类成员引用变量是否具有内置的“ const-correctness”？

Question

struct A {
  int &r; 
  A (int &i) : r(i) {}
  void foo () const {
    r = 5;  // <--- ok
  }
};

The compiler doesn't generate any error at r = 5; .
Does it mean that &r is already const-correct being a reference (logical equivalent of int* const ) ? [Here is one related question.]

Answer 1

I'm not sure exactly what you mean by "already const-correct", but:

Assigning to r is the same as assigning to whatever thing was passed into the constructor of A . You're not modifying anything in the instance of A when you do this, so the fact that foo is declared const isn't an obstacle. It's very much as if you'd done this:

struct A {
  int * r;
  A (int * i) : r(i) {}
  void foo () const { *r = 5; }
}

The fact that foo is const means that it doesn't modify anything in the A instance it's called on. There's no conflict between that and having it modify other data it was supplied with.

Of course if you happened to arrange for r to be a reference to some member of A then calling foo would modify the instance of A after all. The compiler can't catch all possible ways in which const ness of a member function might be violated; when you declare a member function const you're promising that it doesn't engage in any such subterfuge.

Answer 2

Yes, it's the logical equivalent of int* const .

You may want to create and use appropriately qualified accessors in this case to prevent unwanted alterations to the value r references.

Answer 3

I interpret a const member function as implicitly inserting const just to the left of every data member that doesn't already have such a qualifier. That const is already there implicitly for references ( int & const r; is illegal syntax). In other words, references are "already const-correct" to use your nomenclature.

It would be nice if the const qualifier on a member function had the affect of inserting const in every possible valid position for every data member (eg, data member int ** foo; acts like int const * const * const foo; in a const member function), but that isn't what happens, and it isn't what the standard says will happen.