Abstraction of C++

[vineethkumart]

What abstraction C++ gives??

Please see the danger of initializing an object with another of the same class

class A
{
int m,*p;
public:
A(){m=12;p=&m;}
inline void print(){std::cout<<m<<*p;}
inline void modify(){*p=34;}
};

int main()
{
A a,b;
b=a;
b.modify();
a.print();
b.print();
return 0;
}

when we called modify of b, a's m got modified instead of b's. if another object can change members in the private section the whole abstraction fails right?

please tell ur views.

[webJose]

I see indeed the danger in using the compiler-provided copy constructor and assignment operator in classes that hold pointers, yes. I don't know, however, your actual point. Can you please be more specific?

[vineethkumart]

Ya indeed its a danger of def. copy constructor provided by the compiler.

[S6k9GNh0]

If you run this, m is given a separate (since m is allocated per class) allocation therefor p in b shouldn't point to the same location as p in a. This is a poor example of saying C++ doesn't provide decent abstraction. You're statement of calling modify() with b changes a should be incorrect in this example.

Also, please note that there's probably always a work around to get into the private members of a class. You can also change const variables via a const cast. That doesn't mean you should. C++ isn't very restrictive with anything.

If you'd like more advice, please provide a more productive question.

[webJose]

computerquip, the example does demonstrate what the OP describes. The line b = a; calls for the assignment operator. Since none was explicitly written, the compiler-provided one is used. This one will assign the address of a.m to b.p because that's the value of a.p.

[anonymous23323124]

It cannot change the private members of another class, and it does not. It only changes what is in the memory the member points to.

Okay this is an issue, but C++ lets you fix it with custom copy constructors.