Why can derived member function not access protected member of a baseclass object?

[blangela]

If I pass a base class object by reference (likely does not make a
difference here that it is passed by reference) as a parameter to a
derived class member function, the member function is not allowed to
access the protected data members of the base object. This surprises
me.

Can someone explain why this is? I suspect there is a good reason and
I am just having a slow day to not come up with it myself.

Bob

[blangela]

Here is a sample program to show what I mean:

// Base.h
#ifndef Base_incl
#define Base_incl

class Derived; // forward declaration - required because Method2()
// of the Base class references the Derived class
class Base
{
public:
Base(); // constructor

virtual void Method1(Base &);
virtual void Method2(Base &);
virtual void Method3(Derived &); //it is unusual to have a
// derived class object as a parameter for a base
// class method, but as you see, it can be done.
protected:
int mBase;
};

#endif

// Derived.h
#ifndef Derived_incl
#define Derived_incl

#include "Base.h"

class Derived: public Base
{
public:
Derived(); // constructor

virtual void Method2(Base &); // this method
// overrides Base class Method2()
virtual void Method4(Base &); // a new method
// which is not defined for the Base class

private:
int mDerived;
};

#endif

// Base.cpp
#include "Base.h"
#include "Derived.h"

#include <iostream>
using std::cout;
using std::endl;

Base::Base()
{
this->mBase = 2;
}

void Base::Method1(Base & B_Param)
{
cout << (B_Param.mBase) * (this->mBase) << endl;
}

void Base::Method2(Base & B_Param)
{
cout << this->mBase << endl;
}

void Base::Method3(Derived & D_Param)
{
Base & BRef = static_cast<Base &>(D_Param); // This cast creates
// a Base class reference to the D_Param object.
cout << BRef.mBase << endl;
}

//Derived.cpp
#include "Derived.h"
#include <iostream>
using std::cout;
using std::endl;

Derived::Derived()
{
this->mBase = 3;
this->mDerived = 5;
}

void Derived::Method2(Base & B_Param)
{
cout << this->mDerived << endl;
}

void Derived::Method4(Base & B_Param)
{
Derived * DPtr = dynamic_cast <Derived *(&B_Param); // this cast
// creates a Derived class pointer to B_Param, if and only if,
// B_Param actually is Derived class object, otherwise the
// pointer will be set to 0

if (DPtr != 0) // if B_Param is actually a Derived class object
cout << DPtr->mDerived << endl;
else
cout << B_Param.mBase << endl;
}

If I compile Derived.cpp I get the following errors:

1>c:\documents and settings\blangela\desktop\polymorphismtest
\derived.cpp(27) : error C2248: 'Base::mBase' : cannot access
protected member declared in class 'Base'
1 c:\documents and settings\blangela\desktop\polymorphismtest
\base.h(17) : see declaration of 'Base::mBase'
1 c:\documents and settings\blangela\desktop\polymorphismtest
\base.h(7) : see declaration of 'Base'

[blangela]

Sorry, it is the line:

cout << B_Param.mBase << endl;

in the member function below that causes the errors.

void Derived::Method4(Base & B_Param)
{
Derived * DPtr = dynamic_cast <Derived *(&B_Param); // this
cast
// creates a Derived class pointer to B_Param, if and
only if,
// B_Param actually is Derived class object, otherwise
the
// pointer will be set to 0

if (DPtr != 0) // if B_Param is actually a Derived class
object
cout << DPtr->mDerived << endl;
else
cout << B_Param.mBase << endl;
}

[Victor Bazarov]

blangela wrote:

If I pass a base class object by reference (likely does not make a
difference here that it is passed by reference) as a parameter to a
derived class member function, the member function is not allowed to
access the protected data members of the base object. This surprises
me.

Can someone explain why this is? I suspect there is a good reason and
I am just having a slow day to not come up with it myself.

Isn't this in the FAQ? It should be.

In short, the reason is to prevent access to members of a different type
(across the hierarchy). If D1 and D2 derive from B, an instance of D1
is not allowed to access any non-public members of D2 (unless they are
friends, of course). When you pass a reference to B to a member
function of D1, the access to members of that class is blocked because
it *can* be a subobject of a D2 object, and not necessarily of another
D1. Need an example? Look in the archives, we had that topic discussed
several times over the past years.

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

[blangela]

On Sep 25, 12:55*pm, Victor Bazarov <[e-mail]:

blangela wrote:

If I pass a base class object by reference (likely does not make a
difference here that it is passed by reference) as a parameter to a
derived class member function, the member function is not allowed to
access the protected data members of the base object. This surprises
me.

Can someone explain why this is? *I suspect there is a good reason and
I am just having a slow day to not come up with it myself.

Isn't this in the FAQ? *It should be.

In short, the reason is to prevent access to members of a different type
(across the hierarchy). *If D1 and D2 derive from B, an instance of D1
is not allowed to access any non-public members of D2 (unless they are
friends, of course). *When you pass a reference to B to a member
function of D1, the access to members of that class is blocked because
it *can* be a subobject of a D2 object, and not necessarily of another
D1. *Need an example? *Look in the archives, we had that topic discussed
several times over the past years.

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

Even with the example you supply, what would be the harm of allowing
access to a member ( a member that was declared in the B class in your
example above) that we know must exist, no matter what subclass the
object actually belongs to (D1, D2, a subclass of D2, etc.)?

[blangela]

On Sep 25, 1:25*pm, blangela <[e-mail]:

On Sep 25, 12:55*pm, Victor Bazarov <[e-mail]:

blangela wrote:

If I pass a base class object by reference (likely does not make a
difference here that it is passed by reference) as a parameter to a
derived class member function, the member function is not allowed to
access the protected data members of the base object. This surprises
me.

Can someone explain why this is? *I suspect there is a good reason and
I am just having a slow day to not come up with it myself.

Isn't this in the FAQ? *It should be.

In short, the reason is to prevent access to members of a different type
(across the hierarchy). *If D1 and D2 derive from B, an instance of D1
is not allowed to access any non-public members of D2 (unless they are
friends, of course). *When you pass a reference to B to a member
function of D1, the access to members of that class is blocked because
it *can* be a subobject of a D2 object, and not necessarily of another
D1. *Need an example? *Look in the archives, we had that topic discussed
several times over the past years.

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

Even with the example you supply, what would be the harm of allowing
access to a member ( a member that was declared in the B class in your
example above) that we know must exist, no matter what subclass the
object actually belongs to (D1, D2, a subclass of D2, etc.)?- Hide quotedtext -

- Show quoted text -

Also, I changed the member function to:

void Derived::Method4(Base & B_Param)
{
Derived * DPtr = dynamic_cast <Derived *(&B_Param); // this cast
// creates a Derived class pointer to B_Param, if and only if,
// B_Param actually is Derived class object, otherwise the
// pointer will be set to 0

if (DPtr != 0) // if B_Param is actually a Derived class object
cout << DPtr->mDerived << endl;
else
{
Base B_obj = B_Param;
cout << B_obj.mBase << endl;
}
}

And I still have the same errors:

1>c:\documents and settings\blangela\desktop\polymorphismtest
\derived.cpp(29) : error C2248: 'Base::mBase' : cannot access
protected member declared in class 'Base'
1 c:\documents and settings\blangela\desktop\polymorphismtest
\base.h(17) : see declaration of 'Base::mBase'
1 c:\documents and settings\blangela\desktop\polymorphismtest
\base.h(7) : see declaration of 'Base'

I would have thought that B_obj can only be a Base object now (and not
some subclass of Base), that it would now be allowed?

[news.aioe.org]

blangela wrote:

If I pass a base class object by reference (likely does not make a
difference here that it is passed by reference) as a parameter to a
derived class member function, the member function is not allowed to
access the protected data members of the base object. This surprises
me.

Can someone explain why this is? I suspect there is a good reason and
I am just having a slow day to not come up with it myself.

Bob

I know this is a C++ group and I hope I won't offend anyone by saying unlike in
C++ this kind of access is perfectly allowed in Java.

[Jeff Schwab]

news.aioe.org wrote:

blangela wrote:

>If I pass a base class object by reference (likely does not make a
difference here that it is passed by reference) as a parameter to a
derived class member function, the member function is not allowed to
access the protected data members of the base object. This surprises
me.

Can someone explain why this is? I suspect there is a good reason and
I am just having a slow day to not come up with it myself.

I know this is a C++ group and I hope I won't offend anyone by saying
unlike in C++ this kind of access is perfectly allowed in Java.

Er... You sure about that? Is it possible that you just tried a toy
example with two classes in the same (e.g. top-level) package, and
forgot that "protected" in Java grants access to all classes in the same
package?

[news.aioe.org]

Jeff Schwab wrote:

news.aioe.org wrote:

>blangela wrote:

>>If I pass a base class object by reference (likely does not make a
difference here that it is passed by reference) as a parameter to a
derived class member function, the member function is not allowed to
access the protected data members of the base object. This surprises
me.

Can someone explain why this is? I suspect there is a good reason and
I am just having a slow day to not come up with it myself.

>I know this is a C++ group and I hope I won't offend anyone by saying
unlike in C++ this kind of access is perfectly allowed in Java.

Er... You sure about that? Is it possible that you just tried a toy
example with two classes in the same (e.g. top-level) package, and
forgot that "protected" in Java grants access to all classes in the same
package?

Yikes, I am embarrassed. That is exactly what I did! So it is consistent
between the two languages. When I moved the base class to a different package,
compiler throws an error:
x has protected access in package1.Base

Thanks for pointing out my mistake.

[Victor Bazarov]

blangela wrote:

[..]
Also, I changed the member function to:

void Derived::Method4(Base & B_Param)
{
Derived * DPtr = dynamic_cast <Derived *(&B_Param); // this cast
// creates a Derived class pointer to B_Param, if and only if,
// B_Param actually is Derived class object, otherwise the
// pointer will be set to 0

if (DPtr != 0) // if B_Param is actually a Derived class object
cout << DPtr->mDerived << endl;
else
{
Base B_obj = B_Param;
cout << B_obj.mBase << endl;
}
}

And I still have the same errors:

1>c:\documents and settings\blangela\desktop\polymorphismtest
\derived.cpp(29) : error C2248: 'Base::mBase' : cannot access
protected member declared in class 'Base'
1 c:\documents and settings\blangela\desktop\polymorphismtest
\base.h(17) : see declaration of 'Base::mBase'
1 c:\documents and settings\blangela\desktop\polymorphismtest
\base.h(7) : see declaration of 'Base'

I would have thought that B_obj can only be a Base object now (and not
some subclass of Base), that it would now be allowed?

You changes the run-time behaviour of your program, but not the logic
for that particular statement. The rules of the language do not depend
on the run-time behaviour. The compiler is not going to verify the
run-time conditions when it checks the semantics of an expression.
Access is prohibited, period. It does not matter whether the *actual*
type of the super-object is the same or not because in general it isn't.

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

[=?UTF-8?B?RXJpayBXaWtzdHLDtm0=?=]

On 2008-09-25 22:25, blangela wrote:

On Sep 25, 12:55 pm, Victor Bazarov <[e-mail]:

>blangela wrote:

If I pass a base class object by reference (likely does not make a
difference here that it is passed by reference) as a parameter to a
derived class member function, the member function is not allowed to
access the protected data members of the base object. This surprises
me.

Can someone explain why this is? I suspect there is a good reason and
I am just having a slow day to not come up with it myself.

Isn't this in the FAQ? It should be.

In short, the reason is to prevent access to members of a different type
(across the hierarchy). If D1 and D2 derive from B, an instance of D1
is not allowed to access any non-public members of D2 (unless they are
friends, of course). When you pass a reference to B to a member
function of D1, the access to members of that class is blocked because
it *can* be a subobject of a D2 object, and not necessarily of another
D1. Need an example? Look in the archives, we had that topic discussed
several times over the past years.

Please do not quota signatures.

Even with the example you supply, what would be the harm of allowing
access to a member ( a member that was declared in the B class in your
example above) that we know must exist, no matter what subclass the
object actually belongs to (D1, D2, a subclass of D2, etc.)?

Since the members are not public it means that the only means of
changing them should be through the interface (public member functions)
of the class. If arbitrarily could change some member just of a class
just because that class had the same ancestor as you there would be not
way for that class to ensure it does not end up in an invalid state.
Just because the value of a member of the base-class is valid in one
derived class does not mean that it is valid in another derived class.

--
Erik Wikström