New submission from Fredrik Johansson <[EMAIL PROTECTED]>: Python 3.0b2 (r30b2:65106, Jul 18 2008, 18:44:17) [MSC v.1500 32 bit (Intel)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> import math >>> math.frexp(10**100) (0.5714936956411375, 333) >>> math.frexp(10**1000) Traceback (most recent call last): File "<stdin>", line 1, in <module> OverflowError: Python int too large to convert to C double >>>
(Same behavior in Python 2.5.2, and presumably 2.6 although I haven't checked the latter.) I think it should be easy to make frexp work for large integers by calling PyLong_AsScaledDouble and adding the exponents. It would be logical to fix this since math.log(n) already works for large integers. My reason for requesting this change is that math.frexp is the fastest way I know of to accurately count the number of bits in a Python integer (it is more robust than math.log(n,2) and makes it easy to verify that the computed size is exact) and this is something I need to do a lot. Actually, it would be even more useful to have a standard function to directly obtain the bit size of an integer. If I'm not mistaken, PyLong_NumBits does precisely this, and would just have to be wrapped. Aside from my own needs (which don't reflect those of the Python community), there is at least one place in the standard library where this would be useful: decimal.py contains an inefficient implementation (_nbits) that could removed. ---------- components: Library (Lib) messages: 70216 nosy: fredrikj severity: normal status: open title: math.frexp and obtaining the bit size of a large integer type: feature request versions: Python 2.6, Python 3.0 _______________________________________ Python tracker <[EMAIL PROTECTED]> <http://bugs.python.org/issue3439> _______________________________________ _______________________________________________ Python-bugs-list mailing list Unsubscribe: http://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
