A very basic question. Sorry I graduated from school for too long. I have a class A with lots of private data members. Class B is derived from class A, and has its own private data members. Now I have an object of A (OA), and an object of B (OB) as well. I want to copy the values of all variables from OA to OB, can I use the following line to finish the task? OB = OA; If I want to copy the values of those common variables from OB to OA, can I use the following line? OA = (Class A) (OB); BTW, I am using Visual Studio 2003. Thank you in advance. Johnson

It depends on how you implement operator= for your class(es). However, you will need accessor or friend methods to get to A's private data from B and vise versa.

So really dont you have three classes?
The Base class which holds the common variables.
A which inherits from Base and
B which inherits from Base?

I agree that using a base class to hold the common variables as protected or public can solve the problem, or overloading the = operator might also work (yes there are functions for reading/writing private members), but I am working on an existing software, and I really do not want to change the class structure. So any other method to directly copy the private members from its parent class or derived class?

Johnson

Sorry I seem to have missed

Quote:I have a class A with lots of private data members. Class B is derived
from class A, and has its own private data members.

:)

B is a type of A, so maybe this is what you are looking for?

Quote:
A* a;
B b;
a = &b

FYI, I just got a reply saying we can assign the base object to the derived object directly (with all data members), but not vice versa. (see the attach below)

Since we cannot assign a Base to the derived object directly, and we know the derived object has a Base object inside, so can we assign the Base object to the inside Base Object of the derived object? Can we modify the construction function this way.

class Derived : public Base
{
public:
Derived (Base * b)
{ (Base *) this = b};

private:
int x,y,z;
};

Basically I do not know how to get a pointer to the inside Base object. I do not believe (Base *) this will work. Any idea?

Johnson

____________________________________________________________________________
niklasb@microsoft.com wrote:
> Johnson wrote:
> > A very basic question. Sorry I graduated from school for too long.
> >
> > I have a class A with lots of private data members. Class B is derived
> > from class A, and has its own private data members.
> >
> > Now I have an object of A (OA), and an object of B (OB) as well. I want
> > to copy the values of all variables from OA to OB, can I use the
> > following line to finish the task?
> > OB = OA;
> >
> > If I want to copy the values of those common variables from OB to OA,
> > can I use the following line?
> > OA = (Class A) (OB);
>
> Allow me to express this in code, with the names changed to make
> it easier (for me) to remember:
>
> class Base
> {
> int i,j,k;
> };
>
> class Derived : public Base
> {
> int x,y,z;
> };
>
> int main()
> {
> // Ignore the fact that none of these variables are initialized;
> // the point here is just to talk about assignment semantics.
> Base b,b2;
> Derived d;
>
> b = b2; // Ok
> b = d; // Ok
> d = b; // Error!
>
> return 0;
> }
>
> The first assignment is OK because the types are the same. Base
> doesn't define an assignment operator so the default meaning is
> member-wise assignment.
>
> The second assignemnt likewise invokes the compiler-generated
> assignment operator for Base (per the left-hand operand). Of
> course a Base expected for the right-hand operand whereas d is
> a Derived, but that's OK because conceptually a Derived "is a"
> Base (due to public inheritance); it can be implicitly converted
> to a Base and used in places where a Base is required. The effect
> here is to assign the Base part of d to b. In other words, the i, j,
> and k members of d are assigned to the corresponding members
> of b.
>
> The third assignment is an error. Here the compiler looks for an
> assignment operator for Derived (per the left-hand operand). There
> is only the default, compiler-generated assignment which expects a
> Derived as its right-hand operand. But b is a Base. A Derived "is a"
> Base but the reverse is not true. There is no implicit conversion
> Base to Derived (or from reference-to-Base to reference-to-Derived,
> or from pointer-to-Base to pointer-to-Derived) for reasons which are
> obvious if you think about it. Mind you, a class can define
> conversions, e.g., if Derived had a conversion constructor that
> took a Base (or reference to const Base), but that's not the
> case here.

Quote:Original post by JohnsonGPS
Basically I do not know how to get a pointer to the inside Base object. I do not believe (Base *) this will work. Any idea?

Remeber, when dealing with inheritance, the derived object is a base object. It is perfectly legal to do something like this:



Similarly, you can pass a B object to any function expecting an object of type A. For more information, look up object slicing.

It is also perfectly legal to "down cast" pointers to derived objects:



Note that when casting pointers, no slicing occurs.

Quote:Original post by JohnsonGPS

class Derived : public Base
{
public:
Derived (Base * b)
{ (Base *) this = b};

private:
int x,y,z;
};


I don’t even think that’s valid C++, IIRC you can’t modify the this pointer. The this pointer should point to "this" object. What you have there makes it point to some other object.

I think what you want to do is cast this to Bace* and then dereference it, THEN assign it a value.

class Derived : public Base{public:    Derived (Base * b)     { *((Base*)this) =  *b};}

Or even use a the copy constructer.

class Derived : public Base{public:    Derived (Base * b)     { *((Base*)this) =  Base(*b)};}

Try this. It works both way.



class BaseMine

{

int i,j,k;

public:

void Setj(int t)

{j=t;}

int Getj()

{return j;}

};



class DerivedMine : public BaseMine

{

int x,y,z;

};





BaseMine b;



DerivedMine d1;



d1.Setj(100);



b = (BaseMine) d1;



int y = b.Getj();





//It does not work, but can pass the compilor

b.Setj(10);

(BaseMine) d1 = b;

y = d1.Getj();





//It can pass the compiler and works fine.

d1.BaseMine::operator =(b);

y = d1.Getj();