Different Pointer Arithmetic Results when Taking Address of Array

Question

Program:

#include<stdio.h>  int main(void) {     int x[4];     printf("%p\n", x);     printf("%p\n", x + 1);     printf("%p\n", &x);     printf("%p\n", &x + 1); } 

Output:

$ ./a.out 0xbff93510 0xbff93514 0xbff93510 0xbff93520 $ 

I expect that the following is the output of the above program. For example:

x        // 0x100 x+1      // 0x104  Because x is an integer array &x       // 0x100  Address of array &x+1     // 0x104 

But the output of the last statement is different from whast I expected. &x is also the address of the array. So incrementing 1 on this will print the address incremented by 4. But &x+1 gives the address incremented by 10. Why?

Answer 1

x -> Points to the first element of the array. &x ->Points to the entire array. 

Stumbled upon a descriptive explanation here: http://arjunsreedharan.org/post/69303442896/the-difference-between-arr-and-arr-how-to-find

SO link: Why is arr and &arr the same?