What's the output of this and how it's come

Question

#include <stdio.h>
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;
}

Answer 1

The ouput of the code is i = 4, j = 9

fun(&i,&j) -> means you are sending the address of i and j

in the function fun(int* i, int* j) you're pointing to the address of i and j so every operation you will do it will be saved in the main i and j and you don't need the return in the function finally for the operation you only raised the number to its power like: n to the power of n.

Answer 2

Ok we are sending (&i,&i) both same address not &j

Answer 3

Can even more explain confuseing