Whats the significance of return by reference?

Question

In C++,

function() = 10;

works if the function returns a variable by reference.

What are the use cases of it?

Answer 1

Another classic case:

class Foo {
  Foo();
public:
  static Foo& getSingleton();
};

Answer 2

Getters/setters for instance

class C
{
    int some_param_;
public:
    int& param() { return some_param_; }
    int const& param() const { return some_param_; }
};

but here you should go with some_param being a public int. Containers provide functions that return by reference, eg. vector<T>::operator[] so that you can write v[k] = x.

Answer 3

The named parameter idiom is a another use case. Consider

class Foo
{
public:
    Foo(
        int lions,
        float tigers,
        double bears,
        std::string zookeeper
    );
};

users of this class need to remember the position of each parameter

Foo foo( 1, 2.0, 5, "Fred" );

which can be non-obvious without looking at the header. Compared to a creator class like so

class CreateFoo
{
friend class Foo;
public:
    CreateFoo();

    CreateFoo& lions(int lions) {
        _lions = lions;
         return *this;
    }

    CreateFoo& tigers(float tigers) {
        _tigers = tigers;
        return *this;
    }

    CreateFoo& bears(double bears) {
        _bears = bears;
        return *this;
    }

    CreateFoo& zookeeper(const std::string& zookeeper) {
        _zookeeper = zookeeper;
        return *this;
    }

private:
    int _lions;
    float _tigers;
    double _bears;
    std::string _zookeeper;
};

which can then be used by clients like so

Foo foo = CreateFoo().
    lions(1).
    tigers(2.0).
    zookeeper("Fred").
    bears(5)
    ;

assuming Foo has a constructor taking a const CreateFoo&.

Answer 4

A very normal use case is when you write an array like class. Here you want to overload the operator [] so as you can do a[0] = 10; In that case you would want the signature to be like int& operator[](int index);

Answer 5

In case you have a class that contains another structure, it can be useful to directly modify the contained structure:

struct S
{
    int value;
};

class C
{
    public:

        S& ref() { return m_s; }

    private:

        S m_s;
};

Allows you to write something like:

void foo()
{
    C c;

    // Now you can do that:

    c.ref().value = 1;
}

Note: in this example it might be more straightforward to directly make m_s public rather than returning a reference.

Answer 6

Consider the following code, MyFunction returns a pointer to an int, and you set a value to the int.

int  *i;
i = MyFunction();
*i = 10;

Now shorten that to

*(MyFunction()) = 10;

It does exactly the same thing as the first code block.

You can look at a reference as just a pointer that's always dereferenced. So if my function returned a reference - not a pointer - to an int the frist code block would become

int  &i;
i = MyFunction();
i = 10;

and the second would become

MyFunction() = 10;

This is what i was looking for