CODEWITHSUNDEEP
c++classc++11
Eiron
Eiron

Reputation: 503

Return type for the assignment operator

What is the difference between an operator defined with its class name:

class RefObj {
public:
    RefObj &operator=(const RefObj &) { return *this; }
private:
    int m_Counter = 0;
};

and an operator with void:

template<class T> class SmartPtr {
public:
    void operator=(T* pointer) { m_SmartPtr = pointer; }
private:
    T* m_SmartPtr;
}

When should I use the first one and when should I use the second one?

Upvotes: 1

Views: 84

Answers (1)

user1978011
user1978011

Reputation: 3589

The first version allows to do further operations on the returned object in one line as in

(refobj = a).do_something();

When you don't return a reference to the object, this is not possible. You may think this is silly, but think about output operators.

void operator<<(std::ostream &out, const Obj1 &obj1);
std::ostream& operator<<(std::ostream &out, const Obj2 &obj2);

Obj1 obj1;
Obj2 obj2;
std::cout << obj1 << '\t' << obj1 << std::endl;  // compiler error
//               ^^^^ '<<' can't operate on void
std::cout << obj2 << '\t' << obj2 << std::endl;  // works!
//               ^^^^ return type is std::ostream, '<<' work on that

Given that returning a reference is really cheap, I recommend to always return one. This will save you the pain of searching for strange bugs where otherwise perfectly sane syntax will break your code.

Upvotes: 2

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