different between sizeof(p) and sizeof(*p) while p has base type of a matrice

Question

I'm learning pointers, and I'm kind of confused because of the output given by the following program:

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

output : 4 800

why the output is 800? we know that p is a pointer that has a base type of a 2-D array of 10 rows and 20 columns, that means *p points to the 0th array of the matrice, which means sizeof(*p)=20*sizeof(int)=80 and sizeof(p) should be equivalent to 800 but the output is way different from my calculations!

can I get an explanation ? much appreciated.

Answer 1

TL;DR - Always follow the data type!

• p is a pointer type variable, so sizeof(p) is the size of a pointer on your platform, which happens to be 4.

• sizeof(*p), is the same as sizeof(int[R][C]), which is the size of an int, multiplied by R and C. The result is sizeof(int) == 4 (on your platform), 10and 20, multiplied together, 800.

Note: In case of sizeof operator, the array does not decay to a pointer to the first element, it retains the type.

That said, sizeof yields a result of type size_t, use %zu format specifier to print the result.