Output of C++ code

Question

Can anyone explain the output of this C++ code?
#include <iostream>
using namespace std;

void fun(int* i, int* j)
{
*i = *i**i;
cout << *i << endl;
*j = *j**j;
cout << *i << " " << *j << endl;
}

int main() {
int i=2, j=3;
fun(&i,&i);
printf("%d, %d",i,j);
return 0;
}
https://code.sololearn.com/c1ta1BwEW8kK/#cpp

Answer 1

fun(&I, &I) passing I refference so changes in function in i variable location reflect back in original value..
In function:
*i stores *i;
*j stores *i both stores same value i.e 2
*i = *i * *i now *i =4. Original actual argument value in main function also changes, and j also changes since j contains *i location and now *j also 4.
Hence *j = *j * *j =4*4 =16.
Now *i, *j both 16. And actual argument of main i also 16. Notice that j in main, not passed its values and not modified anywhere. So I, j in main prints 16,3

Point to be remembered &I address passing and in function works like *i (farmal) = &i (actual value).. For not to confuse, use different variable names...
Hope it helps.......

Answer 2

Thank you, I get it now.