Output of a code- behaviour of & in c

Question

I came across a code whoes output I'm not able to understand.The code is-

int main() 
{       
  int a[] = {1, 2, 3, 4, 5, 6};   

  int *ptr = (int*)(&a+1); 
  printf("%d ", *(ptr-1) ); 
  return 0; 
} 

The output of above code is coming out 6, but i think that it should be 1. Please explain why it is 6.

Answer 1

In your question "&a" is address of the whole array a[]. If we add 1 to &a, we get “base address of a[] + sizeof(a)”. And this value is typecasted to int *. So ptr points to the memory just after 6 . ptr is typecasted to "int *" and value of *(ptr-1) is printed. Since ptr points memory after 6,so ptr – 1 points to 6.

Answer 2

Yes, because a is an array having a type int[6]. Therefore, &a gives you the type int (*)[6]. It is not same as pointer to int, it is a pointer to int array.

So, &a + 1 increments the pointer by one of the the array size, pointing past the last element of the array.

Then, taking the address in ptr and doing a -1, decreases the address by a sizeof(*ptr) which is sizeof(int), which gives you the address of last element on the array.

Finally, de-referencing that address, you get the value of the last element , 6. Success.

Answer 3

&a is an address of array a. Adding 1 to it will increment it to one past the array (adding 24-bytes). Cast operator (int*) cast &a+1 to pointer to int type. ptr-1 will decrement ptr by 4 bytes only and therefore it is the address of last element of array a. Dereferencing ptr - 1 will give the last element which is 6.

Answer 4

Because (int*)(&a + 1) uses the whole array as the base type and adds 1 whole size of an array to the address , i.e. adding 24 bytes.

And when you do ptr-1 since ptr type is int it will subtract only 4 bytes. Its very important to remember the type of the pointer while doing pointer arithmetic.