Issues on using pointers

Question

I just got started learning about pointers and was executing some code:

#include <stdio.h>

int main(void){

    int num = 12;  // initialize num to 12
    int *pnum = NULL;  // initialize the pointer

    pnum = &num;  // stores the address of num in pnum

    printf("the address of the number is %p\n", &num);  // display the address of the number
    printf("the address of the pointer is %p\n", &pnum);  // display the address of the pointer
    printf("the value of the pointer is %p\n", pnum);  // display the value of the pointer
    printf("the value the pointer is pointing to is %d\n", *pnum);  // display the value the pointer is pointing to

    return 0;
}

In this code above, It prints out 0xffffcbec for the address of the number and value of the pointer, and 0xffffcbe0 for the address of the pointer. I want to know the reason. I feel like this is related to some incorrect inputs of data types.

I use VScode btw.

Answer 1

The value of pnum is the value of &num, i.e. the location of the variable num. Therefore &num and pnum will print the same.

To make it easier to understand and visualize, I recommend you draw it all out:

+-------+      +------+      +-----+
| &pnum | ---> | pnum | ---> | num |
+-------+      +------+      +-----+

That is:

• &pnum is pointing to pnum
• pnum (which is the same as &num) is pointing to num

---

Also, the pointer-to operator & and the dereference operator * are each others opposites.

So pnum is the value of &num. And *pnum is the value of num.

And *&num is plain num, as the operators cancel out each other.