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float_richcompare(): Use the new Py_IS_NAN macro to ensure that, on

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platforms where that macro works, NaN compared to an int or long works
the same as NaN compared to a finite float.
This commit is contained in:
Tim Peters 2004-09-23 19:22:41 +00:00
parent 862f0593d8
commit e1c69b3f6f
1 changed files with 9 additions and 11 deletions
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20
Objects/floatobject.c
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@ -384,13 +384,11 @@ float_richcompare(PyObject *v, PyObject *w, int op)
if (PyFloat_Check(w))
j = PyFloat_AS_DOUBLE(w);
else if (Py_IS_INFINITY(i)) {
/* XXX If we had a reliable way to check whether i is a
* XXX NaN, it would belong in this branch too.
*/
else if (Py_IS_INFINITY(i) || Py_IS_NAN(i)) {
if (PyInt_Check(w) || PyLong_Check(w))
/* The magnitude of i exceeds any finite integer,
* so it doesn't matter which int we compare i with.
/* If i is an infinity, its magnitude exceeds any
* finite integer, so it doesn't matter which int we
* compare i with. If i is a NaN, similarly.
*/
j = 0.0;
else
@ -403,7 +401,7 @@ float_richcompare(PyObject *v, PyObject *w, int op)
* Cray single with 48 bits of precision, and long has 64
* bits.
*/
#if SIZEOF_LONG > 4
#if SIZEOF_LONG > 6
unsigned long abs = (unsigned long)(jj < 0 ? -jj : jj);
if (abs >> 48) {
/* Needs more than 48 bits. Make it take the
@ -443,10 +441,10 @@ float_richcompare(PyObject *v, PyObject *w, int op)
nbits = _PyLong_NumBits(w);
if (nbits == (size_t)-1 && PyErr_Occurred()) {
/* This long is so large that size_t isn't big enough
* to hold the # of Python digits. Replace with
* little doubles that give the same outcome --
* w is so large that its magnitude must exceed
* the magnitude of any finite float.
* to hold the # of bits. Replace with little doubles
* that give the same outcome -- w is so large that
* its magnitude must exceed the magnitude of any
* finite float.
*/
PyErr_Clear();
i = (double)vsign;