Question

#include<stdio.h>
int main(){
    int a[5] = {0,1,2,3,4};
    int * ptr;
    ptr =(int *) &a;
    printf("\n&a:%u,&a[0]:%u,ptr:%u\n",&a,&a[0],ptr);
    ptr = (int*)(&a+1);
    printf("\n&a:%u,&a[0]:%u,ptr:%u\n",&a,&a[0],ptr);
    ptr = (int*)(&a);
    ptr = (int*)(&a[0]+4);
    printf("\n&a:%u,&a[0]:%u,ptr:%u,*ptr:%d\n",&a,&a[0],ptr,*ptr);
    return 0;
}

o/p:

&a:3213284540,&a[0]:3213284540,ptr:3213284540

&a:3213284540,&a[0]:3213284540,ptr:3213284560

&a:3213284540,&a[0]:3213284540,ptr:3213284556,*ptr:4

In the above code &a and &a[0] gives the same address 3213284540. But the two cases when added with 1 gives different address.

&a[0]+1 => 3213284540 + 4     = 3213284544   [The value stored in this address is '1']
&a+1    => 3213284540 + (5*4) = 3213284560   [Goes Out of bounds of the array] 

&a+1 is equivalent to sizeof(array)+1. 

But how the compiler interprets this &a[0]+1 and &a+1 ?

Answer 1

But how the compiler interprets this &a[0]+1 and &a+1

It’s pointer arithmetic, so it’s always important to know the pointed types and one basic thing: adding 1 to a pointer makes it point to some “next” element.

• In your example &a[0] is of type int * so adding 1 moves the
  pointer to the next int. So the address should increase by 4/8
  bytes or so, depending on sizeof(int)

• However, &a is of type int (*)[5]. So adding 1 to it moves the
  pointer to the next array. In effect, the address should increase
  by sizeof(a).

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Side note: use %p when printing pointer values.