[Python-checkins] r77062 - in python/trunk: Lib/test/test_long_future.py Misc/NEWS Objects/intobject.c Objects/longobject.c
mark.dickinson
python-checkins at python.org
Sun Dec 27 15:55:57 CET 2009
Author: mark.dickinson
Date: Sun Dec 27 15:55:57 2009
New Revision: 77062
Log:
Issue #1811: Improve accuracy and consistency of true division for integers.
Modified:
python/trunk/Lib/test/test_long_future.py
python/trunk/Misc/NEWS
python/trunk/Objects/intobject.c
python/trunk/Objects/longobject.c
Modified: python/trunk/Lib/test/test_long_future.py
==============================================================================
--- python/trunk/Lib/test/test_long_future.py (original)
+++ python/trunk/Lib/test/test_long_future.py Sun Dec 27 15:55:57 2009
@@ -3,9 +3,52 @@
# test_long.py instead. In the meantime, it's too obscure to try to
# trick just part of test_long into using future division.
+import sys
+import random
import unittest
from test.test_support import run_unittest
+# decorator for skipping tests on non-IEEE 754 platforms
+requires_IEEE_754 = unittest.skipUnless(
+ float.__getformat__("double").startswith("IEEE"),
+ "test requires IEEE 754 doubles")
+
+DBL_MAX = sys.float_info.max
+DBL_MAX_EXP = sys.float_info.max_exp
+DBL_MIN_EXP = sys.float_info.min_exp
+DBL_MANT_DIG = sys.float_info.mant_dig
+DBL_MIN_OVERFLOW = 2**DBL_MAX_EXP - 2**(DBL_MAX_EXP - DBL_MANT_DIG - 1)
+
+# pure Python version of correctly-rounded true division
+def truediv(a, b):
+ """Correctly-rounded true division for integers."""
+ negative = a^b < 0
+ a, b = abs(a), abs(b)
+
+ # exceptions: division by zero, overflow
+ if not b:
+ raise ZeroDivisionError("division by zero")
+ if a >= DBL_MIN_OVERFLOW * b:
+ raise OverflowError("int/int too large to represent as a float")
+
+ # find integer d satisfying 2**(d - 1) <= a/b < 2**d
+ d = a.bit_length() - b.bit_length()
+ if d >= 0 and a >= 2**d * b or d < 0 and a * 2**-d >= b:
+ d += 1
+
+ # compute 2**-exp * a / b for suitable exp
+ exp = max(d, DBL_MIN_EXP) - DBL_MANT_DIG
+ a, b = a << max(-exp, 0), b << max(exp, 0)
+ q, r = divmod(a, b)
+
+ # round-half-to-even: fractional part is r/b, which is > 0.5 iff
+ # 2*r > b, and == 0.5 iff 2*r == b.
+ if 2*r > b or 2*r == b and q % 2 == 1:
+ q += 1
+
+ result = float(q) * 2.**exp
+ return -result if negative else result
+
class TrueDivisionTests(unittest.TestCase):
def test(self):
huge = 1L << 40000
@@ -45,6 +88,131 @@
with self.assertRaises(ZeroDivisionError):
eval(zero, namespace)
+ def check_truediv(self, a, b, skip_small=True):
+ """Verify that the result of a/b is correctly rounded, by
+ comparing it with a pure Python implementation of correctly
+ rounded division. b should be nonzero."""
+
+ a, b = long(a), long(b)
+
+ # skip check for small a and b: in this case, the current
+ # implementation converts the arguments to float directly and
+ # then applies a float division. This can give doubly-rounded
+ # results on x87-using machines (particularly 32-bit Linux).
+ if skip_small and max(abs(a), abs(b)) < 2**DBL_MANT_DIG:
+ return
+
+ try:
+ # use repr so that we can distinguish between -0.0 and 0.0
+ expected = repr(truediv(a, b))
+ except OverflowError:
+ expected = 'overflow'
+ except ZeroDivisionError:
+ expected = 'zerodivision'
+
+ try:
+ got = repr(a / b)
+ except OverflowError:
+ got = 'overflow'
+ except ZeroDivisionError:
+ got = 'zerodivision'
+
+ if expected != got:
+ self.fail("Incorrectly rounded division {}/{}: expected {!r}, "
+ "got {!r}.".format(a, b, expected, got))
+
+ @requires_IEEE_754
+ def test_correctly_rounded_true_division(self):
+ # more stringent tests than those above, checking that the
+ # result of true division of ints is always correctly rounded.
+ # This test should probably be considered CPython-specific.
+
+ # Exercise all the code paths not involving Gb-sized ints.
+ # ... divisions involving zero
+ self.check_truediv(123, 0)
+ self.check_truediv(-456, 0)
+ self.check_truediv(0, 3)
+ self.check_truediv(0, -3)
+ self.check_truediv(0, 0)
+ # ... overflow or underflow by large margin
+ self.check_truediv(671 * 12345 * 2**DBL_MAX_EXP, 12345)
+ self.check_truediv(12345, 345678 * 2**(DBL_MANT_DIG - DBL_MIN_EXP))
+ # ... a much larger or smaller than b
+ self.check_truediv(12345*2**100, 98765)
+ self.check_truediv(12345*2**30, 98765*7**81)
+ # ... a / b near a boundary: one of 1, 2**DBL_MANT_DIG, 2**DBL_MIN_EXP,
+ # 2**DBL_MAX_EXP, 2**(DBL_MIN_EXP-DBL_MANT_DIG)
+ bases = (0, DBL_MANT_DIG, DBL_MIN_EXP,
+ DBL_MAX_EXP, DBL_MIN_EXP - DBL_MANT_DIG)
+ for base in bases:
+ for exp in range(base - 15, base + 15):
+ self.check_truediv(75312*2**max(exp, 0), 69187*2**max(-exp, 0))
+ self.check_truediv(69187*2**max(exp, 0), 75312*2**max(-exp, 0))
+
+ # overflow corner case
+ for m in [1, 2, 7, 17, 12345, 7**100,
+ -1, -2, -5, -23, -67891, -41**50]:
+ for n in range(-10, 10):
+ self.check_truediv(m*DBL_MIN_OVERFLOW + n, m)
+ self.check_truediv(m*DBL_MIN_OVERFLOW + n, -m)
+
+ # check detection of inexactness in shifting stage
+ for n in range(250):
+ # (2**DBL_MANT_DIG+1)/(2**DBL_MANT_DIG) lies halfway
+ # between two representable floats, and would usually be
+ # rounded down under round-half-to-even. The tiniest of
+ # additions to the numerator should cause it to be rounded
+ # up instead.
+ self.check_truediv((2**DBL_MANT_DIG + 1)*12345*2**200 + 2**n,
+ 2**DBL_MANT_DIG*12345)
+
+ # 1/2731 is one of the smallest division cases that's subject
+ # to double rounding on IEEE 754 machines working internally with
+ # 64-bit precision. On such machines, the next check would fail,
+ # were it not explicitly skipped in check_truediv.
+ self.check_truediv(1, 2731)
+
+ # a particularly bad case for the old algorithm: gives an
+ # error of close to 3.5 ulps.
+ self.check_truediv(295147931372582273023, 295147932265116303360)
+ for i in range(1000):
+ self.check_truediv(10**(i+1), 10**i)
+ self.check_truediv(10**i, 10**(i+1))
+
+ # test round-half-to-even behaviour, normal result
+ for m in [1, 2, 4, 7, 8, 16, 17, 32, 12345, 7**100,
+ -1, -2, -5, -23, -67891, -41**50]:
+ for n in range(-10, 10):
+ self.check_truediv(2**DBL_MANT_DIG*m + n, m)
+
+ # test round-half-to-even, subnormal result
+ for n in range(-20, 20):
+ self.check_truediv(n, 2**1076)
+
+ # largeish random divisions: a/b where |a| <= |b| <=
+ # 2*|a|; |ans| is between 0.5 and 1.0, so error should
+ # always be bounded by 2**-54 with equality possible only
+ # if the least significant bit of q=ans*2**53 is zero.
+ for M in [10**10, 10**100, 10**1000]:
+ for i in range(1000):
+ a = random.randrange(1, M)
+ b = random.randrange(a, 2*a+1)
+ self.check_truediv(a, b)
+ self.check_truediv(-a, b)
+ self.check_truediv(a, -b)
+ self.check_truediv(-a, -b)
+
+ # and some (genuinely) random tests
+ for _ in range(10000):
+ a_bits = random.randrange(1000)
+ b_bits = random.randrange(1, 1000)
+ x = random.randrange(2**a_bits)
+ y = random.randrange(1, 2**b_bits)
+ self.check_truediv(x, y)
+ self.check_truediv(x, -y)
+ self.check_truediv(-x, y)
+ self.check_truediv(-x, -y)
+
def test_main():
run_unittest(TrueDivisionTests)
Modified: python/trunk/Misc/NEWS
==============================================================================
--- python/trunk/Misc/NEWS (original)
+++ python/trunk/Misc/NEWS Sun Dec 27 15:55:57 2009
@@ -12,6 +12,11 @@
Core and Builtins
-----------------
+- Issue #1811: improve accuracy and cross-platform consistency for
+ true division of integers: the result of a/b is now correctly
+ rounded for ints a and b (at least on IEEE 754 platforms), and in
+ particular does not depend on the internal representation of a long.
+
- Issue #6108: unicode(exception) and str(exception) should return the same
message when only __str__ (and not __unicode__) is overridden in the subclass.
Modified: python/trunk/Objects/intobject.c
==============================================================================
--- python/trunk/Objects/intobject.c (original)
+++ python/trunk/Objects/intobject.c Sun Dec 27 15:55:57 2009
@@ -645,16 +645,35 @@
}
static PyObject *
-int_true_divide(PyObject *v, PyObject *w)
+int_true_divide(PyIntObject *x, PyIntObject *y)
{
+ long xi, yi;
/* If they aren't both ints, give someone else a chance. In
particular, this lets int/long get handled by longs, which
underflows to 0 gracefully if the long is too big to convert
to float. */
- if (PyInt_Check(v) && PyInt_Check(w))
- return PyFloat_Type.tp_as_number->nb_true_divide(v, w);
- Py_INCREF(Py_NotImplemented);
- return Py_NotImplemented;
+ CONVERT_TO_LONG(x, xi);
+ CONVERT_TO_LONG(y, yi);
+ if (yi == 0) {
+ PyErr_SetString(PyExc_ZeroDivisionError,
+ "division by zero");
+ return NULL;
+ }
+ if (xi == 0)
+ return PyFloat_FromDouble(yi < 0 ? -0.0 : 0.0);
+
+#define WIDTH_OF_ULONG (CHAR_BIT*SIZEOF_LONG)
+#if DBL_MANT_DIG < WIDTH_OF_ULONG
+ if ((xi >= 0 ? 0UL + xi : 0UL - xi) >> DBL_MANT_DIG ||
+ (yi >= 0 ? 0UL + yi : 0UL - yi) >> DBL_MANT_DIG)
+ /* Large x or y. Use long integer arithmetic. */
+ return PyLong_Type.tp_as_number->nb_true_divide(
+ (PyObject *)x, (PyObject *)y);
+ else
+#endif
+ /* Both ints can be exactly represented as doubles. Do a
+ floating-point division. */
+ return PyFloat_FromDouble((double)xi / (double)yi);
}
static PyObject *
@@ -1355,7 +1374,7 @@
0, /*nb_inplace_xor*/
0, /*nb_inplace_or*/
(binaryfunc)int_div, /* nb_floor_divide */
- int_true_divide, /* nb_true_divide */
+ (binaryfunc)int_true_divide, /* nb_true_divide */
0, /* nb_inplace_floor_divide */
0, /* nb_inplace_true_divide */
(unaryfunc)int_int, /* nb_index */
Modified: python/trunk/Objects/longobject.c
==============================================================================
--- python/trunk/Objects/longobject.c (original)
+++ python/trunk/Objects/longobject.c Sun Dec 27 15:55:57 2009
@@ -3037,50 +3037,271 @@
return (PyObject *)div;
}
+/* PyLong/PyLong -> float, with correctly rounded result. */
+
+#define MANT_DIG_DIGITS (DBL_MANT_DIG / PyLong_SHIFT)
+#define MANT_DIG_BITS (DBL_MANT_DIG % PyLong_SHIFT)
+
static PyObject *
long_true_divide(PyObject *v, PyObject *w)
{
- PyLongObject *a, *b;
- double ad, bd;
- int failed, aexp = -1, bexp = -1;
+ PyLongObject *a, *b, *x;
+ Py_ssize_t a_size, b_size, shift, extra_bits, diff, x_size, x_bits;
+ digit mask, low;
+ int inexact, negate, a_is_small, b_is_small;
+ double dx, result;
CONVERT_BINOP(v, w, &a, &b);
- ad = _PyLong_AsScaledDouble((PyObject *)a, &aexp);
- bd = _PyLong_AsScaledDouble((PyObject *)b, &bexp);
- failed = (ad == -1.0 || bd == -1.0) && PyErr_Occurred();
- Py_DECREF(a);
- Py_DECREF(b);
- if (failed)
- return NULL;
- /* 'aexp' and 'bexp' were initialized to -1 to silence gcc-4.0.x,
- but should really be set correctly after sucessful calls to
- _PyLong_AsScaledDouble() */
- assert(aexp >= 0 && bexp >= 0);
- if (bd == 0.0) {
+ /*
+ Method in a nutshell:
+
+ 0. reduce to case a, b > 0; filter out obvious underflow/overflow
+ 1. choose a suitable integer 'shift'
+ 2. use integer arithmetic to compute x = floor(2**-shift*a/b)
+ 3. adjust x for correct rounding
+ 4. convert x to a double dx with the same value
+ 5. return ldexp(dx, shift).
+
+ In more detail:
+
+ 0. For any a, a/0 raises ZeroDivisionError; for nonzero b, 0/b
+ returns either 0.0 or -0.0, depending on the sign of b. For a and
+ b both nonzero, ignore signs of a and b, and add the sign back in
+ at the end. Now write a_bits and b_bits for the bit lengths of a
+ and b respectively (that is, a_bits = 1 + floor(log_2(a)); likewise
+ for b). Then
+
+ 2**(a_bits - b_bits - 1) < a/b < 2**(a_bits - b_bits + 1).
+
+ So if a_bits - b_bits > DBL_MAX_EXP then a/b > 2**DBL_MAX_EXP and
+ so overflows. Similarly, if a_bits - b_bits < DBL_MIN_EXP -
+ DBL_MANT_DIG - 1 then a/b underflows to 0. With these cases out of
+ the way, we can assume that
+
+ DBL_MIN_EXP - DBL_MANT_DIG - 1 <= a_bits - b_bits <= DBL_MAX_EXP.
+
+ 1. The integer 'shift' is chosen so that x has the right number of
+ bits for a double, plus two or three extra bits that will be used
+ in the rounding decisions. Writing a_bits and b_bits for the
+ number of significant bits in a and b respectively, a
+ straightforward formula for shift is:
+
+ shift = a_bits - b_bits - DBL_MANT_DIG - 2
+
+ This is fine in the usual case, but if a/b is smaller than the
+ smallest normal float then it can lead to double rounding on an
+ IEEE 754 platform, giving incorrectly rounded results. So we
+ adjust the formula slightly. The actual formula used is:
+
+ shift = MAX(a_bits - b_bits, DBL_MIN_EXP) - DBL_MANT_DIG - 2
+
+ 2. The quantity x is computed by first shifting a (left -shift bits
+ if shift <= 0, right shift bits if shift > 0) and then dividing by
+ b. For both the shift and the division, we keep track of whether
+ the result is inexact, in a flag 'inexact'; this information is
+ needed at the rounding stage.
+
+ With the choice of shift above, together with our assumption that
+ a_bits - b_bits >= DBL_MIN_EXP - DBL_MANT_DIG - 1, it follows
+ that x >= 1.
+
+ 3. Now x * 2**shift <= a/b < (x+1) * 2**shift. We want to replace
+ this with an exactly representable float of the form
+
+ round(x/2**extra_bits) * 2**(extra_bits+shift).
+
+ For float representability, we need x/2**extra_bits <
+ 2**DBL_MANT_DIG and extra_bits + shift >= DBL_MIN_EXP -
+ DBL_MANT_DIG. This translates to the condition:
+
+ extra_bits >= MAX(x_bits, DBL_MIN_EXP - shift) - DBL_MANT_DIG
+
+ To round, we just modify the bottom digit of x in-place; this can
+ end up giving a digit with value > PyLONG_MASK, but that's not a
+ problem since digits can hold values up to 2*PyLONG_MASK+1.
+
+ With the original choices for shift above, extra_bits will always
+ be 2 or 3. Then rounding under the round-half-to-even rule, we
+ round up iff the most significant of the extra bits is 1, and
+ either: (a) the computation of x in step 2 had an inexact result,
+ or (b) at least one other of the extra bits is 1, or (c) the least
+ significant bit of x (above those to be rounded) is 1.
+
+ 4. Conversion to a double is straightforward; all floating-point
+ operations involved in the conversion are exact, so there's no
+ danger of rounding errors.
+
+ 5. Use ldexp(x, shift) to compute x*2**shift, the final result.
+ The result will always be exactly representable as a double, except
+ in the case that it overflows. To avoid dependence on the exact
+ behaviour of ldexp on overflow, we check for overflow before
+ applying ldexp. The result of ldexp is adjusted for sign before
+ returning.
+ */
+
+ /* Reduce to case where a and b are both positive. */
+ a_size = ABS(Py_SIZE(a));
+ b_size = ABS(Py_SIZE(b));
+ negate = (Py_SIZE(a) < 0) ^ (Py_SIZE(b) < 0);
+ if (b_size == 0) {
PyErr_SetString(PyExc_ZeroDivisionError,
- "long division or modulo by zero");
- return NULL;
+ "division by zero");
+ goto error;
}
+ if (a_size == 0)
+ goto underflow_or_zero;
- /* True value is very close to ad/bd * 2**(PyLong_SHIFT*(aexp-bexp)) */
- ad /= bd; /* overflow/underflow impossible here */
- aexp -= bexp;
- if (aexp > INT_MAX / PyLong_SHIFT)
+ /* Fast path for a and b small (exactly representable in a double).
+ Relies on floating-point division being correctly rounded; results
+ may be subject to double rounding on x86 machines that operate with
+ the x87 FPU set to 64-bit precision. */
+ a_is_small = a_size <= MANT_DIG_DIGITS ||
+ (a_size == MANT_DIG_DIGITS+1 &&
+ a->ob_digit[MANT_DIG_DIGITS] >> MANT_DIG_BITS == 0);
+ b_is_small = b_size <= MANT_DIG_DIGITS ||
+ (b_size == MANT_DIG_DIGITS+1 &&
+ b->ob_digit[MANT_DIG_DIGITS] >> MANT_DIG_BITS == 0);
+ if (a_is_small && b_is_small) {
+ double da, db;
+ da = a->ob_digit[--a_size];
+ while (a_size > 0)
+ da = da * PyLong_BASE + a->ob_digit[--a_size];
+ db = b->ob_digit[--b_size];
+ while (b_size > 0)
+ db = db * PyLong_BASE + b->ob_digit[--b_size];
+ result = da / db;
+ goto success;
+ }
+
+ /* Catch obvious cases of underflow and overflow */
+ diff = a_size - b_size;
+ if (diff > PY_SSIZE_T_MAX/PyLong_SHIFT - 1)
+ /* Extreme overflow */
goto overflow;
- else if (aexp < -(INT_MAX / PyLong_SHIFT))
- return PyFloat_FromDouble(0.0); /* underflow to 0 */
- errno = 0;
- ad = ldexp(ad, aexp * PyLong_SHIFT);
- if (Py_OVERFLOWED(ad)) /* ignore underflow to 0.0 */
+ else if (diff < 1 - PY_SSIZE_T_MAX/PyLong_SHIFT)
+ /* Extreme underflow */
+ goto underflow_or_zero;
+ /* Next line is now safe from overflowing a Py_ssize_t */
+ diff = diff * PyLong_SHIFT + bits_in_digit(a->ob_digit[a_size - 1]) -
+ bits_in_digit(b->ob_digit[b_size - 1]);
+ /* Now diff = a_bits - b_bits. */
+ if (diff > DBL_MAX_EXP)
goto overflow;
- return PyFloat_FromDouble(ad);
+ else if (diff < DBL_MIN_EXP - DBL_MANT_DIG - 1)
+ goto underflow_or_zero;
+
+ /* Choose value for shift; see comments for step 1 above. */
+ shift = MAX(diff, DBL_MIN_EXP) - DBL_MANT_DIG - 2;
-overflow:
+ inexact = 0;
+
+ /* x = abs(a * 2**-shift) */
+ if (shift <= 0) {
+ Py_ssize_t i, shift_digits = -shift / PyLong_SHIFT;
+ digit rem;
+ /* x = a << -shift */
+ if (a_size >= PY_SSIZE_T_MAX - 1 - shift_digits) {
+ /* In practice, it's probably impossible to end up
+ here. Both a and b would have to be enormous,
+ using close to SIZE_T_MAX bytes of memory each. */
+ PyErr_SetString(PyExc_OverflowError,
+ "intermediate overflow during division");
+ goto error;
+ }
+ x = _PyLong_New(a_size + shift_digits + 1);
+ if (x == NULL)
+ goto error;
+ for (i = 0; i < shift_digits; i++)
+ x->ob_digit[i] = 0;
+ rem = v_lshift(x->ob_digit + shift_digits, a->ob_digit,
+ a_size, -shift % PyLong_SHIFT);
+ x->ob_digit[a_size + shift_digits] = rem;
+ }
+ else {
+ Py_ssize_t shift_digits = shift / PyLong_SHIFT;
+ digit rem;
+ /* x = a >> shift */
+ assert(a_size >= shift_digits);
+ x = _PyLong_New(a_size - shift_digits);
+ if (x == NULL)
+ goto error;
+ rem = v_rshift(x->ob_digit, a->ob_digit + shift_digits,
+ a_size - shift_digits, shift % PyLong_SHIFT);
+ /* set inexact if any of the bits shifted out is nonzero */
+ if (rem)
+ inexact = 1;
+ while (!inexact && shift_digits > 0)
+ if (a->ob_digit[--shift_digits])
+ inexact = 1;
+ }
+ long_normalize(x);
+ x_size = Py_SIZE(x);
+
+ /* x //= b. If the remainder is nonzero, set inexact. We own the only
+ reference to x, so it's safe to modify it in-place. */
+ if (b_size == 1) {
+ digit rem = inplace_divrem1(x->ob_digit, x->ob_digit, x_size,
+ b->ob_digit[0]);
+ long_normalize(x);
+ if (rem)
+ inexact = 1;
+ }
+ else {
+ PyLongObject *div, *rem;
+ div = x_divrem(x, b, &rem);
+ Py_DECREF(x);
+ x = div;
+ if (x == NULL)
+ goto error;
+ if (Py_SIZE(rem))
+ inexact = 1;
+ Py_DECREF(rem);
+ }
+ x_size = ABS(Py_SIZE(x));
+ assert(x_size > 0); /* result of division is never zero */
+ x_bits = (x_size-1)*PyLong_SHIFT+bits_in_digit(x->ob_digit[x_size-1]);
+
+ /* The number of extra bits that have to be rounded away. */
+ extra_bits = MAX(x_bits, DBL_MIN_EXP - shift) - DBL_MANT_DIG;
+ assert(extra_bits == 2 || extra_bits == 3);
+
+ /* Round by directly modifying the low digit of x. */
+ mask = (digit)1 << (extra_bits - 1);
+ low = x->ob_digit[0] | inexact;
+ if (low & mask && low & (3*mask-1))
+ low += mask;
+ x->ob_digit[0] = low & ~(mask-1U);
+
+ /* Convert x to a double dx; the conversion is exact. */
+ dx = x->ob_digit[--x_size];
+ while (x_size > 0)
+ dx = dx * PyLong_BASE + x->ob_digit[--x_size];
+ Py_DECREF(x);
+
+ /* Check whether ldexp result will overflow a double. */
+ if (shift + x_bits >= DBL_MAX_EXP &&
+ (shift + x_bits > DBL_MAX_EXP || dx == ldexp(1.0, x_bits)))
+ goto overflow;
+ result = ldexp(dx, shift);
+
+ success:
+ Py_DECREF(a);
+ Py_DECREF(b);
+ return PyFloat_FromDouble(negate ? -result : result);
+
+ underflow_or_zero:
+ Py_DECREF(a);
+ Py_DECREF(b);
+ return PyFloat_FromDouble(negate ? -0.0 : 0.0);
+
+ overflow:
PyErr_SetString(PyExc_OverflowError,
- "long/long too large for a float");
+ "integer division result too large for a float");
+ error:
+ Py_DECREF(a);
+ Py_DECREF(b);
return NULL;
-
}
static PyObject *
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