How does this code work?

Question

void fun(int* i, int* j) {
*i = *i * *i;
*j = *j * *j;
}
int main() {
int i = 2, j = 3;
fun(&i, &i);
printf("%d, %d", i, j);
return 0;
}

Got this in sololearn challenge. Output was 16,3. How does this work?
I thought the answer would be 4,9.
https://code.sololearn.com/cGrWtuvxOrbO/?ref=app

Answer 1

The output 16, 3 was because you pass the address of <i> (in main) as argument for both <i> and <j> parameter of `fun()` function.

`fun(&i, &i);` // <= this line

So the variable <j> is left unharmed here because you didn't pass it in as argument to function `fun`.

Let's see what happens in `fun` function. Remember we passed address of <i> as argument for both <i> and <j> parameter when we invoke `fun` function.

● Initially:
*i = 2, *j = 2 // both points to same variable, that is <i> in `main` function.

*i = *i * *i; // *i = 2 * 2 => 4

// parameter <j> is actually the same with parameter <i>.
// So here we are actually changing same variable through the address.

*j = *j * *j; // *j = 4 * 4 => 16

● Return to main function:
<i> is now 16 because its value was changed inside `fun`.

Hth, cmiiw

Answer 2

I'm glad, was afraid because I didn't feel it was clear enough 😁

Answer 3

Ipang Thanks! Well explained! I understood it now.