I was trying to come up with a hack to test if std::isnan
is defined without special casing compilers in the preprocessor, and came up with the following, which I was expecting to work fine.
#include <cmath>
#include <type_traits>
namespace detail {
using namespace std;
struct dummy {};
void isnan(dummy);
//bool isnan(float); // Just adding this declaration makes it work!
template <typename T>
struct is_isnan_available {
template <typename T1>
static decltype(isnan(T1())) test(int);
template <typename>
static void test(...);
enum { value = !std::is_void<decltype(test<T>(0))>::value };
};
}
int main() {
return detail::is_isnan_available<float>::value;
}
Turns out it doesn't detect it. I know for certain std::isnan
is defined on ideone, because I tested that manually.
And when I uncomment the marked line above, it works.
What am I missing here? What explains this behaviour?