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author | Mark Dickinson <dickinsm@gmail.com> | 2010-01-02 15:33:56 (GMT) |
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committer | Mark Dickinson <dickinsm@gmail.com> | 2010-01-02 15:33:56 (GMT) |
commit | 6ecd9e53ce43796a0626d16a89cd6a99b28333cc (patch) | |
tree | 006509589544384205fdb543e214b101e1c71414 /Modules | |
parent | 01f748a8326770f98e514da85b4a42cae672cd80 (diff) | |
download | cpython-6ecd9e53ce43796a0626d16a89cd6a99b28333cc.zip cpython-6ecd9e53ce43796a0626d16a89cd6a99b28333cc.tar.gz cpython-6ecd9e53ce43796a0626d16a89cd6a99b28333cc.tar.bz2 |
Merged revisions 77234 via svnmerge from
svn+ssh://pythondev@svn.python.org/python/trunk
........
r77234 | mark.dickinson | 2010-01-02 14:45:40 +0000 (Sat, 02 Jan 2010) | 7 lines
Refactor some longobject internals: PyLong_AsDouble and _PyLong_AsScaledDouble
(the latter renamed to _PyLong_Frexp) now use the same core code. The
exponent produced by _PyLong_Frexp now has type Py_ssize_t instead of the
previously used int, and no longer needs scaling by PyLong_SHIFT. This
frees the math module from having to know anything about the PyLong
implementation. This closes issue #5576.
........
Diffstat (limited to 'Modules')
-rw-r--r-- | Modules/mathmodule.c | 27 |
1 files changed, 15 insertions, 12 deletions
diff --git a/Modules/mathmodule.c b/Modules/mathmodule.c index a057cac..ee479ea 100644 --- a/Modules/mathmodule.c +++ b/Modules/mathmodule.c @@ -54,7 +54,6 @@ raised for division by zero and mod by zero. #include "Python.h" #include "_math.h" -#include "longintrepr.h" /* just for SHIFT */ #ifdef _OSF_SOURCE /* OSF1 5.1 doesn't make this available with XOPEN_SOURCE_EXTENDED defined */ @@ -1342,11 +1341,12 @@ PyDoc_STRVAR(math_modf_doc, /* A decent logarithm is easy to compute even for huge longs, but libm can't do that by itself -- loghelper can. func is log or log10, and name is - "log" or "log10". Note that overflow isn't possible: a long can contain - no more than INT_MAX * SHIFT bits, so has value certainly less than - 2**(2**64 * 2**16) == 2**2**80, and log2 of that is 2**80, which is + "log" or "log10". Note that overflow of the result isn't possible: a long + can contain no more than INT_MAX * SHIFT bits, so has value certainly less + than 2**(2**64 * 2**16) == 2**2**80, and log2 of that is 2**80, which is small enough to fit in an IEEE single. log and log10 are even smaller. -*/ + However, intermediate overflow is possible for a long if the number of bits + in that long is larger than PY_SSIZE_T_MAX. */ static PyObject* loghelper(PyObject* arg, double (*func)(double), char *funcname) @@ -1354,18 +1354,21 @@ loghelper(PyObject* arg, double (*func)(double), char *funcname) /* If it is long, do it ourselves. */ if (PyLong_Check(arg)) { double x; - int e; - x = _PyLong_AsScaledDouble(arg, &e); + Py_ssize_t e; + x = _PyLong_Frexp((PyLongObject *)arg, &e); + if (x == -1.0 && PyErr_Occurred()) + return NULL; if (x <= 0.0) { PyErr_SetString(PyExc_ValueError, "math domain error"); return NULL; } - /* Value is ~= x * 2**(e*PyLong_SHIFT), so the log ~= - log(x) + log(2) * e * PyLong_SHIFT. - CAUTION: e*PyLong_SHIFT may overflow using int arithmetic, - so force use of double. */ - x = func(x) + (e * (double)PyLong_SHIFT) * func(2.0); + /* 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; return PyFloat_FromDouble(x); } |