为什么在重载+ =运算符时我们必须通过引用返回

Question

While overloading += operator why do we have to return by reference. for eg 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 ie 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;

Answer 1

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 .

Answer 2

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