如何防止赋值运算符左侧的临时

Question

If I define operator+ for a type, in the usual fashion

struct S {};

S operator+(S const &, S const &) { 
  return {}; 
}

users of S can write code like

S s{};
s + s = S{}; // huh

From what I can tell, operator+ returns a temporary object of type S , which is then assigned to. The object then dies at the end of the statement, because there's no name for it, and so the statement is effectively a no-op.

I don't see any use for code like that, so I would like to make that a compile error. Is there a way to do that? Even a warning would be better than nothing.

Answer 1

One simple way to prevent this from happening is to return a constant object:

const S operator+(S const &, S const &) { 
    return {}; 
}

That will now result in a compilation error, in this situation, but

s= s + s;

will still work just fine.

This, of course, has a few other ramifications, and may or may not have undesirable side-effects, which may or may not pose an issue, but that would be a new question, here...

Answer 2

Just found what I need here . Apparently I can make the assignment operator only bind to lvalues.

struct S {
    S& operator=(S const&) & // only binds to lvalues
    { 
      return *this; 
    }
}; 

Now I get exactly the error I want.