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).
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