Question

I have created a C++ program in order to test the functionality of passing parameters by reference for functions.

#include <iostream>

using namespace std;

int f(int &b) {
    b = b + 1;
    cout << b << endl;
    return b;
}

int main() {
    int t = 10;

    cout << f(t) << " " << t << endl;
    //cout << f(&t) << " " << t << endl;

    system("PAUSE");

    return 0;
}

Could you please explain to me why does this program won’t affect the value of t after the execution of the f function? The b parameter in passed be reference, so I thought that it’s value would change after the execution of the program because I am working with the actual variable from the main function, not a copy of it. In this case, I would expect it to be 11, but it’s not affected by the execution of the program.

Why is this happening?

Answer 1

The value of t does get incremented. You’ll see this if you split the output statement in two:

cout << f(t) << endl;
cout << t << endl;

With your original single output statement:

cout << f(t) << " " << t << endl;

the compiler is free to evaluate t before f(t), producing in the output you’re seeing. For more info, see cout << order of call to functions it prints?