Concepts of & and * in C

Question 1: #include<stdio.h> int main() {   int a[] = {1, 2, 3, 4, 5, 6};   int *ptr = (int*)(&a+1);   printf("%d ", *(ptr-1) );   getchar();   return 0; }

Output: 6
&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 – 1 points to last element of a[]

Question 2:

#include<stdio.h>   #define R 10 #define C 20   int main() {    int (*p)[R][C];    printf("%d",  sizeof(*p));    getchar();    return 0; }

Output: 10*20*sizeof(int) which is “800″ for compilers with integer size as 4 bytes.
The pointer p is de-referenced, hence it yields type of the object. In the present case, it is an array of array of integers. So, it prints R*C*sizeof(int).