mirror of https://github.com/python/cpython
bpo-37986: Improve perfomance of PyLong_FromDouble() (GH-15611)
* bpo-37986: Improve perfomance of PyLong_FromDouble() * Use strict bound check for safety and symmetry * Remove possibly outdated performance claims Co-authored-by: Mark Dickinson <dickinsm@gmail.com>
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Improve performance of :c:func:`PyLong_FromDouble` for values that fit into
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:c:type:`long`.
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@ -862,27 +862,7 @@ static PyObject *
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float___trunc___impl(PyObject *self)
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/*[clinic end generated code: output=dd3e289dd4c6b538 input=591b9ba0d650fdff]*/
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{
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double x = PyFloat_AsDouble(self);
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double wholepart; /* integral portion of x, rounded toward 0 */
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(void)modf(x, &wholepart);
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/* Try to get out cheap if this fits in a Python int. The attempt
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* to cast to long must be protected, as C doesn't define what
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* happens if the double is too big to fit in a long. Some rare
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* systems raise an exception then (RISCOS was mentioned as one,
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* and someone using a non-default option on Sun also bumped into
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* that). Note that checking for >= and <= LONG_{MIN,MAX} would
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* still be vulnerable: if a long has more bits of precision than
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* a double, casting MIN/MAX to double may yield an approximation,
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* and if that's rounded up, then, e.g., wholepart=LONG_MAX+1 would
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* yield true from the C expression wholepart<=LONG_MAX, despite
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* that wholepart is actually greater than LONG_MAX.
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*/
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if (LONG_MIN < wholepart && wholepart < LONG_MAX) {
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const long aslong = (long)wholepart;
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return PyLong_FromLong(aslong);
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}
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return PyLong_FromDouble(wholepart);
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return PyLong_FromDouble(PyFloat_AS_DOUBLE(self));
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}
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/*[clinic input]
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@ -416,6 +416,21 @@ PyLong_FromSize_t(size_t ival)
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PyObject *
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PyLong_FromDouble(double dval)
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{
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/* Try to get out cheap if this fits in a long. When a finite value of real
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* floating type is converted to an integer type, the value is truncated
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* toward zero. If the value of the integral part cannot be represented by
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* the integer type, the behavior is undefined. Thus, we must check that
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* value is in range (LONG_MIN - 1, LONG_MAX + 1). If a long has more bits
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* of precision than a double, casting LONG_MIN - 1 to double may yield an
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* approximation, but LONG_MAX + 1 is a power of two and can be represented
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* as double exactly (assuming FLT_RADIX is 2 or 16), so for simplicity
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* check against [-(LONG_MAX + 1), LONG_MAX + 1).
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*/
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const double int_max = (unsigned long)LONG_MAX + 1;
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if (-int_max < dval && dval < int_max) {
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return PyLong_FromLong((long)dval);
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}
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PyLongObject *v;
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double frac;
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int i, ndig, expo, neg;
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@ -435,8 +450,7 @@ PyLong_FromDouble(double dval)
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dval = -dval;
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}
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frac = frexp(dval, &expo); /* dval = frac*2**expo; 0.0 <= frac < 1.0 */
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if (expo <= 0)
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return PyLong_FromLong(0L);
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assert(expo > 0);
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ndig = (expo-1) / PyLong_SHIFT + 1; /* Number of 'digits' in result */
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v = _PyLong_New(ndig);
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if (v == NULL)
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