Question

This is a follow up question from Calling constructor in return statement.

This a operator overload fun in a class.

const Integer operator+(const Integer& IntObject)
{
    cout << "Data : " << this->data << endl;
    return Integer(this->data + IntObject.data); 
}

What is the relevance of const in the return type for such functions?

int main()
{
    Integer A(1); //Create 2 object of class Integer

    Integer B(2);

    const Integer C = A + B;  //This will work

    Integer D = A + B;        //This will also work

    fun(A + B);               //Will work
}

void fun(Integer F) {}

This is a case temporaries are not created during return step due to NRVO. The object to be returned is directly constructed on the callee’s address.

Answer 1

There is no relevance in your code snippet, because you are making a copy of the returned value.

In general, it is difficult to find good reasons to return a const value. I can only see it having an effect in this type of expression, attempting to call a non-const method on a const temporary:

(someObject.someMethodReturningConstValue()).someNonConstMethod(); // error, calls non const method on const temporary

so you should only use it if you want to disallow calling non-const methods on temporaries. On the other hand, it kills move-semantics in C++11 so is discouraged.