C++ preprocessor __VA_ARGS__ number of arguments

Question

Simple question for which I could not find answer on the net. In variadic argument macros, how to find the number of arguments? I am okay with boost preprocessor, if it has

Answer 1

I usually use this macro to find a number of params:

#define NUMARGS(...)  (sizeof((int[]){__VA_ARGS__})/sizeof(int))

Full example:

#include 
#include 
#include 

#define NUMARGS(...)  (sizeof((int[]){__VA_ARGS__})/sizeof(int))
#define SUM(...)  (sum(NUMARGS(__VA_ARGS__), __VA_ARGS__))

void sum(int numargs, ...);

int main(int argc, char *argv[]) {

    SUM(1);
    SUM(1, 2);
    SUM(1, 2, 3);
    SUM(1, 2, 3, 4);

    return 1;
}

void sum(int numargs, ...) {
    int     total = 0;
    va_list ap;

    printf("sum() called with %d params:", numargs);
    va_start(ap, numargs);
    while (numargs--)
        total += va_arg(ap, int);
    va_end(ap);

    printf(" %d\n", total);

    return;
}

It is completely valid C99 code. It has one drawback, though - you cannot invoke the macro SUM() without params, but GCC has a solution to it - see here.

So in case of GCC you need to define macros like this:

#define       NUMARGS(...)  (sizeof((int[]){0, ##__VA_ARGS__})/sizeof(int)-1)
#define       SUM(...)  sum(NUMARGS(__VA_ARGS__), ##__VA_ARGS__)

and it will work even with empty parameter list