Merged revisions 71772 via svnmerge from
svn+ssh://pythondev@svn.python.org/python/trunk
........
r71772 | mark.dickinson | 2009-04-20 22:13:33 +0100 (Mon, 20 Apr 2009) | 5 lines
Issue #3166: Make long -> float (and int -> float) conversions
correctly rounded, using round-half-to-even. This ensures that the
value of float(n) doesn't depend on whether we're using 15-bit digits
or 30-bit digits for Python longs.
........
diff --git a/Lib/test/test_long.py b/Lib/test/test_long.py
index 92285b2..53d3e6b 100644
--- a/Lib/test/test_long.py
+++ b/Lib/test/test_long.py
@@ -620,6 +620,65 @@
else:
self.assertRaises(TypeError, pow,longx, longy, int(z))
+ @unittest.skipUnless(float.__getformat__("double").startswith("IEEE"),
+ "test requires IEEE 754 doubles")
+ def test_float_conversion(self):
+ import sys
+ DBL_MAX = sys.float_info.max
+ DBL_MAX_EXP = sys.float_info.max_exp
+ DBL_MANT_DIG = sys.float_info.mant_dig
+
+ exact_values = [0, 1, 2,
+ 2**53-3,
+ 2**53-2,
+ 2**53-1,
+ 2**53,
+ 2**53+2,
+ 2**54-4,
+ 2**54-2,
+ 2**54,
+ 2**54+4]
+ for x in exact_values:
+ self.assertEqual(float(x), x)
+ self.assertEqual(float(-x), -x)
+
+ # test round-half-even
+ for x, y in [(1, 0), (2, 2), (3, 4), (4, 4), (5, 4), (6, 6), (7, 8)]:
+ for p in range(15):
+ self.assertEqual(int(float(2**p*(2**53+x))), 2**p*(2**53+y))
+
+ for x, y in [(0, 0), (1, 0), (2, 0), (3, 4), (4, 4), (5, 4), (6, 8),
+ (7, 8), (8, 8), (9, 8), (10, 8), (11, 12), (12, 12),
+ (13, 12), (14, 16), (15, 16)]:
+ for p in range(15):
+ self.assertEqual(int(float(2**p*(2**54+x))), 2**p*(2**54+y))
+
+ # behaviour near extremes of floating-point range
+ int_dbl_max = int(DBL_MAX)
+ top_power = 2**DBL_MAX_EXP
+ halfway = (int_dbl_max + top_power)//2
+ self.assertEqual(float(int_dbl_max), DBL_MAX)
+ self.assertEqual(float(int_dbl_max+1), DBL_MAX)
+ self.assertEqual(float(halfway-1), DBL_MAX)
+ self.assertRaises(OverflowError, float, halfway)
+ self.assertEqual(float(1-halfway), -DBL_MAX)
+ self.assertRaises(OverflowError, float, -halfway)
+ self.assertRaises(OverflowError, float, top_power-1)
+ self.assertRaises(OverflowError, float, top_power)
+ self.assertRaises(OverflowError, float, top_power+1)
+ self.assertRaises(OverflowError, float, 2*top_power-1)
+ self.assertRaises(OverflowError, float, 2*top_power)
+ self.assertRaises(OverflowError, float, top_power*top_power)
+
+ for p in range(100):
+ x = 2**p * (2**53 + 1) + 1
+ y = 2**p * (2**53 + 2)
+ self.assertEqual(int(float(x)), y)
+
+ x = 2**p * (2**53 + 1)
+ y = 2**p * 2**53
+ self.assertEqual(int(float(x)), y)
+
def test_float_overflow(self):
import math
diff --git a/Misc/NEWS b/Misc/NEWS
index 523166c..ca4a864 100644
--- a/Misc/NEWS
+++ b/Misc/NEWS
@@ -12,6 +12,9 @@
Core and Builtins
-----------------
+- Issue #3166: Make long -> float (and int -> float) conversions
+ correctly rounded.
+
- Issue #1869 (and many duplicates): make round(x, n) correctly
rounded for a float x, by using the decimal <-> binary conversions
from Python/dtoa.c. As a consequence, (e.g.) round(x, 2) now
diff --git a/Objects/longobject.c b/Objects/longobject.c
index 598c873..81af4ea 100644
--- a/Objects/longobject.c
+++ b/Objects/longobject.c
@@ -6,6 +6,7 @@
#include "longintrepr.h"
#include "structseq.h"
+#include <float.h>
#include <ctype.h>
#include <stddef.h>
@@ -100,6 +101,9 @@
if (PyErr_CheckSignals()) PyTryBlock \
}
+/* forward declaration */
+static int bits_in_digit(digit d);
+
/* Normalize (remove leading zeros from) a long int object.
Doesn't attempt to free the storage--in most cases, due to the nature
of the algorithms used, this could save at most be one word anyway. */
@@ -962,33 +966,166 @@
#undef NBITS_WANTED
}
-/* Get a C double from a long int object. */
+/* Get a C double from a long int object. Rounds to the nearest double,
+ using the round-half-to-even rule in the case of a tie. */
double
PyLong_AsDouble(PyObject *vv)
{
- int e = -1;
+ PyLongObject *v = (PyLongObject *)vv;
+ Py_ssize_t rnd_digit, rnd_bit, m, n;
+ digit lsb, *d;
+ int round_up = 0;
double x;
if (vv == NULL || !PyLong_Check(vv)) {
PyErr_BadInternalCall();
- return -1;
- }
- x = _PyLong_AsScaledDouble(vv, &e);
- if (x == -1.0 && PyErr_Occurred())
return -1.0;
- /* 'e' initialized to -1 to silence gcc-4.0.x, but it should be
- set correctly after a successful _PyLong_AsScaledDouble() call */
- assert(e >= 0);
- if (e > INT_MAX / PyLong_SHIFT)
- goto overflow;
- errno = 0;
- x = ldexp(x, e * PyLong_SHIFT);
- if (Py_OVERFLOWED(x))
- goto overflow;
- return x;
+ }
-overflow:
+ /* Notes on the method: for simplicity, assume v is positive and >=
+ 2**DBL_MANT_DIG. (For negative v we just ignore the sign until the
+ end; for small v no rounding is necessary.) Write n for the number
+ of bits in v, so that 2**(n-1) <= v < 2**n, and n > DBL_MANT_DIG.
+
+ Some terminology: the *rounding bit* of v is the 1st bit of v that
+ will be rounded away (bit n - DBL_MANT_DIG - 1); the *parity bit*
+ is the bit immediately above. The round-half-to-even rule says
+ that we round up if the rounding bit is set, unless v is exactly
+ halfway between two floats and the parity bit is zero.
+
+ Write d[0] ... d[m] for the digits of v, least to most significant.
+ Let rnd_bit be the index of the rounding bit, and rnd_digit the
+ index of the PyLong digit containing the rounding bit. Then the
+ bits of the digit d[rnd_digit] look something like:
+
+ rounding bit
+ |
+ v
+ msb -> sssssrttttttttt <- lsb
+ ^
+ |
+ parity bit
+
+ where 's' represents a 'significant bit' that will be included in
+ the mantissa of the result, 'r' is the rounding bit, and 't'
+ represents a 'trailing bit' following the rounding bit. Note that
+ if the rounding bit is at the top of d[rnd_digit] then the parity
+ bit will be the lsb of d[rnd_digit+1]. If we set
+
+ lsb = 1 << (rnd_bit % PyLong_SHIFT)
+
+ then d[rnd_digit] & (PyLong_BASE - 2*lsb) selects just the
+ significant bits of d[rnd_digit], d[rnd_digit] & (lsb-1) gets the
+ trailing bits, and d[rnd_digit] & lsb gives the rounding bit.
+
+ We initialize the double x to the integer given by digits
+ d[rnd_digit:m-1], but with the rounding bit and trailing bits of
+ d[rnd_digit] masked out. So the value of x comes from the top
+ DBL_MANT_DIG bits of v, multiplied by 2*lsb. Note that in the loop
+ that produces x, all floating-point operations are exact (assuming
+ that FLT_RADIX==2). Now if we're rounding down, the value we want
+ to return is simply
+
+ x * 2**(PyLong_SHIFT * rnd_digit).
+
+ and if we're rounding up, it's
+
+ (x + 2*lsb) * 2**(PyLong_SHIFT * rnd_digit).
+
+ Under the round-half-to-even rule, we round up if, and only
+ if, the rounding bit is set *and* at least one of the
+ following three conditions is satisfied:
+
+ (1) the parity bit is set, or
+ (2) at least one of the trailing bits of d[rnd_digit] is set, or
+ (3) at least one of the digits d[i], 0 <= i < rnd_digit
+ is nonzero.
+
+ Finally, we have to worry about overflow. If v >= 2**DBL_MAX_EXP,
+ or equivalently n > DBL_MAX_EXP, then overflow occurs. If v <
+ 2**DBL_MAX_EXP then we're usually safe, but there's a corner case
+ to consider: if v is very close to 2**DBL_MAX_EXP then it's
+ possible that v is rounded up to exactly 2**DBL_MAX_EXP, and then
+ again overflow occurs.
+ */
+
+ if (Py_SIZE(v) == 0)
+ return 0.0;
+ m = ABS(Py_SIZE(v)) - 1;
+ d = v->ob_digit;
+ assert(d[m]); /* v should be normalized */
+
+ /* fast path for case where 0 < abs(v) < 2**DBL_MANT_DIG */
+ if (m < DBL_MANT_DIG / PyLong_SHIFT ||
+ (m == DBL_MANT_DIG / PyLong_SHIFT &&
+ d[m] < (digit)1 << DBL_MANT_DIG%PyLong_SHIFT)) {
+ x = d[m];
+ while (--m >= 0)
+ x = x*PyLong_BASE + d[m];
+ return Py_SIZE(v) < 0 ? -x : x;
+ }
+
+ /* if m is huge then overflow immediately; otherwise, compute the
+ number of bits n in v. The condition below implies n (= #bits) >=
+ m * PyLong_SHIFT + 1 > DBL_MAX_EXP, hence v >= 2**DBL_MAX_EXP. */
+ if (m > (DBL_MAX_EXP-1)/PyLong_SHIFT)
+ goto overflow;
+ n = m * PyLong_SHIFT + bits_in_digit(d[m]);
+ if (n > DBL_MAX_EXP)
+ goto overflow;
+
+ /* find location of rounding bit */
+ assert(n > DBL_MANT_DIG); /* dealt with |v| < 2**DBL_MANT_DIG above */
+ rnd_bit = n - DBL_MANT_DIG - 1;
+ rnd_digit = rnd_bit/PyLong_SHIFT;
+ lsb = (digit)1 << (rnd_bit%PyLong_SHIFT);
+
+ /* Get top DBL_MANT_DIG bits of v. Assumes PyLong_SHIFT <
+ DBL_MANT_DIG, so we'll need bits from at least 2 digits of v. */
+ x = d[m];
+ assert(m > rnd_digit);
+ while (--m > rnd_digit)
+ x = x*PyLong_BASE + d[m];
+ x = x*PyLong_BASE + (d[m] & (PyLong_BASE-2*lsb));
+
+ /* decide whether to round up, using round-half-to-even */
+ assert(m == rnd_digit);
+ if (d[m] & lsb) { /* if (rounding bit is set) */
+ digit parity_bit;
+ if (lsb == PyLong_BASE/2)
+ parity_bit = d[m+1] & 1;
+ else
+ parity_bit = d[m] & 2*lsb;
+ if (parity_bit)
+ round_up = 1;
+ else if (d[m] & (lsb-1))
+ round_up = 1;
+ else {
+ while (--m >= 0) {
+ if (d[m]) {
+ round_up = 1;
+ break;
+ }
+ }
+ }
+ }
+
+ /* and round up if necessary */
+ if (round_up) {
+ x += 2*lsb;
+ if (n == DBL_MAX_EXP &&
+ x == ldexp((double)(2*lsb), DBL_MANT_DIG)) {
+ /* overflow corner case */
+ goto overflow;
+ }
+ }
+
+ /* shift, adjust for sign, and return */
+ x = ldexp(x, rnd_digit*PyLong_SHIFT);
+ return Py_SIZE(v) < 0 ? -x : x;
+
+ overflow:
PyErr_SetString(PyExc_OverflowError,
"Python int too large to convert to C double");
return -1.0;