New submission from STINNER Victor <[email protected]>:
Extract of Python 2.7, Modules/audioop.c:
static int
fbound(double val, double minval, double maxval)
{
if (val > maxval)
val = maxval;
else if (val < minval + 1)
val = minval;
return val;
}
Example of usage:
double factor, fval, maxval, minval;
int len, size, val = 0;
fval = (double)val*factor;
val = (int)floor(fbound(fval, minval, maxval));
It seems wrong to me to call floor() with an integer. Why fbound() doesn't
return a double?
fbound() has been modified to explicitly cast its result to an int in the
master branch:
static int
fbound(double val, double minval, double maxval)
{
if (val > maxval)
val = maxval;
else if (val < minval + 1)
val = minval;
return (int)val;
}
But master still uses something like:
val = floor(fbound(val, minval, maxval));
It seems like fbound() result is always passed into floor(). Maybe floor()
should be moved into fbound()?
Attached PR changes fbound() to round correctly: call floor() into fbound().
--
I looked at the code because of the following compiler warning on Python 2.7:
[00:02:21] ..\Modules\audioop.c(38): warning C4244: 'return' : conversion from
'double' to 'int', possible loss of data
[C:\projects\cpython\PCbuild\pythoncore.vcxproj]
----------
components: Extension Modules
messages: 318808
nosy: vstinner
priority: normal
severity: normal
status: open
title: audioop.c: fbound() casts double to int for its return value
versions: Python 2.7, Python 3.6, Python 3.7, Python 3.8
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Python tracker <[email protected]>
<https://bugs.python.org/issue33781>
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