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Accessing a protected member of a member of type BaseClass????

P: n/a
I find the surprising. If I derive Rectangle from Point, I can access the
members of Point inherited by Rectangle _IF_ they are actually members of a
Rectangle. If I have a member of type Point in Rectangle, the compiler
tells me Point::x is protected. I would have expected Rectangle to see the
protected members of any Point. Compiling the following code give me this
error:
g++ -o rectangle main.cc
main.cc: In member function `size_t Rectangle::dx()':
main.cc:22: error: `size_t Point::y' is protected
main.cc:32: error: within this context

#include <iostream>
#include <cstddef> //size_t

class Point {

public:
Point(const size_t& x_=0,
const size_t& y_=0)
: x(x_),
y(y_)
{}

Point& operator+(const Point& p)
{
this->x += p.x;
this->y += p.y;
return *this;
}

protected:
size_t x;
size_t y; //line 22
};

class Rectangle : public Point {
public:
Rectangle(const Point& xy, const Point& dxdy_)
: Point(xy),
dxdy(dxdy_)
{}
size_t X() const { return this->x; }
size_t dx() const { return this->dxdy.y; } //line 32
protected:
Point dxdy;
};
int main(){

}

ISO/IEC 14882 says this:

11.2 Accessibility of base classes and base class members

[class.access.base]

"If a class is declared to be a base class (clause 10) for another class
using the public access specifier, the public members of the base class are
accessible as public members of the derived class and protected members of
the base class are accessible as protected members of the derived class."

The way I read that, I should be able to access Rectangle::dxdy.y from
within Rectangle. Am I missing something, or is g++ wrong?
--
STH
Hatton's Law: "There is only One inviolable Law"
KDevelop: http://www.kdevelop.org SuSE: http://www.suse.com
Mozilla: http://www.mozilla.org
Jul 22 '05 #1
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2 Replies


P: n/a
Steven T. Hatton wrote:
I find the surprising. If I derive Rectangle from Point, I can access the

....

This looks like an answer. I'm not accessing Point::dxdy through a pointer
to Rectangle. I would need to get really brutal with a cast to do so, and
I'm not sure that would even work(correctly/reliably).

http://publib.boulder.ibm.com/infoce...ns.cplr162.htm

A protected nonstatic base class member can be accessed by members and
friends of any classes derived from that base class by using one of the
following:

* A pointer to a directly or indirectly derived class
* A reference to a directly or indirectly derived class
* An object of a directly or indirectly derived class

If a class is derived privately from a base class, all protected base class
members become private members of the derived class.

If you reference a protected nonstatic member x of a base class A in a
friend or a member function of a derived class B, you must access x through
a pointer to, reference to, or object of a class derived from A. However,
if you are accessing x to create a pointer to member, you must qualify x
with a nested name specifier that names the derived class B. The following
example demonstrates this:

class A {
public:
protected:
int i;
};
class B : public A {
friend void f(A*, B*);
void g(A*);
};

void f(A* pa, B* pb) {
// pa->i = 1;
pb->i = 2;

// int A::* point_i = &A::i;
int A::* point_i2 = &B::i;
}

void B::g(A* pa) {
// pa->i = 1;
i = 2;

// int A::* point_i = &A::i;
int A::* point_i2 = &B::i;
}

void h(A* pa, B* pb) {
// pa->i = 1;
// pb->i = 2;
}

int main() { }

Class A contains one protected data member, an integer i. Because B derives
from A, the members of B have access to the protected member of A. Function
f() is a friend of class B:

* The compiler would not allow pa->i = 1 because pa is not a pointer to
the derived class B.
* The compiler would not allow int A::* point_i = &A::i because i has
not been qualified with the name of the derived class B.

Function g() is a member function of class B. The previous list of remarks
about which statements the compiler would and would not allow apply for g()
except for the following:

* The compiler allows i = 2 because it is equivalent to this->i = 2.

Function h() cannot access any of the protected members of A because h() is
neither a friend or a member of a derived class of A.
--
STH
Hatton's Law: "There is only One inviolable Law"
KDevelop: http://www.kdevelop.org SuSE: http://www.suse.com
Mozilla: http://www.mozilla.org
Jul 22 '05 #2

P: n/a
On Mon, 16 Aug 2004 07:15:24 -0400, "Steven T. Hatton"
<su******@setidava.kushan.aa> wrote:
I find the surprising. If I derive Rectangle from Point, I can access the
members of Point inherited by Rectangle _IF_ they are actually members of a
Rectangle. If I have a member of type Point in Rectangle, the compiler
tells me Point::x is protected.
Yes, that's correct behaviour, and is required to maintain
encapsulation.
I would have expected Rectangle to see the
protected members of any Point.
But that would allow it to break instances of completely unrelated
classes that just happen to also derive from Point. For example,
perhaps Trapezium derives from point; we don't want Rectangle to be
able to modify the Point subobject of a Trapezium.
ISO/IEC 14882 says this:

11.2 Accessibility of base classes and base class members

[class.access.base]

"If a class is declared to be a base class (clause 10) for another class
using the public access specifier, the public members of the base class are
accessible as public members of the derived class and protected members of
the base class are accessible as protected members of the derived class."

The way I read that, I should be able to access Rectangle::dxdy.y from
within Rectangle. Am I missing something, or is g++ wrong?


See 11.5/1

Tom
Jul 22 '05 #3

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