[issue18739] math.log of a long returns a different value of math.log of an int

[Tim Peters]

Tim Peters added the comment:

Yup - 2.7 evaluates this in a less precise way, as

log(10L) = log(10./16 * 2**4) = log(0.625) + log(2)*4

>>> log(10L) == log(0.625) + log(2)*4
True

This patterns works well even for longs that are far too large to represent as a double; e.g.,

>>> log(1L << 50000)
34657.35902799726

which is evaluated internally as log(0.5) + log(2) * 50001:

>>> log(1L << 50000) == log(0.5) + log(2) * 50001
True

Python 3 is more careful, falling back to this pattern _only_ if converting the long to a double overflows.  Of course 10L can be represented exactly as a double, so Python 3 evaluates it directly as log(float(10L)) = log(10.0).  It's minor difference overall, but definitely visible ;-)

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nosy: +tim_one

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