What is the purpose of & before a function declaration here?

August 6, 2024

#include <simplecpp>

int &f(int &x, int &y) {
    if (x > y) return x;
    else return y;
}

main_program{
    int p=5, q=4;
    f(p,q) = 2;
    cout << p << ' ' << q << endl;
    cout << f(p, q) << endl;
}

(It’s a little simple version of C++ used to teach students). What I am not able to understand is the & before the function declaration. What exactly is it doing?

>Solution :

The & before a function declaration in C++ indicates that the function returns a reference. In the provided code:

#include <simplecpp>

int &f(int &x, int &y) {
    if (x > y) return x;
    else return y;
}

main_program{
    int p=5, q=4;
    f(p,q) = 2;
    cout << p << ' ' << q << endl;
    cout << f(p, q) << endl;
}

Here, int &f(int &x, int &y) means that the function f returns a reference to an integer. The parameters int &x and int &y also mean that x and y are passed to the function by reference.

The purpose of returning a reference from the function is to allow the function to return a reference to one of the variables passed to it so that the caller can modify it directly. This is demonstrated in the following line:

f(p, q) = 2;

In this line, f(p, q) returns a reference to either p or q, depending on their values. Since p is greater than q (5 > 4), f(p, q) returns a reference to p. Therefore, f(p, q) = 2; effectively sets p to 2.

Let’s break down the output step by step:

Initially, p is 5, and q is 4.
f(p, q) returns a reference to p (since p > q).
f(p, q) = 2; sets p to 2.
cout << p << ' ' << q << endl; prints the values of p and q, which are now 2 4.
cout << f(p, q) << endl; calls f(p, q) again, and since p (now 2) is not greater than q (4), it returns a reference to q. Thus, it prints the value of q, which is 4.

So, the output of the program will be:

2 4
4

The use of a reference return type allows the function to provide direct access to the variables passed as arguments, enabling the caller to modify these variables through the function’s return value.