Calling a printf with __VA_ARGS__ not working in a function but works in a macro

Question

The following macro works:

#define DEBUG(msg, ...) printf(msg, __VA_ARGS__)

But when I add my own function, it says error: '__VA_ARGS__' was not declared in this scope. My code:

void Debug(const char* msg, ...) {
    printf(msg, __VA_ARGS__);
}

#define DEBUG(msg, ...) Debug(msg, __VA_ARGS__)

Is there any way to do so?

`__VA_ARGS__` is a variadic macro *only* thing. There are plenty of tutorials and examples for variadic functions online if you want to make one. — Some programmer dude, Apr 07 '21 at 07:29

score 2 · Answer 1 · answered Apr 07 '21 at 07:30

2

Variadic parameter pack is your friend in this case:

template< typename ... Args >
void Debug( const char * msg, Args ... args ) {
    printf( msg, args ... );
}

answered Apr 07 '21 at 07:30

NutCracker

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score 2 · Answer 2 · answered Apr 07 '21 at 07:42

__VA_ARGS__ simply does not exist outside of a variadic macro. For what you are attempting, use vprintf() instead of printf(), eg:

void Debug(const char* msg, ...) {
    va_list args;
    va_start(args, msg);
    vprintf(msg, args);
    va_end(args);
}

#define DEBUG(msg, ...) Debug(msg, __VA_ARGS__)

Calling a printf with __VA_ARGS__ not working in a function but works in a macro

2 Answers2