printf function explanation

int printf(const char *format, ...);

Why printf can receive more than 2 arguments although there is no overloading in c.

Anyone can give me some brief explanation the use of ...(3 dots) in printf function argument? :confused:

... in the parameter list means that 0 or more arguments could be passed. You access these arguments with the stdarg.h functions.

Code:

---

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

void myprintf(const char* lpszText, ...)
{
        va_list args;
        char szMessage[256];

        va_start(args,lpszText);
        vsprintf(szMessage,lpszText,args);
        va_end(args);

        fputs(szMessage,stdout);       
}


int main(void)
{
        myprintf("The numbers are %d and %d\n",5,1);
        return EXIT_SUCCESS;
}

---

lpszThisIsMyFormatString. i like that.

vice, you need to read a book.
"The C programming Language" by Kernighan & Ritchie,
chapter 7.3, "Variable-length Argument Lists"

thanks for the explanation