Le Sun, 15 Jan 2012 06:48:28 +0100, Ivan Timofeev <timofeev....@gmail.com>
a écrit:
Hi Mat,
cppcheck reported this :
core/vcl/generic/fontmanager/fontconfig.cxx
196 duplicateExpression style Same expression on both sides of '-'
class SortFont : public ::std::binary_function< const FcPattern*, const
FcPattern*, bool >
{
public:
bool operator()(const FcPattern *a, const FcPattern *b)
{
int comp = compareFontNames(a, b);
if (comp != 0)
return comp < 0;
int nVersionA=0, nVersionB=0;
bool bHaveA = FcPatternGetInteger(a, FC_FONTVERSION, 0,
&nVersionA) == FcResultMatch;
bool bHaveB = FcPatternGetInteger(b, FC_FONTVERSION, 0,
&nVersionB) == FcResultMatch;
if (bHaveA && bHaveB)
return nVersionA > nVersionB;
196: return bHaveA - bHaveA;<--- Same expression on both sides of
'-'.
}
};
it seems it should be :
return bHaveA - bHaveB;
Any idea ?
Subtracting booleans ??? Is there some magic here ?
Moreover, if we state that boolean difference returns a boolean, is
there an automatic cast, as the function should return an int ?
AFAIK in arithmetic bools are concerted to int so that true->1 and
false->0.
Then, the result is converted back to bool (because the function returns
bool) so that 0->false and 'any other value'->true.
So "return bool1 - bool2" can be replaced with not-so-magic "return
bool1 != bool2", right?
For operator(), absolutely. I was actually talking about compareFontNames.
But knowing that C++ does "on-the-fly" cast explains all.
By the way, the author of the code is Caolan, let's ask him. :)
Ivan
Mat
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