Question

I have done type casting with int and char but not with pointers so I posted this question.

#include <stdio.h>
int main() {
    int a[4] = { 1, 2, 6, 4 };
    char *p;
    p = (char *)a;   // what does this statement mean?
    printf("%d\n",*p);
    p = p + 1;        
    printf("%d",*p);  // after incrementing it gives 0 why?
}

The first call to printf gives the first element of the array. And after p=p+1 it gives 0. Why?

Answer 1

int a[4] = { 1, 2, 6, 4 };

declares a as array. a at this point of time stores the address of the first element of the array.

char *p;

declares p as a pointer to character

p = (char *)a;

Now since p & a both stores addresses. So the address stored at a (address of first element of the array) is assigned to p. The typecasting is done as p was declared as char *

what it does is that, assuming address stored at a is say 100 and assuming that int takes 2 bytes and char takes 1 bytes in memory

a+1 would return a+size of int(2) = 102

and

p+1 would return p+size of char(1) = 101 

and that will explain the output as

A. two bytes stored at a contains the first element of the array.

B. one byte stored at p is the character representation of the first byte of integer 1, which is 0.

Hope this helps.