Re: Three Classes Share Data???

From:
Salt_Peter <pj_hern@yahoo.com>
Newsgroups:
comp.lang.c++
Date:
Sun, 24 Aug 2008 00:05:23 -0700 (PDT)
Message-ID:
<34509317-ceb7-4f18-953e-10f30c84e397@d1g2000hsg.googlegroups.com>
On Aug 23, 11:50 pm, Immortal Nephi <Immortal_Ne...@satx.rr.com>
wrote:

First class is the base class. It has two data: m_Base1 and m_Base2.
Second class and third class are derived classes and they are derived
from first class. m_Base1 and m_Base2 are inherited into two derived
classes.

Second class has its own m_Base1 and m_Base2 and third class does the
same. I am curious. How can second class and third class share the
same m_Base1 and m_Base2?


You have a single instance of a given class.

class Base { ... };
class Derived : Base { ... };
class DerivedAgain : Derived { ... };

DerivedAgain da;
Derived& d = da;

d is a reference to the Derived portion of instance da.
Assuming that both Derived and DerivedAgain have access to Base's
members, whether you used da or d to access those is irrelevent, both
are accesing the same object.

You define second class first and enter data into m_Base1 and
m_Base2. Then, you define third class and how can you modify a
pointer to second class' m_Base1 and m_Base2.

For example:

class Base
{
     Base() {}
     ~Base() {}
     int m_Base1;
     int m_Base2;

};


Your ctors above are private, try...

class Base
{
  int m_Base1;
  int m_Base2;
publc:
  Base() : m_Base1(0), m_Base2(0) { }
  Base(int b1, int b2) : m_Base1(b1)
                         m_Base2(b2) { }
};

class Derive1 : public Base
{
     Derive1() {}


public:
  Derive1() : Base(20,30) { }

     ~Derive1() {}
     void Set()
     {
          m_Base1 = 20;
          m_Base2 = 30;
     }


Set() is not needed, and neither would it be alowed to set Base's
private parts.

  void Print()
  {
    std::cout << m_Base1 << std::endl;
    ...
  }

};

class Derive2 : public Base
{
     Derive2() {}
     ~Derive2() {}
     void Print()
     {
// How can you do this below?
          cout << "Derive1::m_Base1 -> " << Derive1::m_Base1 << endl;
          cout << "Derive1::m_Base2 -> " << Derive1::m_Base2 << endl;
     }


You could pass a reference to a Derived1 instance however i don't see
any logical reason.
Basicly, If you want X to do something, you don't ask Y to do it for
you.

    void Print(const Derived1& r_d1)
    {
      r_dl.Print();
    }

Which brings us back to the idea that what you really want is one
object, if that isn't your case, you have a design problem. Does it
really make sense to instantiate some type Derived1 and then
instantiate another derivative of Base just to print the former's
contents? I don't think so.

};

int main(void)
{
     Derive1 d1;
     Derive2 d2;
     d1.Set();
     d2.Print();

     return 0;

}

Nephi

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