Pointer arithmetic

I was looking at what the ‘+’ and ‘-‘ operators actually performs when applied to pointers and arrays, as well as other tricks, and wrote some programs to test it:

#include <stdio.h>

main()
{
    int Arr[16];
    unsigned int a0 = (unsigned int) &Arr[0];
    unsigned int a3 = (unsigned int) &Arr[3];
    printf("%u\n", a3 - a0);
    printf("%u\n", &Arr[3] - &Arr[0]);
}

The result is this:

12
3

This other program explores this feature and some others:

#include <stdio.h>

int main()
{
    int Arr[16];
    Arr[0] = 9;
    Arr[3] = 5;
    printf("Arr[3]:      \t%d\n",Arr[3]);
    printf("*(Arr + 3):  \t%d\n",*(Arr + 3));
    printf("*(3 + Arr):  \t%d\n",*(3 + Arr));
    printf("*(&Arr[0] + 3):\t%d\n",*(&Arr[0] + 3)); // &Arr[0] is the same as Arr
    printf("3[Arr]:      \t%d\n",3[Arr]); // gets converted by compiler to *(3 + Arr)
    printf("\n");

    // showing how + operation behaves when used with addresses of pointers or arrays
    printf("Arr:         \t%u\n", Arr);
    printf("&Arr[0]:     \t%u\n", &Arr[0]);
    printf("&Arr[0] + 1: \t%u\n", &Arr[0] + 1);
    printf("&*(&Arr[0] + 1):%u\n", &*(&Arr[0] + 1));
    printf("sizeof(Arr): \t%d\n", sizeof(Arr));
    printf("sizeof(Arr[0]):\t%d\n", sizeof(Arr[0]));
    printf("sizeof(int):\t%d\n", sizeof(int));
    printf("\n");

    printf("sizeof(&Arr[0]):%d\n", sizeof(&Arr[0]));
    printf("sizeof(unsigned long):%d\n", sizeof(unsigned long));
    printf("\n");


    unsigned long ArrAddr = (unsigned long)&Arr[0];
    printf("ArrAddr: %u\n", ArrAddr);
    int *p;
    p = ArrAddr;
    printf("*p: %d\n", *p);
    printf("*(p+3): %d\n", *(p+3));
    printf("\n");

    // breaking operator overloading of +
    unsigned long ArrAddr3 = ArrAddr + 3;
    p = ArrAddr3;
    printf("ArrAddr3: %u\n", ArrAddr3);
    printf("*p: %d\n", *p);
    printf("\n");

    // doing manually what overloading of + does automatically
    ArrAddr3 = ArrAddr + 3*sizeof(Arr[0]);
    p = ArrAddr3;
    printf("ArrAddr3: %u\n", ArrAddr3);
    printf("*p: %d\n", *p);
    printf("\n");



    return 0;

}

The result is this:

Arr[3]:         5
*(Arr + 3):     5
*(3 + Arr):     5
*(&Arr[0] + 3): 5
3[Arr]:         5

Arr:            1587159312
&Arr[0]:        1587159312
&Arr[0] + 1:    1587159316
&*(&Arr[0] + 1):1587159316
sizeof(Arr):    64
sizeof(Arr[0]): 4
sizeof(int):    4

sizeof(&Arr[0]):8
sizeof(unsigned long):8

ArrAddr: 1587159312
*p: 9
*(p+3): 5

ArrAddr3: 1587159315
*p: 0

ArrAddr3: 1587159324
*p: 5