why is it equal to 7?

Question

#include <iostream>
#include <string>
using namespace std;
void foo(int &k){
k*=2;
}
int main() {
int x=2,y=3;
foo(x);
cout << x + y;

return 0;
}

Answer 1

Because &k means that you are passing the memory address of k, in this way is possible to change the actual value of k.
Without ampersand you pass a copy of k, so foo acts on this copy and not on the actual value.

Answer 2

Its simple I'll explain you how:

When you are calling function foo() with parameter x which has a value 2.
Then it stored in k but k is not simple variable of int type but it's taking value of that address where 2 is stored.
Then you do simple maths now k = 2 and k*=2 is k= k*2 , k=4 now it changed the value of 2 which was stored at that address to 4 now you don't need to return because it changes the value of x =2 to x= 4
Then cout<<x+y; gives you 4+3 = 7

Therefore, it gives you output 7.

Answer 3

Thx~

Answer 4

i was thinking that k* was refering to the value of that address instead of k = k*2
xd

Answer 5

it should be *k instead of k* for the value xd