Why is void returning a value?

Question

I can't understand one weird thing. Here is my program :

#include <iostream>
using namespace std;

void Swap(int *a , int *b)
{
    int temp;
    temp = *a;
    *a = *b;
    *b = temp;
}

int main()
{
    int a=0;
    int b=0;

    cout<<"Please enter integer A: ";
    cin>>a;

    cout<<"Please enter integer B: ";
    cin>>b;

    cout<<endl;

    cout<<"************Before Swap************\n";
    cout<<"Value of A ="<<a<<endl;
    cout<<"Value of B ="<<b<<endl;

    Swap (&a , &b);

    cout<<endl;

    cout<<"************After Swap*************\n";
    cout<<"Value of A ="<<a<<endl;
    cout<<"Value of B ="<<b<<endl;

    return 0;
}

Now if you look at the function "Swap", I have used "void Swap". Therefore it must not return any value to main function (only "int" returns a value (at least that's what my teacher has taught me)). But if you execute it, the values are swaped in main function! How come ? Can anyone tell me how its possible ?

Answer 1

Swap function in your example just swaps two integers, but does NOT return anything.

In order to check what function has retuned, you have to assign it to some variable, like this

int a = swap(&a, &b);

but this piece of code has an error because swap function doesn't return anything.

Another example:

int func() {
    return 18;
}

int main() {
    int a =  func();
    cout << a;
}

Is fine, cause variable a is int and function func returns an int.