CODEWITHSUNDEEP
c++referencereturn-value
pablozoani
pablozoani

Reputation: 33

return reference in cpp function returns a copy

I am trying to return the reference of the argument in the function test, but when I assign another value to the returned reference, the original doesn't change:

#include <iostream>

using namespace std;

int& test(int& n) {
    return n;
}

int main() {
    int n = 1;
    int m = test(n);
    m = 2;
    cout << n << endl; // n = 1
    return 0;
}

How can I return this reference as I expect?

This testing is just for studying purposes.

Upvotes: 1

Views: 152

Answers (2)

digito_evo
digito_evo

Reputation: 3672

Make m an lvalue reference:

 int& m = test(n);

You can also bind to temporaries like this:

int test(int& n) { // return by value
    return n;
}

int main() {
    int n = 1;
    int&& m = test(n); // rvalue reference binding to a copy of `n`
    m = 2;
    std::cout << n << std::endl;
    return 0;
}

Although the above code is not recommended.

Or even this:

int&& test( int& n ) // return by rvalue reference
{
    return std::move( n ); // cast `n` from lvalue to xvalue
}

int main( )
{
    int n = 1;
    int&& m = test(n);
    m = 2;
    std::cout << n << std::endl;
}

This one is also not recommended. Instead of rvalue reference, you should return by value.

Upvotes: 2

dtell
dtell

Reputation: 2568

The line

int m = test(n);

creates a copy of the value test(n) returns (in this case a copy of n).

If you want a reference to the return value of test(n) you have to use one:

int& m = test(n);

A common caveat occurs with type inference using auto: Even in this case you would have to use

auto& m = test(n);

Upvotes: 4

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