Why overloaded operators cannot be defined as static members of a class?

Question

C++ syntax allows defining overloaded operators either inside the struct/class like:

struct X
{
   void operator+(X);
}

or outside of the struct/class like:

void operator+(X, X);

but not as

struct X
{
   static void operator+(X, X);
}

Does any body know reasons for this decision? Why the third form is not allowed? (MSVC gives a syntax error). Maybe there is some story behind this?

p.s. Presence of the first and the second definitions at the same time creates ambiguity:

1>CppTest1.cxx
1>c:\ballerup\misc\cf_html\cpptest1.cxx(39) : error C2593: 'operator +' is ambiguous
1>        c:\ballerup\misc\cf_html\cpptest1.cxx(13): could be 'void B1::operator +(B1 &)'
1>        c:\ballerup\misc\cf_html\cpptest1.cxx(16): or       'void operator +(B1 &,B1 &)'
1>        while trying to match the argument list '(B1, B1)'

I do not understand why this ambiguity is any better than between 1,3 or 2,3.

Answer 1

I have no specific knowledge of any C++ discussion of this concept, so feel free to ignore this.

But to me, you've got the question backwards. The question should be, "why would this syntax be allowed?"

It provides no advantages at all over the current syntax. The non-static member function version has the same access to private members as your proposed static version. So if you need to access the privates to implement it, just make it a non-static member, exactly as you generally do with most members of a class.

It doesn't make it easier to implement asymmetric operators (ie: operator+(const X &x, const Y &y)). If you need private access to implement this, you'd still need a friend declaration for them in one of the classes.

So I would say that the reason it doesn't exist is that it isn't necessary. Between non-member functions and non-static members, all of the necessary use cases are covered.

---

Or, to put it another way:

Free functions can do everything that the static function system can, and more.

Through the use of free functions, you can get argument-dependent lookup happening for operators used in templates. You can't do that with static functions, because those would have to be a member of a particular class. And you cannot add to a class from outside of the class, while you can add to a namespace. So if you need to put an operator in a particular namespace in order to make some ADL code work, you can. You can't do that with static function operators.

Thus, free functions are a superset of everything that your proposed static function system would provide. Since there is no benefit to allowing it, there is no reason to allow it, and therefore it is not allowed.

---

which would make possible to use functors without instantiating them?

That is a contradiction in terms. A "functor" is a "function object". A type is not an object; therefore, it cannot be a functor. It can be a type that, when instantiated, will result in a functor. But the type alone will not be a functor.

Furthermore, being able to declare Typename::operator() static would not mean that Typename() would do what you want. That syntax already has an actual meaning: instantiate a Typename temporary by calling the default constructor.

Lastly, even if all that weren't the case, what good would that actually be? Most template functions that take a callable of some type work just as well with a function pointer as with a functor. Why would you want to restrict your interface, not merely to just functors, but to functors which cannot have internal data? That means you wouldn't be able to pass capturing lambdas and so forth.

What good is a functor that cannot possibly contain state? Why do you want to force the user into passing "functors" that don't have state? And why do you want to prevent the user from being able to use lambdas?

So your question is derived from a false assumption: even if we had it, it wouldn't give you what you want.