Calling protected base class function on other (same typed) object

Discussion in 'C++' started by Rick, Feb 6, 2007.

  1. Rick

    Rick Guest

    Hi,

    Can anyone explain to me why the below fails to compile - seeing
    otherA->f(); as a call to a inaccessible function, while otherB->f();
    is ok?

    It seems you can happily access protected functions of another (same
    type) - but not via a base class pointer.... I've checked the FAQs,
    Meyers etc but nothing obvious I can find explains it.

    I wanted to implement a recursion down a tree (via parent/child base
    class pointers) and keep the virtual functions protected - but this
    behaviour seems to force me to make them public.

    Any info would be very helpful.

    Thanks

    Rick


    class A
    {
    protected:
    virtual void f() const {};
    };

    class B : public A
    {
    protected:
    virtual void f() const
    {
    A::f(); // OK

    // OK
    B * otherB = new B();
    otherB->f();

    // Not OK - f() protected....
    A * otherA = new B();
    otherA->f(); };
    };
    Rick, Feb 6, 2007
    #1
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  2. Rick wrote:
    > Can anyone explain to me why the below fails to compile - seeing
    > otherA->f(); as a call to a inaccessible function, while otherB->f();
    > is ok?
    >
    > It seems you can happily access protected functions of another (same
    > type) - but not via a base class pointer.... I've checked the FAQs,
    > Meyers etc but nothing obvious I can find explains it.


    Access to protected members is allowed only through the object of
    the same type. It's a limitation imposed by the language.

    > I wanted to implement a recursion down a tree (via parent/child base
    > class pointers) and keep the virtual functions protected - but this
    > behaviour seems to force me to make them public.


    Consider the Visitor pattern. Keep virtual functions private and
    have a public non-virtual function in the base class which will call
    your virtual function. Or, make the visiting type a friend and keep
    everything private.

    > [..]


    V
    --
    Please remove capital 'A's when replying by e-mail
    I do not respond to top-posted replies, please don't ask
    Victor Bazarov, Feb 6, 2007
    #2
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  3. Rick

    Rick Guest

    On Feb 6, 3:56 pm, "Victor Bazarov" <> wrote:
    > Rick wrote:
    > > Can anyone explain to me why the below fails to compile - seeing
    > > otherA->f(); as a call to a inaccessible function, while otherB->f();
    > > is ok?

    >
    > > It seems you can happily access protected functions of another (same
    > > type) - but not via a base class pointer.... I've checked the FAQs,
    > > Meyers etc but nothing obvious I can find explains it.

    >
    > Access to protected members is allowed only through the object of
    > the same type. It's a limitation imposed by the language.
    >

    I guess this is a static versus dynamic binding issue? The access
    specifiers having to be checked at compile time...?

    > > I wanted to implement a recursion down a tree (via parent/child base
    > > class pointers) and keep the virtual functions protected - but this
    > > behaviour seems to force me to make them public.

    >
    > Consider the Visitor pattern. Keep virtual functions private and
    > have a public non-virtual function in the base class which will call
    > your virtual function. Or, make the visiting type a friend and keep
    > everything private.


    Looks like a good solution.

    Thanks for the advice.

    Rick
    Rick, Feb 6, 2007
    #3
  4. Rick wrote:
    > On Feb 6, 3:56 pm, "Victor Bazarov" <> wrote:
    >> Rick wrote:
    >>> Can anyone explain to me why the below fails to compile - seeing
    >>> otherA->f(); as a call to a inaccessible function, while
    >>> otherB->f(); is ok?

    >>
    >>> It seems you can happily access protected functions of another (same
    >>> type) - but not via a base class pointer.... I've checked the FAQs,
    >>> Meyers etc but nothing obvious I can find explains it.

    >>
    >> Access to protected members is allowed only through the object of
    >> the same type. It's a limitation imposed by the language.
    >>

    > I guess this is a static versus dynamic binding issue? The access
    > specifiers having to be checked at compile time...?


    I know that the rule of allowing access to protected functions only
    through an object of the same type stems from the fact that you're
    not supposed to be able to use it across a hierarchy. Example:

    class A {
    protected:
    void foo();
    };

    class B : public A {
    public:
    void bar(B& b) {
    b.foo(); // OK
    }
    };

    class C : public A {
    public:
    void bar(A& a) {
    a.foo(); // NOT OK! Line 42
    }
    };

    int main() {
    B b;
    C c;
    c.bar(b);
    }

    If a call on line 42 were allowed, then you could call a protected
    member function for an object (b) which has no relation to you (C).
    I am not sure why it is bad, nothing immediately comes to mind.

    [..]

    V
    --
    Please remove capital 'A's when replying by e-mail
    I do not respond to top-posted replies, please don't ask
    Victor Bazarov, Feb 7, 2007
    #4
  5. * Victor Bazarov:
    > I am not sure why it is bad, nothing immediately comes to mind.


    If it were allowed you could access any protected member of any class
    simply by deriving from that class.

    As it is, you need at least a cast when using the derive-from-it method.

    --
    A: Because it messes up the order in which people normally read text.
    Q: Why is it such a bad thing?
    A: Top-posting.
    Q: What is the most annoying thing on usenet and in e-mail?
    Alf P. Steinbach, Feb 7, 2007
    #5
  6. Rick

    Grizlyk Guest

    Rick wrote:
    >>
    >> > It seems you can happily access protected functions of another (same
    >> > type) - but not via a base class pointer.... I've checked the FAQs,
    >> > Meyers etc but nothing obvious I can find explains it.

    >>
    >> Access to protected members is allowed only through the object of
    >> the same type. It's a limitation imposed by the language.
    >>

    > I guess this is a static versus dynamic binding issue? The access
    > specifiers having to be checked at compile time...?


    The "protected" keyword allows to access to base class members of own
    object, not base class members of other objects. Other objects of the same
    class is exception due to copy and assign wants the access, but the
    protected access to members of other objects of the same class is better to
    use only for copy and assign purpose.

    If you want to access to protected for other objects, use "friend".

    --
    Maksim A. Polyanin

    "In thi world of fairy tales rolls are liked olso"
    /Gnume/
    Grizlyk, Feb 10, 2007
    #6
  7. Rick

    Grizlyk Guest

    Grizlyk wrote:
    >>>
    >>> > It seems you can happily access protected functions of another (same
    >>> > type) - but not via a base class pointer.... I've checked the FAQs,
    >>> > Meyers etc but nothing obvious I can find explains it.
    >>>
    >>> Access to protected members is allowed only through the object of
    >>> the same type. It's a limitation imposed by the language.
    >>>

    >> I guess this is a static versus dynamic binding issue? The access
    >> specifiers having to be checked at compile time...?

    >
    > The "protected" keyword allows to access to base class members of own
    > object, not base class members of other objects. Other objects of the same
    > class is exception due to copy and assign wants the access, but the
    > protected access to members of other objects of the same class is better
    > to use only for copy and assign purpose.
    >
    > If you want to access to protected for other objects, use "friend".


    I have recallected how can access to other object. In the case of two
    non-inherited different classes we can declare one as frient to other

    class B;
    class A{ friend class B; int i; };

    but derived from other (derived from from B) will lose friend acssess, so
    friend is not inherited, and if we wants to get access to one from derived
    from other, we need to declare the access interface in other (in B).

    class B
    {
    protected:
    int& i(A& a)const {return a.i;}
    };

    class derived_B: public B
    {
    protected:
    print(cosnt int);

    public:
    print(A& a){print( i(a) );}
    };

    By analogy, we need to declare in base forwarding interface to access from
    derived from base to other base

    template<class T>
    class A
    {
    //logical protected interface
    protected:
    T member;

    //protected access interface
    //to logical protected interface
    protected:
    T& get_member(A& a)const{return a.member;}
    };

    template<class T>
    class B: public A<T>
    {
    public:

    void test()
    {
    // Not OK - f() protected....
    //A * otherA = new B();
    //otherA->f(); };

    A<T> *otherA = new B;
    T tmp( get_member(*otherA) );
    }
    };

    --
    Maksim A. Polyanin

    "In thi world of fairy tales rolls are liked olso"
    /Gnume/
    Grizlyk, Feb 11, 2007
    #7
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