How does printf handle its arguments?

Question

How does printf handle its arguments? I know that in C# I can use params keyword to do something similar but I can't get it done in C ?

Answer 1

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.

Answer 2

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