how to use private inheritance

[zhangyefei.yefei]

i read book <effective c++>,it tell me that public inheritance means
is-a ,and private inheritance means is-implemented-in-terms-of.
but today i am puzzled by some strange codes.

the following program can not pass compiling , bailing :
g++ d.cpp -o d
d.cpp: In function `int main()':
d.cpp:27: error: `a' is an inaccessible base of `b'

it is obviously okay to understand,because private inheritance is-
implemented-in-terms-of.

#include <iostream>
using namespace std;

class a
{
public:
virtual void doit() {cout<<"a\n";};
};

class b: private a
{
public:
void doit() {cout<<"b\n";}

};
class c
{
public:
void set(a * pa) { m_a =pa;m_a->doit();};
a * m_a;

};

int main ()
{
c cc;
cc.set(new b);
return 0;
}



but when i change source code sightly ,still with private
inheritance, everything is ok,this surprise me.

#include <iostream>
using namespace std;

class a
{
public:
virtual void doit() {cout<<"a\n";};
};

class c
{
public:
void set(a * pa) { m_a =pa;m_a->doit();};
a * m_a;

};

class b: a
{
public:
void doit() {cout<<"b\n";}
void go() { c cc;cc.set(this);};

};
int main ()
{
b bb;
bb.go();
return 0;
}

the above two program seem same to me,but the results arte complete
different.
why ? can anyone do me a favor of giving any hints ?
thanks.

 

[guxiaozhu1]

i read book <effective  c++>,it tell me that public inheritance means
is-a  ,and  private inheritance means is-implemented-in-terms-of.
but today i am puzzled by some strange codes.

the following program can not pass compiling , bailing :
g++     d.cpp   -o d
d.cpp: In function `int main()':
d.cpp:27: error: `a' is an inaccessible base of `b'

it is obviously okay to understand,because  private inheritance is-
implemented-in-terms-of.

#include <iostream>
using namespace std;

class a
{
public:
        virtual void doit() {cout<<"a\n";};

};

class b: private a
{
public:
        void doit() {cout<<"b\n";}

};

class c
{
public:
        void set(a * pa) { m_a =pa;m_a->doit();};
        a * m_a;

};

int main ()
{
        c cc;
        cc.set(new b);
  return 0;

}

but when  i change source code sightly ,still with  private
inheritance, everything is ok,this surprise me.

#include <iostream>
using namespace std;

class a
{
public:
        virtual void doit() {cout<<"a\n";};

};

class c
{
public:
        void set(a * pa) { m_a =pa;m_a->doit();};
        a * m_a;

};

class b: a
{
public:
        void doit() {cout<<"b\n";}
        void go() {  c cc;cc.set(this);};

};

int main ()
{
        b bb;
        bb.go();
  return 0;

}

the above two program seem  same to me,but the results arte complete
different.
why ? can anyone do me a favor of  giving  any hints ?
thanks.

For private inheritage, all public and protected members of base class
automatically become private members of derived class.

Compile code below, you'll get same result.
class A
{
public:
A()
{
}
virtual void Test2(){}
};

class B
{
public:
B()
{
}
private:
void Test2()
{
}
};
int main( )
{

A* a = new B;
a->Test2();
return 0;
}

 

[prasadshetty]

i read book <effective  c++>,it tell me that public inheritance means
is-a  ,and  private inheritance means is-implemented-in-terms-of.
but today i am puzzled by some strange codes.

the following program can not pass compiling , bailing :
g++     d.cpp   -o d
d.cpp: In function `int main()':
d.cpp:27: error: `a' is an inaccessible base of `b'

it is obviously okay to understand,because  private inheritance is-
implemented-in-terms-of.

#include <iostream>
using namespace std;

class a
{
public:
        virtual void doit() {cout<<"a\n";};

};

class b: private a
{
public:
        void doit() {cout<<"b\n";}

};

class c
{
public:
        void set(a * pa) { m_a =pa;m_a->doit();};
        a * m_a;

};

int main ()
{
        c cc;
        cc.set(new b);
  return 0;

}

but when  i change source code sightly ,still with  private
inheritance, everything is ok,this surprise me.

#include <iostream>
using namespace std;

class a
{
public:
        virtual void doit() {cout<<"a\n";};

};

class c
{
public:
        void set(a * pa) { m_a =pa;m_a->doit();};
        a * m_a;

};

class b: a
{
public:
        void doit() {cout<<"b\n";}
        void go() {  c cc;cc.set(this);};

};

int main ()
{
        b bb;
        bb.go();
  return 0;

}

the above two program seem  same to me,but the results arte complete
different.
why ? can anyone do me a favor of  giving  any hints ?
thanks.

In the first program you are
1. Creating an object of type b.
2. Using type b to access function of type a via the inheritance .
Now since a has been privately inherited the compiler does not
allow you to do so.

In the second program you are
1. Creating an object of type b.
2. Calling a public member of class b (i.e. go())
3. In member go() you have created an object of type c (i.e. cc) and
calling member set() for this object of cc.
4. When the object of cc is created it already has a pointer to type a
and the compiler will copy the contents of
object b (i.e. formal argument of set()) to the actual argument
(i.e pointer to a).
Here you are not trying to access the contents of a via the
inheritance and hence the compiler is not complaining.

I hope my explanation was clear.

Regards,
Prasad

 

[Markus Moll]

Hi

#include <iostream>
using namespace std;

class a
{
public:
virtual void doit() {cout<<"a\n";};
};

class b: private a
{
public:
void doit() {cout<<"b\n";}

};
class c
{
public:
void set(a * pa) { m_a =pa;m_a->doit();};
a * m_a;

};

int main ()
{
c cc;
cc.set(new b);
return 0;
}



but when i change source code sightly ,still with private
inheritance, everything is ok,this surprise me.
[...]

class b: a
{
public:
void doit() {cout<<"b\n";}
void go() { c cc;cc.set(this);};

};
[...]

the above two program seem same to me,but the results arte complete
different.
why ? can anyone do me a favor of giving any hints ?

The question is who is converting the pointer to b to a pointer to a.
In the first program, the function main is trying to do so, in the second,
it's a member function of b. Just like private members can only be accessed
by members of the same class, private base-classes can only be accessed by
members of the class. This means that any member of b is allowed to convert
a pointer to b to a pointer to a.

Markus

 

[zhangyefei.yefei]

Hi

#include <iostream>
using namespace std;

class a
{
public:
        virtual void doit() {cout<<"a\n";};
};

class b: private a
{
public:
        void doit() {cout<<"b\n";}

};
class c
{
public:
        void set(a * pa) { m_a =pa;m_a->doit();};
        a * m_a;

int main ()
{
        c cc;
        cc.set(new b);
  return 0;
}

but when  i change source code sightly ,still with  private
inheritance, everything is ok,this surprise me.
[...]

class b: a
{
public:
        void doit() {cout<<"b\n";}
        void go() {  c cc;cc.set(this);};

};

[...]

the above two program seem  same to me,but the results arte complete
different.
why ? can anyone do me a favor of  giving  any hints ?

The question is who is converting the pointer to b to a pointer to a.
In the first program, the function main is trying to do so, in the second,
it's a member function of b. Just like private members can only be accessed
by members of the same class, private base-classes can only be accessed by
members of the class. This means that any member of b is allowed to convert
a pointer to b to a pointer to a.

Markus

it seems to be best answer resolving questions about this usage of
private inheritance .
my last question is , who is converting the pointer to b to a pointer
to a ? i thinks it is compiler doing this ,is not it ?

thanks.