How does printf handle its arguments?

Such a function is called a variadic function. You may declare one in C using ..., like so:

int f(int, ... );

You may then use va_start, va_arg, and va_end to work with the argument list. Here is an example:

#include <stdlib.h>#include <stdarg.h>#include <stdio.h>void f(void);main(){        f();}int maxof(int n_args, ...){        register int i;        int max, a;        va_list ap;        va_start(ap, n_args);        max = va_arg(ap, int);        for(i = 2; i <= n_args; i++) {            if((a = va_arg(ap, int)) > max)                max = a;        }        va_end(ap);        return max;}void f(void) {        int i = 5;        int j[256];        j[42] = 24;        printf("%d\n",maxof(3, i, j[42], 0));}

For more information, please see The C Book and stdarg.h.

This feature is called Variable numbers of arguments in a function. You have to include stdarg.h header file; then use va_list type and va_start, va_arg, and va_end functions within the body of your function:

void print_arguments(int number_of_arguments, ...){  va_list list;  va_start(list, number_of_arguments);  printf("I am first element of the list: %d \n", va_arg(list, int));  printf("I am second element of the list: %d \n", va_arg(list, int));  printf("I am third element of the list: %d \n", va_arg(list, int));  va_end(list);}

Then call your function like this:

print_arguments(3,1,2,3);

which will print out following:

    I am first element of the list: 1    I am second element of the list: 2    I am third element of the list: 3

The way this is done in C is called "varargs". There's a tutorial for it here: http://c-faq.com/~scs/cclass/int/sx11b.html