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authorMark Dickinson <dickinsm@gmail.com>2010-09-28 07:22:27 (GMT)
committerMark Dickinson <dickinsm@gmail.com>2010-09-28 07:22:27 (GMT)
commitfa41e60c9d359b0e24486b884a87fe97505b9153 (patch)
tree7b193854da452541dae589287415ced508b380b2 /Modules/mathmodule.c
parentd057cd62f7336220ea4d51bc36fcfcffba4879e8 (diff)
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Issue #9599: Tweak loghelper algorithm to return slightly improved results for powers of 2.
Diffstat (limited to 'Modules/mathmodule.c')
-rw-r--r--Modules/mathmodule.c14
1 files changed, 8 insertions, 6 deletions
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);
}