On 10 Jul 2013, at 17:33, "O. Hartmann" <ohartman_at_zedat.fu-berlin.de> wrote: > Hi David, > > thanks for the fast response. > > The code I was told to check with is this: > > #include <iostream> > #include <typeinfo> > #include <cmath> > > int > main(void) > { > > std::cout << typeid(isnan(1.0)).name() << "\n"; > > } > > > If I compile it with > > c++ -o testme -std=c++11 -stdlib=libc++ source.cc > > and run the binary, the result is "i" which I interpret as "INT". I believe there is a bug, which is that the math.h things are being exposed but shouldn't be, however it is not the bug that you think it is. Try this line instead: std::cout << typeid(std::isnan(1.0)).name() << "\n"; We have a libm function, isnan(), and a libc++ function, std::isnan(). The former is detected if you do not specify a namespace. I am not sure what will happen if you do: #include <iostream> #include <typeinfo> #include <cmath> using namespace std; int main(void) { cout << typeid(isnan(1.0)).name() << "\n"; } This is considered bad form, but does happen in some code. I am not certain what the precedence rules are in this case and so I don't know what happens. To properly fix this, we'd need to namespace the libm functions when including math.h in C++. This would also include fixing tweaking the macros. A fix for your code is to ensure isnan() and isinf() are explicitly namespaced. Potentially, this may also work: using std::isinf; using std::isnan; David
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