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

The commonest case is to implement things like operator[].

struct A {
    int data[10];
    int & operator[]( int i ) {
         return data[i];
    }
};

Another is to return a big object from a class via an accesor function:

struct b {
    SomeBigThing big;
    const SomeBigThing & MyBig() const {
         return big;
    }
};

in order to avoid the copying overhead.

Answer 2

std::vector has operator[] which would not allow vec[n] = m otherwise.

Answer 3

You can also achieve method chaining (if you so desire) using return by reference.

class A
{
public:
    A& method1()
    {
        //do something
        return *this;   //return ref to the current object
    }
    A& method2(int i);
    A& method3(float f);  //other bodies omitted for brevity
};

int main()
{
    A aObj;
    aObj.method1().method2(5).method3(0.75);

    //or use it like this, if you prefer
    aObj.method1()
        .method2(5)
        .method3(0.75);
}

Answer 4

It can be usefull when implementing accessors

class Matrix
{
   public:
      //I skip constructor, destructor etc

      int & operator ()(int row, int col)
      {
         return m_arr[row + col * size];
      }

   private:
      int size;
      int * m_arr;
}

Matrix m(10);
m(1,0) = 10;  //assign a value to row 1, col 0

Answer 5

SO screwed up my answer

You don't even need to return a reference:

struct C { };

C f() {
  return C();
}

int main() {
  C a;
  f() = a;  // compiles fine
}

Because this behavior is quite surprising, you should normally return a const value or a const reference unless the user has a sensible intent to modify the result.