STINNER Victor <vstin...@python.org> added the comment:
The following result is a little bit surprising:
>>> nan=float("nan"); ([nan]*5).count(nan)
5
>>> nan == nan
False
But in fact, the optimization doesn't change the value. It was already 5
previously.
In fact, PyObject_RichCompareBool() has a fast path if the two object pointers
are equal:
/* Quick result when objects are the same.
Guarantees that identity implies equality. */
if (v == w) {
if (op == Py_EQ)
return 1;
else if (op == Py_NE)
return 0;
}
In short, the optimization is good: thank you ;-)
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Python tracker <rep...@bugs.python.org>
<https://bugs.python.org/issue39425>
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