Rationale of enforcing some operators to be member

2020-01-25 09:30发布

There are 4 operators in C++ which can be overloaded but cannot be overloaded as freestanding (aka nonmember, standalone) functions. These operators are:

  • operator =
  • operator ()
  • operator ->
  • operator []

This thread explains perfectly well the rationale behind prohibiting operator = to be a nonmember function. Any ideas about the other three?

2条回答
Bombasti
2楼-- · 2020-01-25 09:48

The four operators mentioned in the original posting, =, (), -> and [], must indeed be implemented as non-static member functions (by respectively C++98 §13.5.3/1, §13.5.4/1, §13.5.5/1 and §13.5.6/1).

Bjarne Stroustrup's rationale was, as I recall from earlier debates on the subject, to retain some sanity in the language, i.e. having at least some things you could rely on no matter how much Someone Else has screwed up by defining non-member operators for existing classes.

I'm not sure I completely agree that the restriction really helps with that, but.

EDIT: I consulted Bjarne Stroustrup about this (he's always helpful) but it appears that the apparent inconsistencies of the rules are no more than a case of frozen historical accident. He notes that "It looks worse now than it was then because our rules for lvalues and references have changed since the overloading rules were formulated. I tried to look into this issue again a couple of years ago, but ran out of time before producing a complete proposal."

Cheers & hth.,

PS: "The Design and Evolution of C++" book is great for this kind of question, but unfortunately I don't have it.

查看更多
我欲成王,谁敢阻挡
3楼-- · 2020-01-25 10:02

This thread on comp.std.c++ discusses the question.

Francis Glassborow, who was on the committee, stated:

The language designers did not want to support conversions and promotions on the left-hand operand of operator =, nor such on the operand of () and [].

Trying to avoid the situation where:

class A {};

class B { B(A& a) {} };

int operator()(B const& b) { return 0; }


int main(void)
{
    A a;

    // This works even though A doesn't have a () operator
    // It creates a temporary B and calls operator()(B& b)
    return a();                   
}
查看更多
登录 后发表回答