How to make a variadic macro (variable number of arguments)

hasen_j_35364 0
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I want to write a macro in C that accepts any number of parameters, not a specific number

example:

#define macro( X )  something_complicated( whatever( X ) )

where X is any number of parameters

I need this because whatever is overloaded and can be called with 2 or 4 parameters.

I tried defining the macro twice, but the second definition overwrote the first one!

The compiler I'm working with is g++ (more specifically, mingw)

variadic g++ c-preprocessor c
6 Answers

C99 way, also supported by VC++ compiler.

#define FOO(fmt, ...) printf(fmt, ##__VA_ARGS__)
Checkers_236430

I don't think that's possible, you could fake it with double parens ... just as long you don't need the arguments individually.

#define macro(ARGS) some_complicated (whatever ARGS)
// ...
macro((a,b,c))
macro((d,e))
user_75360 
#define DEBUG

#ifdef DEBUG
  #define PRINT print
#else
  #define PRINT(...) ((void)0) //strip out PRINT instructions from code
#endif 

void print(const char *fmt, ...) {

    va_list args;
    va_start(args, fmt);
    vsprintf(str, fmt, args);
        va_end(args);

        printf("%s\n", str);

}

int main() {
   PRINT("[%s %d, %d] Hello World", "March", 26, 2009);
   return 0;
}

If the compiler does not understand variadic macros, you can also strip out PRINT with either of the following:

#define PRINT //

or

#define PRINT if(0)print

The first comments out the PRINT instructions, the second prevents PRINT instruction because of a NULL if condition. If optimization is set, the compiler should strip out never executed instructions like: if(0) print("hello world"); or ((void)0);

Admin1
user_104680

• Variable number of arguments is denoted by an ellipsis (...) • The syntax of ISO C requires at least one fixed argument before the ‘...’

For example, you can type:

#define DEBUGMSG ( int, ...)
user_142007170
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