What does __VA_ARGS__ in a macro mean?

Question

/* Debugging */
#ifdef DEBUG_THRU_UART0
#   define DEBUG(...)  printString (__VA_ARGS__)
#else
void dummyFunc(void);
#   define DEBUG(...)  dummyFunc()   
#endif
         


        

Answer 1

The dots are called, together with the __VA_ARGS__, variadic macros

When the macro is invoked, all the tokens in its argument list [...], including any commas, become the variable argument. This sequence of tokens replaces the identifier VA_ARGS in the macro body wherever it appears.

source, bold emphasis of mine.

A sample of usage:

#ifdef DEBUG_THRU_UART0
#   define DEBUG(...)  printString (__VA_ARGS__)
#else
void dummyFunc(void);
#   define DEBUG(...)  dummyFunc()   
#endif
DEBUG(1,2,3); //calls printString(1,2,3) or dummyFunc() depending on
              //-DDEBUG_THRU_UART0 compiler define was given or not, when compiling.

Answer 2

It's a variadic macro. It means you can call it with any number of arguments. The three ... is similar to the same construct used in a variadic function in C

That means you can use the macro like this

DEBUG("foo", "bar", "baz");

Or with any number of arguments.

The __VA_ARGS__ refers back again to the variable arguments in the macro itself.

#define DEBUG(...)  printString (__VA_ARGS__)
               ^                     ^
               +-----<-refers to ----+

So DEBUG("foo", "bar", "baz"); would be replaced with printString ("foo", "bar", "baz")