Re: Alternative to Abtract Class?

From:
Victor Bazarov <v.Abazarov@comAcast.net>
Newsgroups:
comp.lang.c++
Date:
Thu, 16 Apr 2009 15:49:40 -0400
Message-ID:
<gs824q$588$1@news.datemas.de>
Immortal Nephi wrote:

I want to know. Is first version of abtract class the same as second
version of class with protected constructor() and destructor()?


Of course not. An abstract class cannot be instantiated, no matter
where you are. A class with protected constructor/destructor can still
be instantiated (perhaps by mistake) in a member function of itself or
any of its descendants.

 > Here

is an example. You can't use class A so class B is used to derive
from class A. Also, you can't use class A2 with protected constructor
() so class B2 is used to derive from class A2.

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

class A
{
public:
    A() : m_x(0) { cout << "Constructor A" << endl; }
    ~A() { cout << "Destructor A" << endl; }

    void set(int x) { m_x = x; }
    int get() { return m_x; }

    virtual void print() = 0;

private:
    int m_x;
};

class B : public A
{
public:
    B() : A() { cout << "Constructor B" << endl; }
    ~B() { cout << "Constructor B" << endl; }

    void print() { cout << "m_x: " << get() << endl; }
};

class A2
{
protected:
    A2() : m_x(0) { cout << "Constructor A2" << endl; }
    ~A2() { cout << "Destructor A2" << endl; }

    int m_x;
};

class B2 : public A2
{
public:
    B2() : A2() { cout << "Constructor B2" << endl; }
    ~B2() { cout << "Destructor B2" << endl; }

    void set(int x) { A2::m_x = x; }
    int get() { return m_x; }

    void print() { cout << "m_x: " << get() << endl; }
};

int main5()
{
// Compilation Error because of abtract class
// A a;
// a.set( 5 );
// a.print();

    B b;
    b.set( 5 );
    b.print();

// Compilation Error because of protected constructor
// A2 a2;

    B2 b2;
    b2.set( 10 );
    b2.print();

    return 0;
}

Thanks....


V
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