Why does this *ptr NOT give the actual value stored at the memory address contained in ptr variable?

Question

Here is a basic C program about pointers:

#include <stdio.h>

int main() {
    int variable = 20;
    int *pointerToVariable;

    pointerToVariable = &variable;

    printf("Address of variable: %x\n", &variable);
    printf("Address of variable: %x\n", pointerToVariable);
    printf("Address of variable: %x\n", *pointerToVariable); // * is the DEREFERENCING OPERATOR/INDIRECTION OPERATOR in C. 
                                                                //It gives the value stored at the memory address 
                                                                //stored in the variable which is its operand.
    getchar();
    return 0;
}

This produces the following output:

Address of variable: 8ffbe4
Address of variable: 8ffbe4
Address of variable: 14

But *pointerToVariable should print 20, shouldn't it? Because * gives the actual value stored at the memory address stored in its operand, right? What am I missing?

Answer 1

14 is the HEX value of 20.

Change printf format specifier to %d instead of %x to have 20 as output

printf("Address of variable: %d\n", *pointerToVariable);

Moreover correct format specifier for pointers is %p, so

printf("Address of variable: %x\n", pointerToVariable);

must be

printf("Address of variable: %p\n", (void *)pointerToVariable);