Accessing protected members in a derived class

问题:

I ran into an error yesterday and, while it\'s easy to get around, I wanted to make sure that I\'m understanding C++ right.

I have a base class with a protected member:

class Base
{
  protected:
    int b;
  public:
    void DoSomething(const Base& that)
    {
      b+=that.b;
    }
};

This compiles and works just fine. Now I extend Base but still want to use b:

class Derived : public Base
{
  protected:
    int d;
  public:
    void DoSomething(const Base& that)
    {
      b+=that.b;
      d=0;
    }
};

Note that in this case DoSomething is still taking a reference to a Base, not Derived. I would expect that I can still have access to that.b inside of Derived, but I get a cannot access protected member error (MSVC 8.0 - haven\'t tried gcc yet).

Obviously, adding a public getter on b solved the problem, but I was wondering why I couldn\'t have access directly to b. I though that when you use public inheritance the protected variables are still visible to the derived class.

回答1:

You can only access protected members in instances of your type (or derived from your type).
You cannot access protected members of an instance of a parent or cousin type.

In your case, the Derived class can only access the b member of a Derived instance, not of a different Base instance.

Changing the constructor to take a Derived instance will also solve the problem.

回答2:

You have access to the protected members of Derived, but not those of Base (even if the only reason it\'s a protected member of Derived is because it\'s inherited from Base)

回答3:

As mentioned, it\'s just the way the language works.

Another solution is to exploit the inheritance and pass to the parent method:

class Derived : public Base
{
  protected:
    int d;
  public:
    void DoSomething(const Base& that)
    {
      Base::DoSomething(that);
      d=0;
    }
};

回答4:

protected members can be accessed:

• through this pointer
• or to the same type protected members even if declared in base
• or from friend classes, functions

To solve your case you can use one of last two options.

Accept Derived in Derived::DoSomething or declare Derived friend to Base:

class Derived;

class Base
{
  friend class Derived;
  protected:
    int b;
  public:
    void DoSomething(const Base& that)
    {
      b+=that.b;
    }
};

class Derived : public Base
{
  protected:
    int d;
  public:
    void DoSomething(const Base& that)
    {
      b+=that.b;
      d=0;
    }
};

You may also consider public getters in some cases.

回答5:

class Derived : public Base
{
  protected:
    int d;
  public:
    void DoSomething()
    {
      b+=this->b;
      d=0;
    }
};

//this will work

回答6:

You can try with static_cast< const Derived*>(pBase)->protected_member ...

class Base
{
  protected:
    int b;

  public:
    ...
};

class Derived : public Base
{
  protected:
    int d;

  public:
    void DoSomething(const Base* that)
    {
      b += static_cast<const Derived*>(that)->b;
      d=0;
    }
};

回答7:

Use this pointer to access protected members

class Derived : public Base
{
  protected:
    int d;
  public:
    void DoSomething(const Base& that)
    {
      this->b+=that.b;
      d=0;
    }
};