CODEWITHSUNDEEP
c++
anonymous
anonymous

Reputation: 37

Why do we have to return by reference when overloading += operator

While overloading += operator why do we have to return by reference. for e.g. the below code also do the same thing

class Integer
{
  int i;
  public:
    Integer::Integer(int i):i(i){}
    void operator+=(const Integer& arg)
    {
      i = i + arg.i;
    }
};

//main.cpp
int _tmain(int argc, _TCHAR* argv[])
{
   Integer a(10),b(20);
   b += a;
}

Most of the books suggest that for the above operator overloaded function should return by reference i.e as below:

Integer& operator+=(const Integer&)
{
    i = i + arg.i;
    return *this;
}

If we return by reference then what happens to the return object reference when below statement is executed:

b += a;

Upvotes: 2

Views: 218

Answers (2)

Xaqq
Xaqq

Reputation: 4386

If we return by reference then what happens to the return object reference when below statement is executed:

b += a;

Nothing really. The statement gets executed, the reference is unused and b keep going on with its life.

This interesting stuff that returning by reference allows is chaining call: you cannot do (b += b) += a if you don't return by reference. This would looks like this: (void) += const Integer &, because b += b is of type void due to operator+= returning void.

Upvotes: 3

user1804599
user1804599

Reputation:

So that T& x = (y += z); works consistently with fundamental types.

Upvotes: 1

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