Memory allocation for pointer to an array in c?

Question

#include <stdio.h>
int main()
{
 int (*ptr) [5];
 printf ("%lu\n" , sizeof (ptr));
 printf ("%lu\n" , sizeof (*ptr));
}

I am using 64 bit system. When I run this code in gcc compiler it generate output 8 and 20 . My question is how many byte going to allocate for this pointer to an array ? It is 8 or 20 and why?

Answer 1

ptr designates an object that will store the address of a 5-element array of int. On your system, such a pointer takes up 8 bytes of storage.

The expression *ptr can designate an object that is a 5-element array of int; on your system, such an array takes up 20 bytes (5 4-byte elements).

Assume the code:

int (*ptr)[5] = malloc( sizeof *ptr );

This gives you something like the following (each box takes up 4 bytes):

     +---+        +---+
ptr: |   | -----> |   | (*ptr)[0]  // we don't want to index into ptr,
     + - +        +---+            // we want to index into what ptr
     |   |        |   | (*ptr)[1]  // *points to*, hence the explicit
     +---+        +---+            // dereference on ptr.  
                  |   | (*ptr)[2]  //
                  +---+            // Since a[i] is defined as *(a + i),
                  |   | (*ptr)[3]  // (*ptr)[i] == (*(ptr + 0))[i] ==
                  +---+            // (ptr[0])[i] == ptr[0][i]
                  |   | (*ptr)[4]  //
                  +---+

sizeof  ptr == sizeof (int (*)[5]) ==  8 // assuming 8-byte addresses
sizeof *ptr == sizeof (int [5])    == 20 // assuming 4-byte int

This is equivalent to writing

int arr[5];            // sizeof arr == sizeof (int [5])    == 20
int (*ptr)[5] = &arr;  // sizeof ptr == sizeof (int (*)[5]) ==  8

The size of a pointer does not depend on the size of the thing it points to.