C language: addresses and pointers

An example of how addresses are stored in memory:

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
    int* x = NULL;

    printf ("address of new pointer-type variable x: %p\n", &x);
    printf ("value of variable x: %p\n", x);

    x = (int*) malloc(sizeof(int));

    printf ("now pointer-type variable x value exists and it is: %p\n", x);

    printf ("address of new pointer-type variable x: %p\n", &x);

    return 0;
}

A more complex example from cs50, which we analyzed in detail with the help of comments:

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
    int* x = NULL;
    int* y = NULL;

    printf ("address of new pointer-type variable x: %p\n", &x);
    printf ("value of variable x: %p\n", x);

    x = (int*) malloc(sizeof(int));

    printf ("now pointer-type variable x value exists and it is: %p\n", x);

    printf ("Note that address of pointer-type variable x still: %p\n\n", &x);

    *x = 42;

    printf ("So.. variable x contain address %p.\n", x);
    printf ("In bytes which can be found by this address we will write number: %d\n\n", *x);

//  now if we will try...
//  *y = 13;
//  ...we will have CRUSH: cause we try to write a value by address which was not defined;
    printf ("address of pointer-type variable y: %p\n", y);

// now we copy value of a pointer-type variable x to y
    y = x;
    printf ("pointer-type variable y now store address: %p, which has value y: %d\n", y, *y);

// put a number ('13') into allocated space by address which is stored as value in pointer-type variable y
    *y = 13;
    printf ("pointer-type variable y contain address: %p; value y: %d\n", y, *y);
    printf ("x variable still contain the same address: %p, but different value x: %d\n", x, *x);


    return 0;
}