GATE EXAM PREPARATION IN "C" - Assume sizeof an integer and a pointer is 4 byte. Output?

Assume sizeof an integer and a pointer is 4 byte. Output?

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

(A) 200
(B) 4
(C) 800
(D) 80


Answer: (C)

Explanation: Output is 10*20*sizeof(int) which is “800″ for compilers with integer size as 4 bytes.

When a pointer is de-referenced using *, it yields type of the object being pointed. In the present case, it is an array of array of integers. So, it prints R*C*sizeof(int)