Changing the variable's value, by using a pointer

Question

#include <stdio.h>
void functionx(int *ptr) {
    *ptr = *ptr + 100;
}
int main(void) {
    int i = 100;
    int *p = &i;
    printf("i = %d\n", i);
    functionx(p);
    printf("i = %d\n", i);
    return 0;
}

How does the pointer change the value of i variable? Why is i=200?

Answer 1

You have your variable i in memory:

+---+
| i |
+---+

Then you create a pointer p and make it point to i:

+---+     +---+
| p | --> | i |
+---+     +---+

Now you call your function, creating a copy of the pointer p and store it in the variable ptr. That means you have two pointers both pointing to the same variable i:

+---------------+
| p (from main) | ---------\
+---------------+          |     +---+
                            >--> | i |
+----------------------+   |     +---+
| ptr (from functionx) | --/
+----------------------+

---

If you're wondering about how dereferencing with the * operator works, lets get back to the simpler case with only p and i:

+---+     +---+
| p | --> | i |
+---+     +---+

By dereferencing the pointer variable p we follow the arrow and end up with the variable i.

So *p and i are both the same.

In the main function you could to the same as functionx by using p instead:

*p = 300;
printf("i = %d\n", i);  // Will print that i is equal to 300