Josh Rosenberg added the comment: While you're doing this, might it make sense to add a special case to long_pow so it identifies cases where a (digit-sized) value with an absolute value equal to a power of 2 is being raised to a positive integer exponent, and convert said cases to equivalent left shifts?
Identifying powers of 2 can be done in constant time with no memory overhead for register sized values (see http://graphics.stanford.edu/~seander/bithacks.html#DetermineIfPowerOf2 ); if you've already handled the case for 0, then checking if it's a power of 2 is just: (v & (v - 1)) == 0 It adds a little complexity, but even just handling 2 alone specially would at least make it so the semi-common case of making a large power of 2 doesn't have significantly different performance using 2 ** (numbits - 1) instead of 1 << (numbits - 1). ---------- _______________________________________ Python tracker <rep...@bugs.python.org> <http://bugs.python.org/issue21419> _______________________________________ _______________________________________________ Python-bugs-list mailing list Unsubscribe: https://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
