From fa41e60c9d359b0e24486b884a87fe97505b9153 Mon Sep 17 00:00:00 2001 From: Mark Dickinson Date: Tue, 28 Sep 2010 07:22:27 +0000 Subject: Issue #9599: Tweak loghelper algorithm to return slightly improved results for powers of 2. --- Misc/NEWS | 3 +++ Modules/mathmodule.c | 14 ++++++++------ 2 files changed, 11 insertions(+), 6 deletions(-) diff --git a/Misc/NEWS b/Misc/NEWS index 1ed5172..beedb0c 100644 --- a/Misc/NEWS +++ b/Misc/NEWS @@ -2201,6 +2201,9 @@ Library Extension Modules ----------------- +- Issue #9959: Tweak formula used for computing math.log of an integer, + making it marginally more accurate for exact powers of 2. + - Issue #9422: Fix memory leak when re-initializing a struct.Struct object. - Issue #7900: The getgroups(2) system call on MacOSX behaves rather oddly diff --git a/Modules/mathmodule.c b/Modules/mathmodule.c index 69e1423..3cfb5f7 100644 --- a/Modules/mathmodule.c +++ b/Modules/mathmodule.c @@ -1572,12 +1572,14 @@ loghelper(PyObject* arg, double (*func)(double), char *funcname) "math domain error"); return NULL; } - /* Special case for log(1), to make sure we get an - exact result there. */ - if (e == 1 && x == 0.5) - return PyFloat_FromDouble(0.0); - /* Value is ~= x * 2**e, so the log ~= log(x) + log(2) * e. */ - x = func(x) + func(2.0) * e; + /* Value is ~= x * 2**e, so the log ~= log(x) + log(2) * e. + + It's slightly better to compute the log as log(2 * x) + log(2) * (e + - 1): then when 'arg' is a power of 2, 2**k say, this gives us 0.0 + + log(2) * k instead of log(0.5) + log(2)*(k+1), and so marginally + increases the chances of log(arg, 2) returning the correct result. + */ + x = func(2.0 * x) + func(2.0) * (e - 1); return PyFloat_FromDouble(x); } -- cgit v0.12