what does "..." do as a function argument in C?

Question

I have been wondering how C's printf function worked, so I decided to look at gcc's stdio.h for the definition. To my surprise, the printf function in gcc is defined with the arguments "const char*, . . . ". I tried doing this for myself, in a simple program that I made.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int Print(const char *text, ...) {

    printf("%s\n", text);
}

int main() {
    Print("Hello, World!", "a");
}

I can pass however many arguments I want into it, even though those arguments won't have any future access point. I am wondering more about this, and I wish to know if anybody has more information.

Answer 1

This is how ISO C defines a syntax for declaring a function to take a variable number or type of arguments. Using ... notation with stdarg.h you can implement variadic funtion: functions that accept an unlimited number of argument. Its usage is explained in Gnu library manual.

You can go through the parameters using va_args from stdarg.h. Here is good tutorial to start with.

You can also implement variadic macros SOME_MACRO(...). For that you can refer to this thread.