On 10 Jul 2013, at 17:33, "O. Hartmann" <ohart...@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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