Accessing protected members in a derived class

0 votes
asked Jul 14, 2010 by miked

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.

5 Answers

0 votes
answered Jul 14, 2010 by slaks

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.

0 votes
answered Jul 14, 2010 by james-curran

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)

0 votes
answered Jul 14, 2010 by sje397

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;
    }
};
0 votes
answered Jul 23, 2015 by user5595410

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;
    }
};
0 votes
answered Sep 15, 2017 by sergei

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.

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