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svn+ssh://pythondev@svn.python.org/python/trunk ........ r77062 | mark.dickinson | 2009-12-27 14:55:57 +0000 (Sun, 27 Dec 2009) | 2 lines Issue #1811: Improve accuracy and consistency of true division for integers. ........
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Mark Dickinson
parent
99b2c8f811
commit
cbb62745ac
3 changed files with 418 additions and 33 deletions
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@ -3213,47 +3213,267 @@ long_div(PyObject *a, PyObject *b)
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return (PyObject *)div;
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}
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/* PyLong/PyLong -> float, with correctly rounded result. */
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#define MANT_DIG_DIGITS (DBL_MANT_DIG / PyLong_SHIFT)
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#define MANT_DIG_BITS (DBL_MANT_DIG % PyLong_SHIFT)
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static PyObject *
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long_true_divide(PyObject *a, PyObject *b)
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long_true_divide(PyObject *v, PyObject *w)
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{
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double ad, bd;
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int failed, aexp = -1, bexp = -1;
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PyLongObject *a, *b, *x;
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Py_ssize_t a_size, b_size, shift, extra_bits, diff, x_size, x_bits;
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digit mask, low;
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int inexact, negate, a_is_small, b_is_small;
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double dx, result;
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CHECK_BINOP(a, b);
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ad = _PyLong_AsScaledDouble((PyObject *)a, &aexp);
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bd = _PyLong_AsScaledDouble((PyObject *)b, &bexp);
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failed = (ad == -1.0 || bd == -1.0) && PyErr_Occurred();
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if (failed)
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return NULL;
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/* 'aexp' and 'bexp' were initialized to -1 to silence gcc-4.0.x,
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but should really be set correctly after sucessful calls to
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_PyLong_AsScaledDouble() */
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assert(aexp >= 0 && bexp >= 0);
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CHECK_BINOP(v, w);
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a = (PyLongObject *)v;
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b = (PyLongObject *)w;
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if (bd == 0.0) {
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/*
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Method in a nutshell:
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0. reduce to case a, b > 0; filter out obvious underflow/overflow
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1. choose a suitable integer 'shift'
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2. use integer arithmetic to compute x = floor(2**-shift*a/b)
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3. adjust x for correct rounding
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4. convert x to a double dx with the same value
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5. return ldexp(dx, shift).
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In more detail:
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0. For any a, a/0 raises ZeroDivisionError; for nonzero b, 0/b
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returns either 0.0 or -0.0, depending on the sign of b. For a and
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b both nonzero, ignore signs of a and b, and add the sign back in
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at the end. Now write a_bits and b_bits for the bit lengths of a
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and b respectively (that is, a_bits = 1 + floor(log_2(a)); likewise
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for b). Then
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2**(a_bits - b_bits - 1) < a/b < 2**(a_bits - b_bits + 1).
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So if a_bits - b_bits > DBL_MAX_EXP then a/b > 2**DBL_MAX_EXP and
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so overflows. Similarly, if a_bits - b_bits < DBL_MIN_EXP -
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DBL_MANT_DIG - 1 then a/b underflows to 0. With these cases out of
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the way, we can assume that
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DBL_MIN_EXP - DBL_MANT_DIG - 1 <= a_bits - b_bits <= DBL_MAX_EXP.
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1. The integer 'shift' is chosen so that x has the right number of
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bits for a double, plus two or three extra bits that will be used
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in the rounding decisions. Writing a_bits and b_bits for the
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number of significant bits in a and b respectively, a
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straightforward formula for shift is:
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shift = a_bits - b_bits - DBL_MANT_DIG - 2
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This is fine in the usual case, but if a/b is smaller than the
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smallest normal float then it can lead to double rounding on an
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IEEE 754 platform, giving incorrectly rounded results. So we
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adjust the formula slightly. The actual formula used is:
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shift = MAX(a_bits - b_bits, DBL_MIN_EXP) - DBL_MANT_DIG - 2
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2. The quantity x is computed by first shifting a (left -shift bits
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if shift <= 0, right shift bits if shift > 0) and then dividing by
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b. For both the shift and the division, we keep track of whether
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the result is inexact, in a flag 'inexact'; this information is
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needed at the rounding stage.
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With the choice of shift above, together with our assumption that
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a_bits - b_bits >= DBL_MIN_EXP - DBL_MANT_DIG - 1, it follows
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that x >= 1.
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3. Now x * 2**shift <= a/b < (x+1) * 2**shift. We want to replace
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this with an exactly representable float of the form
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round(x/2**extra_bits) * 2**(extra_bits+shift).
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For float representability, we need x/2**extra_bits <
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2**DBL_MANT_DIG and extra_bits + shift >= DBL_MIN_EXP -
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DBL_MANT_DIG. This translates to the condition:
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extra_bits >= MAX(x_bits, DBL_MIN_EXP - shift) - DBL_MANT_DIG
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To round, we just modify the bottom digit of x in-place; this can
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end up giving a digit with value > PyLONG_MASK, but that's not a
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problem since digits can hold values up to 2*PyLONG_MASK+1.
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With the original choices for shift above, extra_bits will always
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be 2 or 3. Then rounding under the round-half-to-even rule, we
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round up iff the most significant of the extra bits is 1, and
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either: (a) the computation of x in step 2 had an inexact result,
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or (b) at least one other of the extra bits is 1, or (c) the least
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significant bit of x (above those to be rounded) is 1.
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4. Conversion to a double is straightforward; all floating-point
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operations involved in the conversion are exact, so there's no
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danger of rounding errors.
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5. Use ldexp(x, shift) to compute x*2**shift, the final result.
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The result will always be exactly representable as a double, except
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in the case that it overflows. To avoid dependence on the exact
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behaviour of ldexp on overflow, we check for overflow before
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applying ldexp. The result of ldexp is adjusted for sign before
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returning.
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*/
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/* Reduce to case where a and b are both positive. */
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a_size = ABS(Py_SIZE(a));
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b_size = ABS(Py_SIZE(b));
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negate = (Py_SIZE(a) < 0) ^ (Py_SIZE(b) < 0);
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if (b_size == 0) {
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PyErr_SetString(PyExc_ZeroDivisionError,
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"integer division or modulo by zero");
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return NULL;
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"division by zero");
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goto error;
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}
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if (a_size == 0)
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goto underflow_or_zero;
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/* Fast path for a and b small (exactly representable in a double).
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Relies on floating-point division being correctly rounded; results
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may be subject to double rounding on x86 machines that operate with
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the x87 FPU set to 64-bit precision. */
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a_is_small = a_size <= MANT_DIG_DIGITS ||
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(a_size == MANT_DIG_DIGITS+1 &&
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a->ob_digit[MANT_DIG_DIGITS] >> MANT_DIG_BITS == 0);
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b_is_small = b_size <= MANT_DIG_DIGITS ||
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(b_size == MANT_DIG_DIGITS+1 &&
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b->ob_digit[MANT_DIG_DIGITS] >> MANT_DIG_BITS == 0);
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if (a_is_small && b_is_small) {
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double da, db;
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da = a->ob_digit[--a_size];
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while (a_size > 0)
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da = da * PyLong_BASE + a->ob_digit[--a_size];
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db = b->ob_digit[--b_size];
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while (b_size > 0)
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db = db * PyLong_BASE + b->ob_digit[--b_size];
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result = da / db;
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goto success;
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}
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/* True value is very close to ad/bd * 2**(PyLong_SHIFT*(aexp-bexp)) */
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ad /= bd; /* overflow/underflow impossible here */
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aexp -= bexp;
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if (aexp > INT_MAX / PyLong_SHIFT)
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/* Catch obvious cases of underflow and overflow */
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diff = a_size - b_size;
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if (diff > PY_SSIZE_T_MAX/PyLong_SHIFT - 1)
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/* Extreme overflow */
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goto overflow;
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else if (aexp < -(INT_MAX / PyLong_SHIFT))
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return PyFloat_FromDouble(0.0); /* underflow to 0 */
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errno = 0;
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ad = ldexp(ad, aexp * PyLong_SHIFT);
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if (Py_OVERFLOWED(ad)) /* ignore underflow to 0.0 */
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else if (diff < 1 - PY_SSIZE_T_MAX/PyLong_SHIFT)
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/* Extreme underflow */
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goto underflow_or_zero;
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/* Next line is now safe from overflowing a Py_ssize_t */
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diff = diff * PyLong_SHIFT + bits_in_digit(a->ob_digit[a_size - 1]) -
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bits_in_digit(b->ob_digit[b_size - 1]);
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/* Now diff = a_bits - b_bits. */
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if (diff > DBL_MAX_EXP)
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goto overflow;
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return PyFloat_FromDouble(ad);
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else if (diff < DBL_MIN_EXP - DBL_MANT_DIG - 1)
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goto underflow_or_zero;
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overflow:
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/* Choose value for shift; see comments for step 1 above. */
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shift = MAX(diff, DBL_MIN_EXP) - DBL_MANT_DIG - 2;
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inexact = 0;
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/* x = abs(a * 2**-shift) */
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if (shift <= 0) {
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Py_ssize_t i, shift_digits = -shift / PyLong_SHIFT;
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digit rem;
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/* x = a << -shift */
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if (a_size >= PY_SSIZE_T_MAX - 1 - shift_digits) {
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/* In practice, it's probably impossible to end up
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here. Both a and b would have to be enormous,
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using close to SIZE_T_MAX bytes of memory each. */
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PyErr_SetString(PyExc_OverflowError,
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"intermediate overflow during division");
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goto error;
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}
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x = _PyLong_New(a_size + shift_digits + 1);
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if (x == NULL)
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goto error;
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for (i = 0; i < shift_digits; i++)
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x->ob_digit[i] = 0;
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rem = v_lshift(x->ob_digit + shift_digits, a->ob_digit,
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a_size, -shift % PyLong_SHIFT);
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x->ob_digit[a_size + shift_digits] = rem;
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}
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else {
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Py_ssize_t shift_digits = shift / PyLong_SHIFT;
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digit rem;
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/* x = a >> shift */
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assert(a_size >= shift_digits);
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x = _PyLong_New(a_size - shift_digits);
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if (x == NULL)
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goto error;
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rem = v_rshift(x->ob_digit, a->ob_digit + shift_digits,
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a_size - shift_digits, shift % PyLong_SHIFT);
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/* set inexact if any of the bits shifted out is nonzero */
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if (rem)
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inexact = 1;
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while (!inexact && shift_digits > 0)
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if (a->ob_digit[--shift_digits])
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inexact = 1;
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}
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long_normalize(x);
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x_size = Py_SIZE(x);
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/* x //= b. If the remainder is nonzero, set inexact. We own the only
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reference to x, so it's safe to modify it in-place. */
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if (b_size == 1) {
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digit rem = inplace_divrem1(x->ob_digit, x->ob_digit, x_size,
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b->ob_digit[0]);
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long_normalize(x);
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if (rem)
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inexact = 1;
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}
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else {
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PyLongObject *div, *rem;
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div = x_divrem(x, b, &rem);
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Py_DECREF(x);
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x = div;
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if (x == NULL)
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goto error;
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if (Py_SIZE(rem))
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inexact = 1;
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Py_DECREF(rem);
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}
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x_size = ABS(Py_SIZE(x));
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assert(x_size > 0); /* result of division is never zero */
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x_bits = (x_size-1)*PyLong_SHIFT+bits_in_digit(x->ob_digit[x_size-1]);
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/* The number of extra bits that have to be rounded away. */
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extra_bits = MAX(x_bits, DBL_MIN_EXP - shift) - DBL_MANT_DIG;
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assert(extra_bits == 2 || extra_bits == 3);
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/* Round by directly modifying the low digit of x. */
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mask = (digit)1 << (extra_bits - 1);
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low = x->ob_digit[0] | inexact;
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if (low & mask && low & (3*mask-1))
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low += mask;
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x->ob_digit[0] = low & ~(mask-1U);
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/* Convert x to a double dx; the conversion is exact. */
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dx = x->ob_digit[--x_size];
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while (x_size > 0)
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dx = dx * PyLong_BASE + x->ob_digit[--x_size];
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Py_DECREF(x);
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/* Check whether ldexp result will overflow a double. */
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if (shift + x_bits >= DBL_MAX_EXP &&
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(shift + x_bits > DBL_MAX_EXP || dx == ldexp(1.0, x_bits)))
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goto overflow;
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result = ldexp(dx, shift);
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success:
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return PyFloat_FromDouble(negate ? -result : result);
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underflow_or_zero:
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return PyFloat_FromDouble(negate ? -0.0 : 0.0);
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overflow:
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PyErr_SetString(PyExc_OverflowError,
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"int/int too large for a float");
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"integer division result too large for a float");
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error:
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return NULL;
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}
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static PyObject *
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