Define macro using _Generic gives typename not allowed

Question

I am trying to define a generic macro, which I intend to use with exception handling while debugging the code. When I try to compile the code below, it says typename not allowed. I am a noob when it comes to the macros, any and all help is highly appreciated.


#define ASSERTEXCP(x) _Generic((x),\
char *: printf( "assertion error line %d, file(%s):-> %s\n", __LINE__, __FILE__, x );
char strMsg[2014] = {'\0'}; \
sprintf(strMsg, "\nassertion error line %d, file(%s):-> %s\n", __LINE__, __FILE__, x); \
OutputDebugString(strMsg););


#endif

https://en.cppreference.com/w/c/language/generic It does work in C++ too. — Qasim Shafi, Oct 22 '19 at 22:09
@QasimShafi where in that page does it say that it works on C++ too? — Amadeus, Oct 22 '19 at 22:10
try this one. working fine in C++11 & 14 http://cpp.sh/75hnig — Qasim Shafi, Oct 22 '19 at 22:14
I know _Generic is C, and for current instance i am unable to use it with C++ somehow, any suggestions about how I can achieve what I am trying to? — Qasim Shafi, Oct 22 '19 at 22:17
@QasimShafi Your link does not use _Generic. Try this one http://coliru.stacked-crooked.com/a/9c11154ed84255ce — Amadeus, Oct 22 '19 at 22:19
MS is doing something like this _GENERIC_MATH1(cbrt, _CRTSPECIAL). with extern C syntax. — Qasim Shafi, Oct 22 '19 at 22:30
`_Generic` is the C approach to C++ overloading. Perhaps use overloading or `if constexpr` — Swift - Friday Pie, Oct 22 '19 at 23:17

score 1 · Accepted Answer · answered Oct 23 '19 at 06:36

_Generic is the C approach to C++ overloading. C++ approach is to use if constexpr or an overloaded function:

    #include <type_traits>

    #define ASSERTEXCP(x) if constexpr (std::is_same<decltype(x), char *>::value) {  \
    printf( "assertion error line %d, file(%s):-> %s\n", __LINE__, __FILE__, x ); \
    char strMsg[2014] = {'\0'}; \
    sprintf(strMsg, "\nassertion error line %d, file(%s):-> %s\n", __LINE__, __FILE__, x); \
    OutputDebugString(strMsg); }

Define macro using _Generic gives typename not allowed

1 Answers1