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erik.corry@gmail.com4a6c3272010-11-18 12:04:40 +00001// Copyright 2010 the V8 project authors. All rights reserved.
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27
28#include <math.h>
29
30#include "v8.h"
31#include "bignum-dtoa.h"
32
33#include "bignum.h"
34#include "double.h"
35
36namespace v8 {
37namespace internal {
38
39static int NormalizedExponent(uint64_t significand, int exponent) {
40 ASSERT(significand != 0);
41 while ((significand & Double::kHiddenBit) == 0) {
42 significand = significand << 1;
43 exponent = exponent - 1;
44 }
45 return exponent;
46}
47
48
49// Forward declarations:
50// Returns an estimation of k such that 10^(k-1) <= v < 10^k.
51static int EstimatePower(int exponent);
52// Computes v / 10^estimated_power exactly, as a ratio of two bignums, numerator
53// and denominator.
54static void InitialScaledStartValues(double v,
55 int estimated_power,
56 bool need_boundary_deltas,
57 Bignum* numerator,
58 Bignum* denominator,
59 Bignum* delta_minus,
60 Bignum* delta_plus);
61// Multiplies numerator/denominator so that its values lies in the range 1-10.
62// Returns decimal_point s.t.
63// v = numerator'/denominator' * 10^(decimal_point-1)
64// where numerator' and denominator' are the values of numerator and
65// denominator after the call to this function.
66static void FixupMultiply10(int estimated_power, bool is_even,
67 int* decimal_point,
68 Bignum* numerator, Bignum* denominator,
69 Bignum* delta_minus, Bignum* delta_plus);
70// Generates digits from the left to the right and stops when the generated
71// digits yield the shortest decimal representation of v.
72static void GenerateShortestDigits(Bignum* numerator, Bignum* denominator,
73 Bignum* delta_minus, Bignum* delta_plus,
74 bool is_even,
75 Vector<char> buffer, int* length);
76// Generates 'requested_digits' after the decimal point.
77static void BignumToFixed(int requested_digits, int* decimal_point,
78 Bignum* numerator, Bignum* denominator,
79 Vector<char>(buffer), int* length);
80// Generates 'count' digits of numerator/denominator.
81// Once 'count' digits have been produced rounds the result depending on the
82// remainder (remainders of exactly .5 round upwards). Might update the
83// decimal_point when rounding up (for example for 0.9999).
84static void GenerateCountedDigits(int count, int* decimal_point,
85 Bignum* numerator, Bignum* denominator,
86 Vector<char>(buffer), int* length);
87
88
89void BignumDtoa(double v, BignumDtoaMode mode, int requested_digits,
90 Vector<char> buffer, int* length, int* decimal_point) {
91 ASSERT(v > 0);
92 ASSERT(!Double(v).IsSpecial());
93 uint64_t significand = Double(v).Significand();
94 bool is_even = (significand & 1) == 0;
95 int exponent = Double(v).Exponent();
96 int normalized_exponent = NormalizedExponent(significand, exponent);
97 // estimated_power might be too low by 1.
98 int estimated_power = EstimatePower(normalized_exponent);
99
100 // Shortcut for Fixed.
101 // The requested digits correspond to the digits after the point. If the
102 // number is much too small, then there is no need in trying to get any
103 // digits.
104 if (mode == BIGNUM_DTOA_FIXED && -estimated_power - 1 > requested_digits) {
105 buffer[0] = '\0';
106 *length = 0;
107 // Set decimal-point to -requested_digits. This is what Gay does.
108 // Note that it should not have any effect anyways since the string is
109 // empty.
110 *decimal_point = -requested_digits;
111 return;
112 }
113
114 Bignum numerator;
115 Bignum denominator;
116 Bignum delta_minus;
117 Bignum delta_plus;
118 // Make sure the bignum can grow large enough. The smallest double equals
119 // 4e-324. In this case the denominator needs fewer than 324*4 binary digits.
120 // The maximum double is 1.7976931348623157e308 which needs fewer than
121 // 308*4 binary digits.
122 ASSERT(Bignum::kMaxSignificantBits >= 324*4);
123 bool need_boundary_deltas = (mode == BIGNUM_DTOA_SHORTEST);
124 InitialScaledStartValues(v, estimated_power, need_boundary_deltas,
125 &numerator, &denominator,
126 &delta_minus, &delta_plus);
127 // We now have v = (numerator / denominator) * 10^estimated_power.
128 FixupMultiply10(estimated_power, is_even, decimal_point,
129 &numerator, &denominator,
130 &delta_minus, &delta_plus);
131 // We now have v = (numerator / denominator) * 10^(decimal_point-1), and
132 // 1 <= (numerator + delta_plus) / denominator < 10
133 switch (mode) {
134 case BIGNUM_DTOA_SHORTEST:
135 GenerateShortestDigits(&numerator, &denominator,
136 &delta_minus, &delta_plus,
137 is_even, buffer, length);
138 break;
139 case BIGNUM_DTOA_FIXED:
140 BignumToFixed(requested_digits, decimal_point,
141 &numerator, &denominator,
142 buffer, length);
143 break;
144 case BIGNUM_DTOA_PRECISION:
145 GenerateCountedDigits(requested_digits, decimal_point,
146 &numerator, &denominator,
147 buffer, length);
148 break;
149 default:
150 UNREACHABLE();
151 }
152 buffer[*length] = '\0';
153}
154
155
156// The procedure starts generating digits from the left to the right and stops
157// when the generated digits yield the shortest decimal representation of v. A
158// decimal representation of v is a number lying closer to v than to any other
159// double, so it converts to v when read.
160//
161// This is true if d, the decimal representation, is between m- and m+, the
162// upper and lower boundaries. d must be strictly between them if !is_even.
163// m- := (numerator - delta_minus) / denominator
164// m+ := (numerator + delta_plus) / denominator
165//
166// Precondition: 0 <= (numerator+delta_plus) / denominator < 10.
167// If 1 <= (numerator+delta_plus) / denominator < 10 then no leading 0 digit
168// will be produced. This should be the standard precondition.
169static void GenerateShortestDigits(Bignum* numerator, Bignum* denominator,
170 Bignum* delta_minus, Bignum* delta_plus,
171 bool is_even,
172 Vector<char> buffer, int* length) {
173 // Small optimization: if delta_minus and delta_plus are the same just reuse
174 // one of the two bignums.
175 if (Bignum::Equal(*delta_minus, *delta_plus)) {
176 delta_plus = delta_minus;
177 }
178 *length = 0;
179 while (true) {
180 uint16_t digit;
181 digit = numerator->DivideModuloIntBignum(*denominator);
182 ASSERT(digit <= 9); // digit is a uint16_t and therefore always positive.
183 // digit = numerator / denominator (integer division).
184 // numerator = numerator % denominator.
185 buffer[(*length)++] = digit + '0';
186
187 // Can we stop already?
188 // If the remainder of the division is less than the distance to the lower
189 // boundary we can stop. In this case we simply round down (discarding the
190 // remainder).
191 // Similarly we test if we can round up (using the upper boundary).
192 bool in_delta_room_minus;
193 bool in_delta_room_plus;
194 if (is_even) {
195 in_delta_room_minus = Bignum::LessEqual(*numerator, *delta_minus);
196 } else {
197 in_delta_room_minus = Bignum::Less(*numerator, *delta_minus);
198 }
199 if (is_even) {
200 in_delta_room_plus =
201 Bignum::PlusCompare(*numerator, *delta_plus, *denominator) >= 0;
202 } else {
203 in_delta_room_plus =
204 Bignum::PlusCompare(*numerator, *delta_plus, *denominator) > 0;
205 }
206 if (!in_delta_room_minus && !in_delta_room_plus) {
207 // Prepare for next iteration.
208 numerator->Times10();
209 delta_minus->Times10();
210 // We optimized delta_plus to be equal to delta_minus (if they share the
211 // same value). So don't multiply delta_plus if they point to the same
212 // object.
213 if (delta_minus != delta_plus) {
214 delta_plus->Times10();
215 }
216 } else if (in_delta_room_minus && in_delta_room_plus) {
217 // Let's see if 2*numerator < denominator.
218 // If yes, then the next digit would be < 5 and we can round down.
219 int compare = Bignum::PlusCompare(*numerator, *numerator, *denominator);
220 if (compare < 0) {
221 // Remaining digits are less than .5. -> Round down (== do nothing).
222 } else if (compare > 0) {
223 // Remaining digits are more than .5 of denominator. -> Round up.
224 // Note that the last digit could not be a '9' as otherwise the whole
225 // loop would have stopped earlier.
226 // We still have an assert here in case the preconditions were not
227 // satisfied.
228 ASSERT(buffer[(*length) - 1] != '9');
229 buffer[(*length) - 1]++;
230 } else {
231 // Halfway case.
232 // TODO(floitsch): need a way to solve half-way cases.
233 // For now let's round towards even (since this is what Gay seems to
234 // do).
235
236 if ((buffer[(*length) - 1] - '0') % 2 == 0) {
237 // Round down => Do nothing.
238 } else {
239 ASSERT(buffer[(*length) - 1] != '9');
240 buffer[(*length) - 1]++;
241 }
242 }
243 return;
244 } else if (in_delta_room_minus) {
245 // Round down (== do nothing).
246 return;
247 } else { // in_delta_room_plus
248 // Round up.
249 // Note again that the last digit could not be '9' since this would have
250 // stopped the loop earlier.
251 // We still have an ASSERT here, in case the preconditions were not
252 // satisfied.
253 ASSERT(buffer[(*length) -1] != '9');
254 buffer[(*length) - 1]++;
255 return;
256 }
257 }
258}
259
260
261// Let v = numerator / denominator < 10.
262// Then we generate 'count' digits of d = x.xxxxx... (without the decimal point)
263// from left to right. Once 'count' digits have been produced we decide wether
264// to round up or down. Remainders of exactly .5 round upwards. Numbers such
265// as 9.999999 propagate a carry all the way, and change the
266// exponent (decimal_point), when rounding upwards.
267static void GenerateCountedDigits(int count, int* decimal_point,
268 Bignum* numerator, Bignum* denominator,
269 Vector<char>(buffer), int* length) {
270 ASSERT(count >= 0);
271 for (int i = 0; i < count - 1; ++i) {
272 uint16_t digit;
273 digit = numerator->DivideModuloIntBignum(*denominator);
274 ASSERT(digit <= 9); // digit is a uint16_t and therefore always positive.
275 // digit = numerator / denominator (integer division).
276 // numerator = numerator % denominator.
277 buffer[i] = digit + '0';
278 // Prepare for next iteration.
279 numerator->Times10();
280 }
281 // Generate the last digit.
282 uint16_t digit;
283 digit = numerator->DivideModuloIntBignum(*denominator);
284 if (Bignum::PlusCompare(*numerator, *numerator, *denominator) >= 0) {
285 digit++;
286 }
287 buffer[count - 1] = digit + '0';
288 // Correct bad digits (in case we had a sequence of '9's). Propagate the
289 // carry until we hat a non-'9' or til we reach the first digit.
290 for (int i = count - 1; i > 0; --i) {
291 if (buffer[i] != '0' + 10) break;
292 buffer[i] = '0';
293 buffer[i - 1]++;
294 }
295 if (buffer[0] == '0' + 10) {
296 // Propagate a carry past the top place.
297 buffer[0] = '1';
298 (*decimal_point)++;
299 }
300 *length = count;
301}
302
303
304// Generates 'requested_digits' after the decimal point. It might omit
305// trailing '0's. If the input number is too small then no digits at all are
306// generated (ex.: 2 fixed digits for 0.00001).
307//
308// Input verifies: 1 <= (numerator + delta) / denominator < 10.
309static void BignumToFixed(int requested_digits, int* decimal_point,
310 Bignum* numerator, Bignum* denominator,
311 Vector<char>(buffer), int* length) {
312 // Note that we have to look at more than just the requested_digits, since
313 // a number could be rounded up. Example: v=0.5 with requested_digits=0.
314 // Even though the power of v equals 0 we can't just stop here.
315 if (-(*decimal_point) > requested_digits) {
316 // The number is definitively too small.
317 // Ex: 0.001 with requested_digits == 1.
318 // Set decimal-point to -requested_digits. This is what Gay does.
319 // Note that it should not have any effect anyways since the string is
320 // empty.
321 *decimal_point = -requested_digits;
322 *length = 0;
323 return;
324 } else if (-(*decimal_point) == requested_digits) {
325 // We only need to verify if the number rounds down or up.
326 // Ex: 0.04 and 0.06 with requested_digits == 1.
327 ASSERT(*decimal_point == -requested_digits);
328 // Initially the fraction lies in range (1, 10]. Multiply the denominator
329 // by 10 so that we can compare more easily.
330 denominator->Times10();
331 if (Bignum::PlusCompare(*numerator, *numerator, *denominator) >= 0) {
332 // If the fraction is >= 0.5 then we have to include the rounded
333 // digit.
334 buffer[0] = '1';
335 *length = 1;
336 (*decimal_point)++;
337 } else {
338 // Note that we caught most of similar cases earlier.
339 *length = 0;
340 }
341 return;
342 } else {
343 // The requested digits correspond to the digits after the point.
344 // The variable 'needed_digits' includes the digits before the point.
345 int needed_digits = (*decimal_point) + requested_digits;
346 GenerateCountedDigits(needed_digits, decimal_point,
347 numerator, denominator,
348 buffer, length);
349 }
350}
351
352
353// Returns an estimation of k such that 10^(k-1) <= v < 10^k where
354// v = f * 2^exponent and 2^52 <= f < 2^53.
355// v is hence a normalized double with the given exponent. The output is an
356// approximation for the exponent of the decimal approimation .digits * 10^k.
357//
358// The result might undershoot by 1 in which case 10^k <= v < 10^k+1.
359// Note: this property holds for v's upper boundary m+ too.
360// 10^k <= m+ < 10^k+1.
361// (see explanation below).
362//
363// Examples:
364// EstimatePower(0) => 16
365// EstimatePower(-52) => 0
366//
367// Note: e >= 0 => EstimatedPower(e) > 0. No similar claim can be made for e<0.
368static int EstimatePower(int exponent) {
369 // This function estimates log10 of v where v = f*2^e (with e == exponent).
370 // Note that 10^floor(log10(v)) <= v, but v <= 10^ceil(log10(v)).
371 // Note that f is bounded by its container size. Let p = 53 (the double's
372 // significand size). Then 2^(p-1) <= f < 2^p.
373 //
374 // Given that log10(v) == log2(v)/log2(10) and e+(len(f)-1) is quite close
375 // to log2(v) the function is simplified to (e+(len(f)-1)/log2(10)).
376 // The computed number undershoots by less than 0.631 (when we compute log3
377 // and not log10).
378 //
379 // Optimization: since we only need an approximated result this computation
380 // can be performed on 64 bit integers. On x86/x64 architecture the speedup is
381 // not really measurable, though.
382 //
383 // Since we want to avoid overshooting we decrement by 1e10 so that
384 // floating-point imprecisions don't affect us.
385 //
386 // Explanation for v's boundary m+: the computation takes advantage of
387 // the fact that 2^(p-1) <= f < 2^p. Boundaries still satisfy this requirement
388 // (even for denormals where the delta can be much more important).
389
390 const double k1Log10 = 0.30102999566398114; // 1/lg(10)
391
392 // For doubles len(f) == 53 (don't forget the hidden bit).
393 const int kSignificandSize = 53;
394 double estimate = ceil((exponent + kSignificandSize - 1) * k1Log10 - 1e-10);
395 return static_cast<int>(estimate);
396}
397
398
399// See comments for InitialScaledStartValues.
400static void InitialScaledStartValuesPositiveExponent(
401 double v, int estimated_power, bool need_boundary_deltas,
402 Bignum* numerator, Bignum* denominator,
403 Bignum* delta_minus, Bignum* delta_plus) {
404 // A positive exponent implies a positive power.
405 ASSERT(estimated_power >= 0);
406 // Since the estimated_power is positive we simply multiply the denominator
407 // by 10^estimated_power.
408
409 // numerator = v.
410 numerator->AssignUInt64(Double(v).Significand());
411 numerator->ShiftLeft(Double(v).Exponent());
412 // denominator = 10^estimated_power.
413 denominator->AssignPowerUInt16(10, estimated_power);
414
415 if (need_boundary_deltas) {
416 // Introduce a common denominator so that the deltas to the boundaries are
417 // integers.
418 denominator->ShiftLeft(1);
419 numerator->ShiftLeft(1);
420 // Let v = f * 2^e, then m+ - v = 1/2 * 2^e; With the common
421 // denominator (of 2) delta_plus equals 2^e.
422 delta_plus->AssignUInt16(1);
423 delta_plus->ShiftLeft(Double(v).Exponent());
424 // Same for delta_minus (with adjustments below if f == 2^p-1).
425 delta_minus->AssignUInt16(1);
426 delta_minus->ShiftLeft(Double(v).Exponent());
427
428 // If the significand (without the hidden bit) is 0, then the lower
429 // boundary is closer than just half a ulp (unit in the last place).
430 // There is only one exception: if the next lower number is a denormal then
431 // the distance is 1 ulp. This cannot be the case for exponent >= 0 (but we
432 // have to test it in the other function where exponent < 0).
433 uint64_t v_bits = Double(v).AsUint64();
434 if ((v_bits & Double::kSignificandMask) == 0) {
435 // The lower boundary is closer at half the distance of "normal" numbers.
436 // Increase the common denominator and adapt all but the delta_minus.
437 denominator->ShiftLeft(1); // *2
438 numerator->ShiftLeft(1); // *2
439 delta_plus->ShiftLeft(1); // *2
440 }
441 }
442}
443
444
445// See comments for InitialScaledStartValues
446static void InitialScaledStartValuesNegativeExponentPositivePower(
447 double v, int estimated_power, bool need_boundary_deltas,
448 Bignum* numerator, Bignum* denominator,
449 Bignum* delta_minus, Bignum* delta_plus) {
450 uint64_t significand = Double(v).Significand();
451 int exponent = Double(v).Exponent();
452 // v = f * 2^e with e < 0, and with estimated_power >= 0.
453 // This means that e is close to 0 (have a look at how estimated_power is
454 // computed).
455
456 // numerator = significand
457 // since v = significand * 2^exponent this is equivalent to
458 // numerator = v * / 2^-exponent
459 numerator->AssignUInt64(significand);
460 // denominator = 10^estimated_power * 2^-exponent (with exponent < 0)
461 denominator->AssignPowerUInt16(10, estimated_power);
462 denominator->ShiftLeft(-exponent);
463
464 if (need_boundary_deltas) {
465 // Introduce a common denominator so that the deltas to the boundaries are
466 // integers.
467 denominator->ShiftLeft(1);
468 numerator->ShiftLeft(1);
469 // Let v = f * 2^e, then m+ - v = 1/2 * 2^e; With the common
470 // denominator (of 2) delta_plus equals 2^e.
471 // Given that the denominator already includes v's exponent the distance
472 // to the boundaries is simply 1.
473 delta_plus->AssignUInt16(1);
474 // Same for delta_minus (with adjustments below if f == 2^p-1).
475 delta_minus->AssignUInt16(1);
476
477 // If the significand (without the hidden bit) is 0, then the lower
478 // boundary is closer than just one ulp (unit in the last place).
479 // There is only one exception: if the next lower number is a denormal
480 // then the distance is 1 ulp. Since the exponent is close to zero
481 // (otherwise estimated_power would have been negative) this cannot happen
482 // here either.
483 uint64_t v_bits = Double(v).AsUint64();
484 if ((v_bits & Double::kSignificandMask) == 0) {
485 // The lower boundary is closer at half the distance of "normal" numbers.
486 // Increase the denominator and adapt all but the delta_minus.
487 denominator->ShiftLeft(1); // *2
488 numerator->ShiftLeft(1); // *2
489 delta_plus->ShiftLeft(1); // *2
490 }
491 }
492}
493
494
495// See comments for InitialScaledStartValues
496static void InitialScaledStartValuesNegativeExponentNegativePower(
497 double v, int estimated_power, bool need_boundary_deltas,
498 Bignum* numerator, Bignum* denominator,
499 Bignum* delta_minus, Bignum* delta_plus) {
500 const uint64_t kMinimalNormalizedExponent =
501 V8_2PART_UINT64_C(0x00100000, 00000000);
502 uint64_t significand = Double(v).Significand();
503 int exponent = Double(v).Exponent();
504 // Instead of multiplying the denominator with 10^estimated_power we
505 // multiply all values (numerator and deltas) by 10^-estimated_power.
506
507 // Use numerator as temporary container for power_ten.
508 Bignum* power_ten = numerator;
509 power_ten->AssignPowerUInt16(10, -estimated_power);
510
511 if (need_boundary_deltas) {
512 // Since power_ten == numerator we must make a copy of 10^estimated_power
513 // before we complete the computation of the numerator.
514 // delta_plus = delta_minus = 10^estimated_power
515 delta_plus->AssignBignum(*power_ten);
516 delta_minus->AssignBignum(*power_ten);
517 }
518
519 // numerator = significand * 2 * 10^-estimated_power
520 // since v = significand * 2^exponent this is equivalent to
521 // numerator = v * 10^-estimated_power * 2 * 2^-exponent.
522 // Remember: numerator has been abused as power_ten. So no need to assign it
523 // to itself.
524 ASSERT(numerator == power_ten);
525 numerator->MultiplyByUInt64(significand);
526
527 // denominator = 2 * 2^-exponent with exponent < 0.
528 denominator->AssignUInt16(1);
529 denominator->ShiftLeft(-exponent);
530
531 if (need_boundary_deltas) {
532 // Introduce a common denominator so that the deltas to the boundaries are
533 // integers.
534 numerator->ShiftLeft(1);
535 denominator->ShiftLeft(1);
536 // With this shift the boundaries have their correct value, since
537 // delta_plus = 10^-estimated_power, and
538 // delta_minus = 10^-estimated_power.
539 // These assignments have been done earlier.
540
541 // The special case where the lower boundary is twice as close.
542 // This time we have to look out for the exception too.
543 uint64_t v_bits = Double(v).AsUint64();
544 if ((v_bits & Double::kSignificandMask) == 0 &&
545 // The only exception where a significand == 0 has its boundaries at
546 // "normal" distances:
547 (v_bits & Double::kExponentMask) != kMinimalNormalizedExponent) {
548 numerator->ShiftLeft(1); // *2
549 denominator->ShiftLeft(1); // *2
550 delta_plus->ShiftLeft(1); // *2
551 }
552 }
553}
554
555
556// Let v = significand * 2^exponent.
557// Computes v / 10^estimated_power exactly, as a ratio of two bignums, numerator
558// and denominator. The functions GenerateShortestDigits and
559// GenerateCountedDigits will then convert this ratio to its decimal
560// representation d, with the required accuracy.
561// Then d * 10^estimated_power is the representation of v.
562// (Note: the fraction and the estimated_power might get adjusted before
563// generating the decimal representation.)
564//
565// The initial start values consist of:
566// - a scaled numerator: s.t. numerator/denominator == v / 10^estimated_power.
567// - a scaled (common) denominator.
568// optionally (used by GenerateShortestDigits to decide if it has the shortest
569// decimal converting back to v):
570// - v - m-: the distance to the lower boundary.
571// - m+ - v: the distance to the upper boundary.
572//
573// v, m+, m-, and therefore v - m- and m+ - v all share the same denominator.
574//
575// Let ep == estimated_power, then the returned values will satisfy:
576// v / 10^ep = numerator / denominator.
577// v's boundarys m- and m+:
578// m- / 10^ep == v / 10^ep - delta_minus / denominator
579// m+ / 10^ep == v / 10^ep + delta_plus / denominator
580// Or in other words:
581// m- == v - delta_minus * 10^ep / denominator;
582// m+ == v + delta_plus * 10^ep / denominator;
583//
584// Since 10^(k-1) <= v < 10^k (with k == estimated_power)
585// or 10^k <= v < 10^(k+1)
586// we then have 0.1 <= numerator/denominator < 1
587// or 1 <= numerator/denominator < 10
588//
589// It is then easy to kickstart the digit-generation routine.
590//
591// The boundary-deltas are only filled if need_boundary_deltas is set.
592static void InitialScaledStartValues(double v,
593 int estimated_power,
594 bool need_boundary_deltas,
595 Bignum* numerator,
596 Bignum* denominator,
597 Bignum* delta_minus,
598 Bignum* delta_plus) {
599 if (Double(v).Exponent() >= 0) {
600 InitialScaledStartValuesPositiveExponent(
601 v, estimated_power, need_boundary_deltas,
602 numerator, denominator, delta_minus, delta_plus);
603 } else if (estimated_power >= 0) {
604 InitialScaledStartValuesNegativeExponentPositivePower(
605 v, estimated_power, need_boundary_deltas,
606 numerator, denominator, delta_minus, delta_plus);
607 } else {
608 InitialScaledStartValuesNegativeExponentNegativePower(
609 v, estimated_power, need_boundary_deltas,
610 numerator, denominator, delta_minus, delta_plus);
611 }
612}
613
614
615// This routine multiplies numerator/denominator so that its values lies in the
616// range 1-10. That is after a call to this function we have:
617// 1 <= (numerator + delta_plus) /denominator < 10.
618// Let numerator the input before modification and numerator' the argument
619// after modification, then the output-parameter decimal_point is such that
620// numerator / denominator * 10^estimated_power ==
621// numerator' / denominator' * 10^(decimal_point - 1)
622// In some cases estimated_power was too low, and this is already the case. We
623// then simply adjust the power so that 10^(k-1) <= v < 10^k (with k ==
624// estimated_power) but do not touch the numerator or denominator.
625// Otherwise the routine multiplies the numerator and the deltas by 10.
626static void FixupMultiply10(int estimated_power, bool is_even,
627 int* decimal_point,
628 Bignum* numerator, Bignum* denominator,
629 Bignum* delta_minus, Bignum* delta_plus) {
630 bool in_range;
631 if (is_even) {
632 // For IEEE doubles half-way cases (in decimal system numbers ending with 5)
633 // are rounded to the closest floating-point number with even significand.
634 in_range = Bignum::PlusCompare(*numerator, *delta_plus, *denominator) >= 0;
635 } else {
636 in_range = Bignum::PlusCompare(*numerator, *delta_plus, *denominator) > 0;
637 }
638 if (in_range) {
639 // Since numerator + delta_plus >= denominator we already have
640 // 1 <= numerator/denominator < 10. Simply update the estimated_power.
641 *decimal_point = estimated_power + 1;
642 } else {
643 *decimal_point = estimated_power;
644 numerator->Times10();
645 if (Bignum::Equal(*delta_minus, *delta_plus)) {
646 delta_minus->Times10();
647 delta_plus->AssignBignum(*delta_minus);
648 } else {
649 delta_minus->Times10();
650 delta_plus->Times10();
651 }
652 }
653}
654
655} } // namespace v8::internal