HI,
I have a small query. In virtual inheritence , the function wil be called
basing on the object we assign for the base class pointer. But how variable
address will be resolved. I observed that even variable are getting resolved
basing on the object we assign (like function). But I think there is no virtual
concept for variables. only for functions.
#include<stdio.h>
#include<conio.h>
class base
{
public:
int var1 , var2;
base(int a,int b)
{
var1=a;
var2=b;
printf("\n I am in base. var1 = %d , var2 = %d",var1 , var2);
};
virtual void show()
{
printf("\n I am in base class virtual method . show. My var1 value is
%d , var 2 value is %d",var1 , var2);
};
void disp()
{
printf("\n I am in base class non virtual method. disp. My var1 is %d
, var2 is %d",var1 , var2);
};
};
class derived:public base
{
public:
int var1 , var2;
derived(int a,int b):base(a+1,b+1)
{
var1=a;
var2=b;
printf("\n I m in derived class. My var1 is %d , var 2 is %d ",var1 ,
var2);
};
void show()
{
printf("\n I am in defrived class virtual method . show. My var1 value
is %d , var 2 value is %d",var1 , var2);
}
void disp()
{
printf("\n I am in derived class non virtual method. disp. My var1 is
%d , var2 is %d",var1 , var2);
}
};
void main()
{
base *x = new derived(1,2);
base *y = new base(10,20);
x->show(); /// show here displaying 1,2 instead of base variable values.
But I think there is no virtual concept here. so I think it should display base
class variable values. please clarify my confusion.
printf("\n \n ]\n");
y->show();
}
Thx,
--Gopi
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