*The prototype of printf is*
*int printf(const char ***format**, ...);*
*
*
*Thus it takes a string and then variable number of arguments.*
*
*
*Every argument passed after (char *format)string is to resolve the"%"
inside the string (which is passed as the first argument)*
*
*
*Thus "anuj" will be printed as normal as printf("anuj");*
*Since "anuj" which is char *format for the printf function does not need to
resolve % because there aren't any.*
*
*
It will be good to read the printf implementation in Dennis Ritchie to get
a better understanding
On Fri, Nov 5, 2010 at 7:58 PM, ANUJ KUMAR <kumar.anuj...@gmail.com> wrote:
> #include<stdio.h>
> int main()
> {
> printf("anuj","kumar");
> return 0;
> }
>
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