Question

When i execute this code

#include<stdio.h>

int main() {
 int (*x)[5];
printf("\nx = %u\nx+1 = %u\n&x = %u\n&x + 1 = %u",x,x+1,&x,&x+1);
}

This is the output in C or C++:

x = 134513520
x+1 = 134513540
&x = 3221191940
&x + 1 = 3221191944

Please explain. Also what is the difference between:

int x[5] and int (*x)[5] ?

Answer 1

• int x[5] is an array of 5 integers
• int (*x)[5] is a pointer to an array of 5 integers

When you increment a pointer, you increment by the size of the pointed to type. x+1 is therefore 5*sizeof(int) bytes larger than just x – giving the 8048370 and 8048384 hex values with a difference of 0x14, or 20.

&x is a pointer to a pointer – so when you increment it you add sizeof(a pointer) bytes – this gives the bf9b08b4 and bf9b08b8 hex values, with a difference of 4.