what is the output of this code

Hi all,

Code:

---

#include<stdio.h>

int main()
{
        char p[]="hello";

        printf("sizeof(p)=%d\n",sizeof(p));
        printf("sizeof(p+1)=%d\n",sizeof(p+1));

        return 0;
}

---

the output is 6 and 4

Can someone explain why is the output of sizeof(p+1)=4

Regards and Thanks
C_Enthusiast

p by itself is an array.

p+1 is an expression. Arrays in expressions become "a pointer to the first element". So what you end up with is sizeof(char*)

Take a look here,

sizeof - Wikipedia, the free encyclopedia