 | fValue /= singleSmall10pow[-exp]; |
| 1860 | return (isNegative)? -fValue : fValue; |
| 1861 | } |
| 1862 | /* |
| 1863 | * Else we have a hard case with a negative exp. |
| 1864 | */ |
| 1865 | } |
| 1866 | } else if ( (decExponent >= nDigits) && (nDigits+decExponent <= maxDecimalDigits) ){ |
| 1867 | /* |
| 1868 | * In double-precision, this is an exact floating integer. |
| 1869 | * So we can compute to double, then shorten to float |
| 1870 | * with one round, and get the right answer. |
| 1871 | * |
| 1872 | * First, finish accumulating digits. |
| 1873 | * Then convert that integer to a double, multiply |
| 1874 | * by the appropriate power of ten, and convert to float. |
| 1875 | */ |
| 1876 | long lValue = (long)iValue; |
| 1877 | for ( int i=kDigits; i < nDigits; i++ ){ |
| 1878 | lValue = lValue*10L + (long)((int)digits[i]-(int)'0'); |
| 1879 | } |
| 1880 | double dValue = (double)lValue; |
| 1881 | exp = decExponent-nDigits; |
| 1882 | dValue *= small10pow[exp]; |
| 1883 | fValue = (float)dValue; |
| 1884 | return (isNegative)? -fValue : fValue; |
| 1885 | |
| 1886 | } |
| 1887 | /* |
| 1888 | * Harder cases: |
| 1889 | * The sum of digits plus exponent is greater than |
| 1890 | * what we think we can do with one error. |
| 1891 | * |
| 1892 | * Start by weeding out obviously out-of-range |
| 1893 | * results, then convert to double and go to |
| 1894 | * common hard-case code. |
| 1895 | */ |
| 1896 | if ( decExponent > singleMaxDecimalExponent+1 ){ |
| 1897 | /* |
| 1898 | * Lets face it. This is going to be |
| 1899 | * Infinity. Cut to the chase. |
| 1900 | */ |
| 1901 | return (isNegative)? Float.NEGATIVE_INFINITY : Float.POSITIVE_INFINITY; |
| 1902 | } else if ( decExponent < singleMinDecimalExponent-1 ){ |
| 1903 | /* |
| 1904 | * Lets face it. This is going to be |
| 1905 | * zero. Cut to the chase. |
| 1906 | */ |
| 1907 | return (isNegative)? -0.0f : 0.0f; |
| 1908 | } |
| 1909 | |
| 1910 | /* |
| 1911 | * Here, we do 'way too much work, but throwing away |
| 1912 | * our partial results, and going and doing the whole |
| 1913 | * thing as double, then throwing away half the bits that computes |
| 1914 | * when we convert back to float. |
| 1915 | * |
| 1916 | * The alternative is to reproduce the whole multiple-precision |
| 1917 | * algorythm for float precision, or to try to parameterize it |
| 1918 | * for common usage. The former will take about 400 lines of code, |
| 1919 | * and the latter I tried without success. Thus the semi-hack |
| 1920 | * answer here. |
| 1921 | */ |
| 1922 | mustSetRoundDir = !fromHex; |
| 1923 | double dValue = doubleValue(); |
| 1924 | return stickyRound( dValue ); |
| 1925 | } |
| 1926 | } |
| 1927 | |
| 1928 | |
| 1929 | /* |
| 1930 | * All the positive powers of 10 that can be |
| 1931 | * represented exactly in double/float. |
| 1932 | */ |
| 1933 | private static final double small10pow[] = { |
| 1934 | 1.0e0, |
| 1935 | 1.0e1, 1.0e2, 1.0e3, 1.0e4, 1.0e5, |
| 1936 | 1.0e6, 1.0e7, 1.0e8, 1.0e9, 1.0e10, |
| 1937 | 1.0e11, 1.0e12, 1.0e13, 1.0e14, 1.0e15, |
| 1938 | 1.0e16, 1.0e17, 1.0e18, 1.0e19, 1.0e20, |
| 1939 | 1.0e21, 1.0e22 |
| 1940 | }; |
| 1941 | |
| 1942 | private static final float singleSmall10pow[] = { |
| 1943 | 1.0e0f, |
| 1944 | 1.0e1f, 1.0e2f, 1.0e3f, 1.0e4f, 1.0e5f, |
| 1945 | 1.0e6f, 1.0e7f, 1.0e8f, 1.0e9f, 1.0e10f |
| 1946 | }; |
| 1947 | |
| 1948 | private static final double big10pow[] = { |
| 1949 | 1e16, 1e32, 1e64, 1e128, 1e256 }; |
| 1950 | private static final double tiny10pow[] = { |
| 1951 | 1e-16, 1e-32, 1e-64, 1e-128, 1e-256 }; |
| 1952 | |
| 1953 | private static final int maxSmallTen = small10pow.length-1; |
| 1954 | private static final int singleMaxSmallTen = singleSmall10pow.length-1; |
| 1955 | |
| 1956 | private static final int small5pow[] = { |
| 1957 | 1, |
| 1958 | 5, |
| 1959 | 5*5, |
| 1960 | 5*5*5, |
| 1961 | 5*5*5*5, |
| 1962 | 5*5*5*5*5, |
| 1963 | 5*5*5*5*5*5, |
| 1964 | 5*5*5*5*5*5*5, |
| 1965 | 5*5*5*5*5*5*5*5, |
| 1966 | 5*5*5*5*5*5*5*5*5, |
| 1967 | 5*5*5*5*5*5*5*5*5*5, |
| 1968 | 5*5*5*5*5*5*5*5*5*5*5, |
| 1969 | 5*5*5*5*5*5*5*5*5*5*5*5, |
| 1970 | 5*5*5*5*5*5*5*5*5*5*5*5*5 |
| 1971 | }; |
| 1972 | |
| 1973 | |
| 1974 | private static final long long5pow[] = { |
| 1975 | 1L, |
| 1976 | 5L, |
| 1977 | 5L*5, |
| 1978 | 5L*5*5, |
| 1979 | 5L*5*5*5, |
| 1980 | 5L*5*5*5*5, |
| 1981 | 5L*5*5*5*5*5, |
| 1982 | 5L*5*5*5*5*5*5, |
| 1983 | 5L*5*5*5*5*5*5*5, |
| 1984 | 5L*5*5*5*5*5*5*5*5, |
| 1985 | 5L*5*5*5*5*5*5*5*5*5, |
| 1986 | 5L*5*5*5*5*5*5*5*5*5*5, |
| 1987 | 5L*5*5*5*5*5*5*5*5*5*5*5, |
| 1988 | 5L*5*5*5*5*5*5*5*5*5*5*5*5, |
| 1989 | 5L*5*5*5*5*5*5*5*5*5*5*5*5*5, |
| 1990 | 5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5, |
| 1991 | 5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5, |
| 1992 | 5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5, |
| 1993 | 5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5, |
| 1994 | 5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5, |
| 1995 | 5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5, |
| 1996 | 5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5, |
| 1997 | 5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5, |
| 1998 | 5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5, |
| 1999 | 5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5, |
| 2000 | 5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5, |
| 2001 | 5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5, |
| 2002 | }; |
| 2003 | |
| 2004 | // approximately ceil( log2( long5pow[i] ) ) |
| 2005 | private static final int n5bits[] = { |
| 2006 | 0, |
| 2007 | 3, |
| 2008 | 5, |
| 2009 | 7, |
| 2010 | 10, |
| 2011 | 12, |
| 2012 | 14, |
| 2013 | 17, |
| 2014 | 19, |
| 2015 | 21, |
| 2016 | 24, |
| 2017 | 26, |
| 2018 | 28, |
| 2019 | 31, |
| 2020 | 33, |
| 2021 | 35, |
| 2022 | 38, |
| 2023 | 40, |
| 2024 | 42, |
| 2025 | 45, |
| 2026 | 47, |
| 2027 | 49, |
| 2028 | 52, |
| 2029 | 54, |
| 2030 | 56, |
| 2031 | 59, |
| 2032 | 61, |
| 2033 | }; |
| 2034 | |
| 2035 | private static final char infinity[] = { 'I', 'n', 'f', 'i', 'n', 'i', 't', 'y' }; |
| 2036 | private static final char notANumber[] = { 'N', 'a', 'N' }; |
| 2037 | private static final char zero[] = { '0', '0', '0', '0', '0', '0', '0', '0' }; |
| 2038 | |
| 2039 | |
| 2040 | /* |
| 2041 | * Grammar is compatible with hexadecimal floating-point constants |
| 2042 | * described in section 6.4.4.2 of the C99 specification. |
| 2043 | */ |
| 2044 | private static Pattern hexFloatPattern = Pattern.compile( |
| 2045 | //1 234 56 7 8 9 |
| 2046 | "([-+])?0[xX](((\\p{XDigit}+)\\.?)|((\\p{XDigit}*)\\.(\\p{XDigit}+)))[pP]([-+])?(\\p{Digit}+)[fFdD]?" |
| 2047 | ); |
| 2048 | |
| 2049 | /* |
| 2050 | * Convert string s to a suitable floating decimal; uses the |
| 2051 | * double constructor and set the roundDir variable appropriately |
| 2052 | * in case the value is later converted to a float. |
| 2053 | */ |
| 2054 | static FormattedFloatingDecimal parseHexString(String s) { |
| 2055 | // Verify string is a member of the hexadecimal floating-point |
| 2056 | // string language. |
| 2057 | Matcher m = hexFloatPattern.matcher(s); |
| 2058 | boolean validInput = m.matches(); |
| 2059 | |
| 2060 | if (!validInput) { |
| 2061 | // Input does not match pattern |
| 2062 | throw new NumberFormatException("For input string: \"" + s + "\""); |
| 2063 | } else { // validInput |
| 2064 | /* |
| 2065 | * We must isolate the sign, significand, and exponent |
| 2066 | * fields. The sign value is straightforward. Since |
| 2067 | * floating-point numbers are stored with a normalized |
| 2068 | * representation, the significand and exponent are |
| 2069 | * interrelated. |
| 2070 | * |
| 2071 | * After extracting the sign, we normalized the |
| 2072 | * significand as a hexadecimal value, calculating an |
| 2073 | * exponent adjust for any shifts made during |
| 2074 | * normalization. If the significand is zero, the |
| 2075 | * exponent doesn't need to be examined since the output |
| 2076 | * will be zero. |
| 2077 | * |
| 2078 | * Next the exponent in the input string is extracted. |
| 2079 | * Afterwards, the significand is normalized as a *binary* |
| 2080 | * value and the input value's normalized exponent can be |
| 2081 | * computed. The significand bits are copied into a |
| 2082 | * double significand; if the string has more logical bits |
| 2083 | * than can fit in a double, the extra bits affect the |
| 2084 | * round and sticky bits which are used to round the final |
| 2085 | * value. |
| 2086 | */ |
| 2087 | |
| 2088 | // Extract significand sign |
| 2089 | String group1 = m.group(1); |
| 2090 | double sign = (( group1 == null ) || group1.equals("+"))? 1.0 : -1.0; |
| 2091 | |
| 2092 | |
| 2093 | // Extract Significand magnitude |
| 2094 | /* |
| 2095 | * Based on the form of the significand, calculate how the |
| 2096 | * binary exponent needs to be adjusted to create a |
| 2097 | * normalized *hexadecimal* floating-point number; that |
| 2098 | * is, a number where there is one nonzero hex digit to |
| 2099 | * the left of the (hexa)decimal point. Since we are |
| 2100 | * adjusting a binary, not hexadecimal exponent, the |
| 2101 | * exponent is adjusted by a multiple of 4. |
| 2102 | * |
| 2103 | * There are a number of significand scenarios to consider; |
| 2104 | * letters are used in indicate nonzero digits: |
| 2105 | * |
| 2106 | * 1. 000xxxx => x.xxx normalized |
| 2107 | * increase exponent by (number of x's - 1)*4 |
| 2108 | * |
| 2109 | * 2. 000xxx.yyyy => x.xxyyyy normalized |
| 2110 | * increase exponent by (number of x's - 1)*4 |
| 2111 | * |
| 2112 | * 3. .000yyy => y.yy normalized |
| 2113 | * decrease exponent by (number of zeros + 1)*4 |
| 2114 | * |
| 2115 | * 4. 000.00000yyy => y.yy normalized |
| 2116 | * decrease exponent by (number of zeros to right of point + 1)*4 |
| 2117 | * |
| 2118 | * If the significand is exactly zero, return a properly |
| 2119 | * signed zero. |
| 2120 | */ |
| 2121 | |
| 2122 | String significandString =null; |
| 2123 | int signifLength = 0; |
| 2124 | int exponentAdjust = 0; |
| 2125 | { |
| 2126 | int leftDigits = 0; // number of meaningful digits to |
| 2127 | // left of "decimal" point |
| 2128 | // (leading zeros stripped) |
| 2129 | int rightDigits = 0; // number of digits to right of |
| 2130 | // "decimal" point; leading zeros |
| 2131 | // must always be accounted for |
| 2132 | /* |
| 2133 | * The significand is made up of either |
| 2134 | * |
| 2135 | * 1. group 4 entirely (integer portion only) |
| 2136 | * |
| 2137 | * OR |
| 2138 | * |
| 2139 | * 2. the fractional portion from group 7 plus any |
| 2140 | * (optional) integer portions from group 6. |
| 2141 | */ |
| 2142 | String group4; |
| 2143 | if( (group4 = m.group(4)) != null) { // Integer-only significand |
| 2144 | // Leading zeros never matter on the integer portion |
| 2145 | significandString = stripLeadingZeros(group4); |
| 2146 | leftDigits = significandString.length(); |
| 2147 | } |
| 2148 | else { |
| 2149 | // Group 6 is the optional integer; leading zeros |
| 2150 | // never matter on the integer portion |
| 2151 | String group6 = stripLeadingZeros(m.group(6)); |
| 2152 | leftDigits = group6.length(); |
| 2153 | |
| 2154 | // fraction |
| 2155 | String group7 = m.group(7); |
| 2156 | rightDigits = group7.length(); |
| 2157 | |
| 2158 | // Turn "integer.fraction" into "integer"+"fraction" |
| 2159 | significandString = |
| 2160 | ((group6 == null)?"":group6) + // is the null |
| 2161 | // check necessary? |
| 2162 | group7; |
| 2163 | } |
| 2164 | |
| 2165 | significandString = stripLeadingZeros(significandString); |
| 2166 | signifLength = significandString.length(); |
| 2167 | |
| 2168 | /* |
| 2169 | * Adjust exponent as described above |
| 2170 | */ |
| 2171 | if (leftDigits >= 1) { // Cases 1 and 2 |
| 2172 | exponentAdjust = 4*(leftDigits - 1); |
| 2173 | } else { // Cases 3 and 4 |
| 2174 | exponentAdjust = -4*( rightDigits - signifLength + 1); |
| 2175 | } |
| 2176 | |
| 2177 | // If the significand is zero, the exponent doesn't |
| 2178 | // matter; return a properly signed zero. |
| 2179 | |
| 2180 | if (signifLength == 0) { // Only zeros in input |
| 2181 | return new FormattedFloatingDecimal(sign * 0.0); |
| 2182 | } |
| 2183 | } |
| 2184 | |
| 2185 | // Extract Exponent |
| 2186 | /* |
| 2187 | * Use an int to read in the exponent value; this should |
| 2188 | * provide more than sufficient range for non-contrived |
| 2189 | * inputs. If reading the exponent in as an int does |
| 2190 | * overflow, examine the sign of the exponent and |
| 2191 | * significand to determine what to do. |
| 2192 | */ |
| 2193 | String group8 = m.group(8); |
| 2194 | boolean positiveExponent = ( group8 == null ) || group8.equals("+"); |
| 2195 | long unsignedRawExponent; |
| 2196 | try { |
| 2197 | unsignedRawExponent = Integer.parseInt(m.group(9)); |
| 2198 | } |
| 2199 | catch (NumberFormatException e) { |
| 2200 | // At this point, we know the exponent is |
| 2201 | // syntactically well-formed as a sequence of |
| 2202 | // digits. Therefore, if an NumberFormatException |
| 2203 | // is thrown, it must be due to overflowing int's |
| 2204 | // range. Also, at this point, we have already |
| 2205 | // checked for a zero significand. Thus the signs |
| 2206 | // of the exponent and significand determine the |
| 2207 | // final result: |
| 2208 | // |
| 2209 | // significand |
| 2210 | // + - |
| 2211 | // exponent + +infinity -infinity |
| 2212 | // - +0.0 -0.0 |
| 2213 | return new FormattedFloatingDecimal(sign * (positiveExponent ? |
| 2214 | Double.POSITIVE_INFINITY : 0.0)); |
| 2215 | } |
| 2216 | |
| 2217 | long rawExponent = |
| 2218 | (positiveExponent ? 1L : -1L) * // exponent sign |
| 2219 | unsignedRawExponent; // exponent magnitude |
| 2220 | |
| 2221 | // Calculate partially adjusted exponent |
| 2222 | long exponent = rawExponent + exponentAdjust ; |
| 2223 | |
| 2224 | // Starting copying non-zero bits into proper position in |
| 2225 | // a long; copy explicit bit too; this will be masked |
| 2226 | // later for normal values. |
| 2227 | |
| 2228 | boolean round = false; |
| 2229 | boolean sticky = false; |
| 2230 | int bitsCopied=0; |
| 2231 | int nextShift=0; |
| 2232 | long significand=0L; |
| 2233 | // First iteration is different, since we only copy |
| 2234 | // from the leading significand bit; one more exponent |
| 2235 | // adjust will be needed... |
| 2236 | |
| 2237 | // IMPORTANT: make leadingDigit a long to avoid |
| 2238 | // surprising shift semantics! |
| 2239 | long leadingDigit = getHexDigit(significandString, 0); |
| 2240 | |
| 2241 | /* |
| 2242 | * Left shift the leading digit (53 - (bit position of |
| 2243 | * leading 1 in digit)); this sets the top bit of the |
| 2244 | * significand to 1. The nextShift value is adjusted |
| 2245 | * to take into account the number of bit positions of |
| 2246 | * the leadingDigit actually used. Finally, the |
| 2247 | * exponent is adjusted to normalize the significand |
| 2248 | * as a binary value, not just a hex value. |
| 2249 | */ |
| 2250 | if (leadingDigit == 1) { |
| 2251 | significand |= leadingDigit << 52; |
| 2252 | nextShift = 52 - 4; |
| 2253 | /* exponent += 0 */ } |
| 2254 | else if (leadingDigit <= 3) { // [2, 3] |
| 2255 | significand |= leadingDigit << 51; |
| 2256 | nextShift = 52 - 5; |
| 2257 | exponent += 1; |
| 2258 | } |
| 2259 | else if (leadingDigit <= 7) { // [4, 7] |
| 2260 | significand |= leadingDigit << 50; |
| 2261 | nextShift = 52 - 6; |
| 2262 | exponent += 2; |
| 2263 | } |
| 2264 | else if (leadingDigit <= 15) { // [8, f] |
| 2265 | significand |= leadingDigit << 49; |
| 2266 | nextShift = 52 - 7; |
| 2267 | exponent += 3; |
| 2268 | } else { |
| 2269 | throw new AssertionError("Result from digit converstion too large!"); |
| 2270 | } |
| 2271 | // The preceding if-else could be replaced by a single |
| 2272 | // code block based on the high-order bit set in |
| 2273 | // leadingDigit. Given leadingOnePosition, |
| 2274 | |
| 2275 | // significand |= leadingDigit << (SIGNIFICAND_WIDTH - leadingOnePosition); |
| 2276 | // nextShift = 52 - (3 + leadingOnePosition); |
| 2277 | // exponent += (leadingOnePosition-1); |
| 2278 | |
| 2279 | |
| 2280 | /* |
| 2281 | * Now the exponent variable is equal to the normalized |
| 2282 | * binary exponent. Code below will make representation |
| 2283 | * adjustments if the exponent is incremented after |
| 2284 | * rounding (includes overflows to infinity) or if the |
| 2285 | * result is subnormal. |
| 2286 | */ |
| 2287 | |
| 2288 | // Copy digit into significand until the significand can't |
| 2289 | // hold another full hex digit or there are no more input |
| 2290 | // hex digits. |
| 2291 | int i = 0; |
| 2292 | for(i = 1; |
| 2293 | i < signifLength && nextShift >= 0; |
| 2294 | i++) { |
| 2295 | long currentDigit = getHexDigit(significandString, i); |
| 2296 | significand |= (currentDigit << nextShift); |
| 2297 | nextShift-=4; |
| 2298 | } |
| 2299 | |
| 2300 | // After the above loop, the bulk of the string is copied. |
| 2301 | // Now, we must copy any partial hex digits into the |
| 2302 | // significand AND compute the round bit and start computing |
| 2303 | // sticky bit. |
| 2304 | |
| 2305 | if ( i < signifLength ) { // at least one hex input digit exists |
| 2306 | long currentDigit = getHexDigit(significandString, i); |
| 2307 | |
| 2308 | // from nextShift, figure out how many bits need |
| 2309 | // to be copied, if any |
| 2310 | switch(nextShift) { // must be negative |
| 2311 | case -1: |
| 2312 | // three bits need to be copied in; can |
| 2313 | // set round bit |
| 2314 | significand |= ((currentDigit & 0xEL) >> 1); |
| 2315 | round = (currentDigit & 0x1L) != 0L; |
| 2316 | break; |
| 2317 | |
| 2318 | case -2: |
| 2319 | // two bits need to be copied in; can |
| 2320 | // set round and start sticky |
| 2321 | significand |= ((currentDigit & 0xCL) >> 2); |
| 2322 | round = (currentDigit &0x2L) != 0L; |
| 2323 | sticky = (currentDigit & 0x1L) != 0; |
| 2324 | break; |
| 2325 | |
| 2326 | case -3: |
| 2327 | // one bit needs to be copied in |
| 2328 | significand |= ((currentDigit & 0x8L)>>3); |
| 2329 | // Now set round and start sticky, if possible |
| 2330 | round = (currentDigit &0x4L) != 0L; |
| 2331 | sticky = (currentDigit & 0x3L) != 0; |
| 2332 | break; |
| 2333 | |
| 2334 | case -4: |
| 2335 | // all bits copied into significand; set |
| 2336 | // round and start sticky |
| 2337 | round = ((currentDigit & 0x8L) != 0); // is top bit set? |
| 2338 | // nonzeros in three low order bits? |
| 2339 | sticky = (currentDigit & 0x7L) != 0; |
| 2340 | break; |
| 2341 | |
| 2342 | default: |
| 2343 | throw new AssertionError("Unexpected shift distance remainder."); |
| 2344 | // break; |
| 2345 | } |
| 2346 | |
| 2347 | // Round is set; sticky might be set. |
| 2348 | |
| 2349 | // For the sticky bit, it suffices to check the |
| 2350 | // current digit and test for any nonzero digits in |
| 2351 | // the remaining unprocessed input. |
| 2352 | i++; |
| 2353 | while(i < signifLength && !sticky) { |
| 2354 | currentDigit = getHexDigit(significandString,i); |
| 2355 | sticky = sticky || (currentDigit != 0); |
| 2356 | i++; |
| 2357 | } |
| 2358 | |
| 2359 | } |
| 2360 | // else all of string was seen, round and sticky are |
| 2361 | // correct as false. |
| 2362 | |
| 2363 | |
| 2364 | // Check for overflow and update exponent accordingly. |
| 2365 | |
| 2366 | if (exponent > DoubleConsts.MAX_EXPONENT) { // Infinite result |
| 2367 | // overflow to properly signed infinity |
| 2368 | return new FormattedFloatingDecimal(sign * Double.POSITIVE_INFINITY); |
| 2369 | } else { // Finite return value |
| 2370 | if (exponent <= DoubleConsts.MAX_EXPONENT && // (Usually) normal result |
| 2371 | exponent >= DoubleConsts.MIN_EXPONENT) { |
| 2372 | |
| 2373 | // The result returned in this block cannot be a |
| 2374 | // zero or subnormal; however after the |
| 2375 | // significand is adjusted from rounding, we could |
| 2376 | // still overflow in infinity. |
| 2377 | |
| 2378 | // AND exponent bits into significand; if the |
| 2379 | // significand is incremented and overflows from |
| 2380 | // rounding, this combination will update the |
| 2381 | // exponent correctly, even in the case of |
| 2382 | // Double.MAX_VALUE overflowing to infinity. |
| 2383 | |
| 2384 | significand = (( ((long)exponent + |
| 2385 | (long)DoubleConsts.EXP_BIAS) << |
| 2386 | (DoubleConsts.SIGNIFICAND_WIDTH-1)) |
| 2387 | & DoubleConsts.EXP_BIT_MASK) | |
| 2388 | (DoubleConsts.SIGNIF_BIT_MASK & significand); |
| 2389 | |
| 2390 | } else { // Subnormal or zero |
| 2391 | // (exponent < DoubleConsts.MIN_EXPONENT) |
| 2392 | |
| 2393 | if (exponent < (DoubleConsts.MIN_SUB_EXPONENT -1 )) { |
| 2394 | // No way to round back to nonzero value |
| 2395 | // regardless of significand if the exponent is |
| 2396 | // less than -1075. |
| 2397 | return new FormattedFloatingDecimal(sign * 0.0); |
| 2398 | } else { // -1075 <= exponent <= MIN_EXPONENT -1 = -1023 |
| 2399 | /* |
| 2400 | * Find bit position to round to; recompute |
| 2401 | * round and sticky bits, and shift |
| 2402 | * significand right appropriately. |
| 2403 | */ |
| 2404 | |
| 2405 | sticky = sticky || round; |
| 2406 | round = false; |
| 2407 | |
| 2408 | // Number of bits of significand to preserve is |
| 2409 | // exponent - abs_min_exp +1 |
| 2410 | // check: |
| 2411 | // -1075 +1074 + 1 = 0 |
| 2412 | // -1023 +1074 + 1 = 52 |
| 2413 | |
| 2414 | int bitsDiscarded = 53 - |
| 2415 | ((int)exponent - DoubleConsts.MIN_SUB_EXPONENT + 1); |
| 2416 | assert bitsDiscarded >= 1 && bitsDiscarded <= 53; |
| 2417 | |
| 2418 | // What to do here: |
| 2419 | // First, isolate the new round bit |
| 2420 | round = (significand & (1L << (bitsDiscarded -1))) != 0L; |
| 2421 | if (bitsDiscarded > 1) { |
| 2422 | // create mask to update sticky bits; low |
| 2423 | // order bitsDiscarded bits should be 1 |
| 2424 | long mask = ~((~0L) << (bitsDiscarded -1)); |
| 2425 | sticky = sticky || ((significand & mask) != 0L ) ; |
| 2426 | } |
| 2427 | |
| 2428 | // Now, discard the bits |
| 2429 | significand = significand >> bitsDiscarded; |
| 2430 | |
| 2431 | significand = (( ((long)(DoubleConsts.MIN_EXPONENT -1) + // subnorm exp. |
| 2432 | (long)DoubleConsts.EXP_BIAS) << |
| 2433 | (DoubleConsts.SIGNIFICAND_WIDTH-1)) |
| 2434 | & DoubleConsts.EXP_BIT_MASK) | |
| 2435 | (DoubleConsts.SIGNIF_BIT_MASK & significand); |
| 2436 | } |
| 2437 | } |
| 2438 | |
| 2439 | // The significand variable now contains the currently |
| 2440 | // appropriate exponent bits too. |
| 2441 | |
| 2442 | /* |
| 2443 | * Determine if significand should be incremented; |
| 2444 | * making this determination depends on the least |
| 2445 | * significant bit and the round and sticky bits. |
| 2446 | * |
| 2447 | * Round to nearest even rounding table, adapted from |
| 2448 | * table 4.7 in "Computer Arithmetic" by IsraelKoren. |
| 2449 | * The digit to the left of the "decimal" point is the |
| 2450 | * least significant bit, the digits to the right of |
| 2451 | * the point are the round and sticky bits |
| 2452 | * |
| 2453 | * Number Round(x) |
| 2454 | * x0.00 x0. |
| 2455 | * x0.01 x0. |
| 2456 | * x0.10 x0. |
| 2457 | * x0.11 x1. = x0. +1 |
| 2458 | * x1.00 x1. |
| 2459 | * x1.01 x1. |
| 2460 | * x1.10 x1. + 1 |
| 2461 | * x1.11 x1. + 1 |
| 2462 | */ |
| 2463 | boolean incremented = false; |
| 2464 | boolean leastZero = ((significand & 1L) == 0L); |
| 2465 | if( ( leastZero && round && sticky ) || |
| 2466 | ((!leastZero) && round )) { |
| 2467 | incremented = true; |
| 2468 | significand++; |
| 2469 | } |
| 2470 | |
| 2471 | FormattedFloatingDecimal fd = new FormattedFloatingDecimal(FpUtils.rawCopySign( |
| 2472 | Double.longBitsToDouble(significand), |
| 2473 | sign)); |
| 2474 | |
| 2475 | /* |
| 2476 | * Set roundingDir variable field of fd properly so |
| 2477 | * that the input string can be properly rounded to a |
| 2478 | * float value. There are two cases to consider: |
| 2479 | * |
| 2480 | * 1. rounding to double discards sticky bit |
| 2481 | * information that would change the result of a float |
| 2482 | * rounding (near halfway case between two floats) |
| 2483 | * |
| 2484 | * 2. rounding to double rounds up when rounding up |
| 2485 | * would not occur when rounding to float. |
| 2486 | * |
| 2487 | * For former case only needs to be considered when |
| 2488 | * the bits rounded away when casting to float are all |
| 2489 | * zero; otherwise, float round bit is properly set |
| 2490 | * and sticky will already be true. |
| 2491 | * |
| 2492 | * The lower exponent bound for the code below is the |
| 2493 | * minimum (normalized) subnormal exponent - 1 since a |
| 2494 | * value with that exponent can round up to the |
| 2495 | * minimum subnormal value and the sticky bit |
| 2496 | * information must be preserved (i.e. case 1). |
| 2497 | */ |
| 2498 | if ((exponent >= FloatConsts.MIN_SUB_EXPONENT-1) && |
| 2499 | (exponent <= FloatConsts.MAX_EXPONENT ) ){ |
| 2500 | // Outside above exponent range, the float value |
| 2501 | // will be zero or infinity. |
| 2502 | |
| 2503 | /* |
| 2504 | * If the low-order 28 bits of a rounded double |
| 2505 | * significand are 0, the double could be a |
| 2506 | * half-way case for a rounding to float. If the |
| 2507 | * double value is a half-way case, the double |
| 2508 | * significand may have to be modified to round |
| 2509 | * the the right float value (see the stickyRound |
| 2510 | * method). If the rounding to double has lost |
| 2511 | * what would be float sticky bit information, the |
| 2512 | * double significand must be incremented. If the |
| 2513 | * double value's significand was itself |
| 2514 | * incremented, the float value may end up too |
| 2515 | * large so the increment should be undone. |
| 2516 | */ |
| 2517 | if ((significand & 0xfffffffL) == 0x0L) { |
| 2518 | // For negative values, the sign of the |
| 2519 | // roundDir is the same as for positive values |
| 2520 | // since adding 1 increasing the significand's |
| 2521 | // magnitude and subtracting 1 decreases the |
| 2522 | // significand's magnitude. If neither round |
| 2523 | // nor sticky is true, the double value is |
| 2524 | // exact and no adjustment is required for a |
| 2525 | // proper float rounding. |
| 2526 | if( round || sticky) { |
| 2527 | if (leastZero) { // prerounding lsb is 0 |
| 2528 | // If round and sticky were both true, |
| 2529 | // and the least significant |
| 2530 | // significand bit were 0, the rounded |
| 2531 | // significand would not have its |
| 2532 | // low-order bits be zero. Therefore, |
| 2533 | // we only need to adjust the |
| 2534 | // significand if round XOR sticky is |
| 2535 | // true. |
| 2536 | if (round ^ sticky) { |
| 2537 | fd.roundDir = 1; |
| 2538 | } |
| 2539 | } |
| 2540 | else { // prerounding lsb is 1 |
| 2541 | // If the prerounding lsb is 1 and the |
| 2542 | // resulting significand has its |
| 2543 | // low-order bits zero, the significand |
| 2544 | // was incremented. Here, we undo the |
| 2545 | // increment, which will ensure the |
| 2546 | // right guard and sticky bits for the |
| 2547 | // float rounding. |
| 2548 | if (round) |
| 2549 | fd.roundDir = -1; |
| 2550 | } |
| 2551 | } |
| 2552 | } |
| 2553 | } |
| 2554 | |
| 2555 | fd.fromHex = true; |
| 2556 | return fd; |
| 2557 | } |
| 2558 | } |
| 2559 | } |
| 2560 | |
| 2561 | /** |
| 2562 | * Return <code>s</code> with any leading zeros removed. |
| 2563 | */ |
| 2564 | static String stripLeadingZeros(String s) { |
| 2565 | return s.replaceFirst("^0+", ""); |
| 2566 | } |
| 2567 | |
| 2568 | /** |
| 2569 | * Extract a hexadecimal digit from position <code>position</code> |
| 2570 | * of string <code>s</code>. |
| 2571 | */ |
| 2572 | static int getHexDigit(String s, int position) { |
| 2573 | int value = Character.digit(s.charAt(position), 16); |
| 2574 | if (value <= -1 || value >= 16) { |
| 2575 | throw new AssertionError("Unxpected failure of digit converstion of " + |
| 2576 | s.charAt(position)); |
| 2577 | } |
| 2578 | return value; |
| 2579 | } |
| 2580 | |
| 2581 | |
| 2582 | } |