Self operator override - Return by reference or return by copy

Consider the f开发者_C百科ollowing code?

I was wondering, if I change it from (Function body still the same)

error_code& operator|=(const error_code &e)

to

error_code operator|=(const error_code &e)

Is there any potential bug that might possible occur? The only difference I see is, it will perform an extra copy operation, other than that, no big deal.

So, should I just stick to return by reference, or it doesn't matter?

---

class error_code {
public:
 error_code() : hi(0), lo(0) {}
 error_code(__int64 lo) : hi(0), lo(lo) {}
 error_code(__int64 hi, __int64 lo) : hi(hi), lo(lo) {}

 // How about return by copy?
 error_code& operator|=(const error_code &e) {
  this->hi |= e.hi;
  this->lo |= e.lo;
  return *this;
 }

 __int64 hi;
 __int64 lo;
};

error_code operator|(const error_code& e0, const error_code& e1) {
 return error_code(e0.hi | e1.hi, e0.lo | e1.lo); 
}

int main() {
 error_code e0(1);
 error_code e1(2);
 e0 |= e1;
}

---

Is there any potential bug that might possible occur?

Yes. This won't work as expected anymore:

(ec |= x) = y;

Now, this is a silly piece of code, no doubt about that, but

1. it does work for other types and
2. there might be other cases where this difference matters.

For example, if your class error_code will have member functions modifying the object they are invoked for, then these will modify a temporary object if you return by copy instead of reference:

(ec |= x).normalize();  // whatever "normalizing" error code means...

---

It would mean that anything you then do with the returned object would not affect the original object any more. And the other way round - the returned object would never reflect changes done to the original object. This is a typical issue of value vs identity. You are keeping a copy of the object's value, but you're losing its identity.