Why a pointer * is inputted for expressions alonem

Question

#include <stdio.h>

int main() {
int x = 5;
int y;
int *p = NULL;
p = &x;

y = *p + 2; /* y is assigned 7 */
y += *p; /* y is assigned 12 */
*p = y; /* x is assigned 12 */
(*p)++; /* x is incremented to 13 */

printf("p is pointing to the value %d\n", *p);
}

Answer 1

What this lesson showed me is that pointers can refer to variables or just values themselves.

int *p=NULL; //is just how you set up a blank interger pointer named p.

p=&x; // assigns the pointer to the address of x (so if x changes, so does p)

but,
*p=y; //is just saying that x (*p is assigned to the address of x) is equal to the current value of y wich was 12.
(*p, otherwise known as x, now equals 12)

So it looks like p would be the address of x and *p would be the x addresse's value.