accessing a protected member of a base class in an

问题:

Why does this compile:

class FooBase
{
protected:
    void fooBase(void);
};

class Foo : public FooBase
{
public:
    void foo(Foo& fooBar)
    {
        fooBar.fooBase();
    }
};

but this does not?

class FooBase
{
protected:
    void fooBase(void);
};

class Foo : public FooBase
{
public:
    void foo(FooBase& fooBar)
    {
        fooBar.fooBase();
    }
};

On the one hand C++ grants access to private/protected members for all instances of that class, but on the other hand it does not grant access to protected members of a base class for all instances of a subclass. This looks rather inconsistent to me.

I have tested compiling with VC++ and with ideone.com and both compile the first but not the second code snippet.

回答1:

When foo receives a FooBase reference, the compiler doesn't know whether the argument is a descendant of Foo, so it has to assume it's not. Foo has access to inherited protected members of other Foo objects, not all other sibling classes.

Consider this code:

class FooSibling: public FooBase { };

FooSibling sib;
Foo f;
f.foo(sib); // calls sib.fooBase()!?

If Foo::foo can call protected members of arbitrary FooBase descendants, then it can call the protected method of FooSibling, which has no direct relationship to Foo. That's not how protected access is supposed to work.

If Foo needs access to protected members of all FooBase objects, not just those that are also known to be Foo descendants, then Foo needs to be a friend of FooBase:

class FooBase
{
protected:
  void fooBase(void);
  friend class Foo;
};

回答2:

The C++ FAQ summarizes this issue nicely:

[You] are allowed to pick your own pockets, but you are not allowed to pick your father's pockets nor your brother's pockets.

回答3:

The key point is that protected grants you access to your own copy of the member, not to those members in any other object. This is a common misconception, as more often than not we generalize and state protected grants access to the member to the derived type (without explicitly stating that only to their own bases...)

Now, that is for a reason, and in general you should not access the member in a different branch of the hierarchy, as you might break the invariants on which other objects depend. Consider a type that performs an expensive calculation on some large data member (protected) and two derived types that caches the result following different strategies:

class base {
protected:
   LargeData data;
// ...
public:
   virtual int result() const;      // expensive calculation
   virtual void modify();           // modifies data
};
class cache_on_read : base {
private:
   mutable bool cached;
   mutable int cache_value;
// ...
   virtual int result() const {
       if (cached) return cache_value;
       cache_value = base::result();
       cached = true;
   }
   virtual void modify() {
       cached = false;
       base::modify();
   }
};
class cache_on_write : base {
   int result_value;
   virtual int result() const {
      return result_value;
   }
   virtual void modify() {
      base::modify();
      result_value = base::result(); 
   }
};

The cache_on_read type captures modifications to the data and marks the result as invalid, so that the next read of the value recalculates. This is a good approach if the number of writes is relatively high, as we only perform the calculation on demand (i.e. multiple modifies will not trigger recalculations). The cache_on_write precalculates the result upfront, which might be a good strategy if the number of writes is small, and you want deterministic costs for the read (think low latency on reads).

Now, back to the original problem. Both cache strategies maintain a stricter set of invariants than the base. In the first case, the extra invariant is that cached is true only if data has not been modified after the last read. In the second case, the extra invariant is that result_value is the value of the operation at all times.

If a third derived type took a reference to a base and accessed data to write (if protected allowed it to), then it would break with the invariants of the derived types.

That being said, the specification of the language is broken (personal opinion) as it leaves a backdoor to achieve that particular result. In particular, if you create a pointer to member of a member from a base in a derived type, access is checked in derived, but the returned pointer is a pointer to member of base, which can be applied to any base object:

class base {
protected:
   int x;
};
struct derived : base {
   static void modify( base& b ) {
      // b.x = 5;                        // error!
      b.*(&derived::x) = 5;              // allowed ?!?!?!
   }
}

回答4:

In both examples Foo inherits a protected method fooBase. However, in your first example you try to access the given protected method from the same class (Foo::foo calls Foo::fooBase), while in the second example you try to access a protected method from another class which isn't declared as friend class (Foo::foo tries to call FooBase::fooBase, which fails, the later is protected).

回答5:

In the first example you pass an object of type Foo, which obviously inherits the method fooBase() and so is able to call it. In the second example you are trying to call a protected function, simply so, regardless in which context you can't call a protected function from a class instance where its declared so. In the first example you inherit the protected method fooBase, and so you have the right to call it WITHIN Foo context

回答6:

I tend to see things in terms of concepts and messages. If your FooBase method was actually called "SendMessage" and Foo was "EnglishSpeakingPerson" and FooBase was SpeakingPerson, your protected declaration is intended to restrict SendMessage to between EnglishSpeakingPersons (and subclasses eg: AmericanEnglishSpeakingPerson, AustralianEnglishSpeakingPerson) . Another type FrenchSpeakingPerson derived from SpeakingPerson would not be able to receive a SendMessage, unless you declared the FrenchSpeakingPerson as a friend, where 'friend' meant that the FrenchSpeakingPerson has a special ability to receive SendMessage from EnglishSpeakingPerson (ie can understand English).

回答7:

In addition to hobo's answer you may seek a workaround.

If you want the subclasses to want to call the fooBase method you can make it static. static protected methods are accessible by subclasses with all arguments.