Copy Inherited Objects

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I'm attempting to learn how to use inheritance in C++, the following
program should print the word 'Orange' on the screen but it prints
'Apple' instead. Can someone explain what is wrong?

#include <iostream>

class Fruit
{
public:
    virtual const char* get_name() {return "(nothing)";}
};

class Apple: public Fruit
{
public:
    const char* get_name() {return "Apple";}
};

class Orange: public Fruit
{
    const char* get_name() {return "Orange";}
};

int main()
{
    std::cout << "Test inheritation" << std::endl;

    Fruit *myfruit1 = new Apple();
    Fruit *myfruit2 = new Orange();
    Fruit *myfruit3 = new Apple();

    std::cout << "myfruit1 = " << myfruit1->get_name() <<
std::endl; // Prints "Apple"
    std::cout << "myfruit2 = " << myfruit2->get_name() <<
std::endl; // Prints "Orange"
    std::cout << "myfruit3 = " << myfruit3->get_name() <<
std::endl; // Prints "Apple"

    // The following does not work
    std::cout << "Now change myfruit3 from Apple to Orange..." <<
std::endl;
    *myfruit3 = *myfruit2;

    // Prints "Apple" but should print "Orange"
    std::cout << "myfruit3 = " << myfruit3->get_name() << std::endl;

    // Wait for a keypress
    std::cin.get();

    delete myfruit1;
    delete myfruit2;
    delete myfruit3;
}
0
Reply visicalcenator1 5/17/2008 9:11:52 PM

timid wrote:

> #include <iostream>
> 
> class Fruit
> {
> public:
>     virtual const char* get_name() {return "(nothing)";}

Use std::string rather than char*.

Also, a function such as "get_name" should usually be declared const.

> };
> 
> class Apple: public Fruit
> {
> public:
>     const char* get_name() {return "Apple";}
> };
> 
> class Orange: public Fruit
> {
>     const char* get_name() {return "Orange";}
> };
> 
> int main()
> {
>     std::cout << "Test inheritation" << std::endl;
> 
>     Fruit *myfruit1 = new Apple();
>     Fruit *myfruit2 = new Orange();
>     Fruit *myfruit3 = new Apple();
> 
>     [...]
> 
>     // The following does not work
>     std::cout << "Now change myfruit3 from Apple to Orange..." <<
> std::endl;
>     *myfruit3 = *myfruit2;

* derefences the pointer, yielding the object it's pointing to. 
Consequently, in your example you don't assign the _pointers_, you 
assign the _objects_ they are pointing to. That's a huge difference. 
Assigning an Orange to an Apple obviously does not do anything because 
you did not specify anything to happen in case of such an assignment. 
The invisible assignment operator automatically generated by the 
compiler is empty. In other words, the Apple is completely unaffected by 
the operation.


If you want to assign a pointer to another pointer, you have to write:

myfruit3 = myfruit2;

However, in your program this would have disastrous results because you 
would lose the old value of myfruit3, which means you cannot delete your 
second Apple anymore. (You'd have to save the old value somewhere, for 
example in a temporary pointer variable.)

What do you try to achieve by having myfruit3 point to a different object?


-- 
Christian Hackl
0
Reply hacki (114) 5/17/2008 9:49:58 PM


On May 17, 10:49 pm, Christian Hackl <ha...@sbox.tugraz.at> wrote:
> timid wrote:
> > #include <iostream>
>
> > class Fruit
> > {
> > public:
> >     virtual const char* get_name() {return "(nothing)";}
>
> Use std::string rather than char*.
>
> Also, a function such as "get_name" should usually be declared const.
>
>
>
> > };
>
> > class Apple: public Fruit
> > {
> > public:
> >     const char* get_name() {return "Apple";}
> > };
>
> > class Orange: public Fruit
> > {
> >     const char* get_name() {return "Orange";}
> > };
>
> > int main()
> > {
> >     std::cout << "Test inheritation" << std::endl;
>
> >     Fruit *myfruit1 = new Apple();
> >     Fruit *myfruit2 = new Orange();
> >     Fruit *myfruit3 = new Apple();
>
> >     [...]
>
> >     // The following does not work
> >     std::cout << "Now change myfruit3 from Apple to Orange..." <<
> > std::endl;
> >     *myfruit3 = *myfruit2;
>
> * derefences the pointer, yielding the object it's pointing to.
> Consequently, in your example you don't assign the _pointers_, you
> assign the _objects_ they are pointing to. That's a huge difference.
> Assigning an Orange to an Apple obviously does not do anything because
> you did not specify anything to happen in case of such an assignment.
> The invisible assignment operator automatically generated by the
> compiler is empty. In other words, the Apple is completely unaffected by
> the operation.
>
> If you want to assign a pointer to another pointer, you have to write:
>
> myfruit3 = myfruit2;
>
> However, in your program this would have disastrous results because you
> would lose the old value of myfruit3, which means you cannot delete your
> second Apple anymore. (You'd have to save the old value somewhere, for
> example in a temporary pointer variable.)
>
> What do you try to achieve by having myfruit3 point to a different object?
>
> --
> Christian Hackl

Thank's for the help so far, it would appear that I need to learn a
lot more.

I'm going to try & rewrite this program so it'll work.
0
Reply visicalcenator1 5/17/2008 10:14:44 PM

Hi!

Christian Hackl schrieb:
>> class Fruit
>> {
>> public:
>>     virtual const char* get_name() {return "(nothing)";}
> 
> Use std::string rather than char*.

Normally I would agree, but in case of /const/ char* there is usually no 
problem. Only the use of char* for strings in C++ applications is nearly 
always a risk. On the other hand, using std::string for this (and 
similar) purposes is a significant runtime overhead. Only if you already 
have a std::string it is more advisable to return std::string, or if 
this causes no threading or aliasing problem const std::string&.

> Also, a function such as "get_name" should usually be declared const.

That's obviously true.


> If you want to assign a pointer to another pointer, you have to write:
> 
> myfruit3 = myfruit2;
> 
> However, in your program this would have disastrous results because you 
> would lose the old value of myfruit3, which means you cannot delete your 
> second Apple anymore. (You'd have to save the old value somewhere, for 
> example in a temporary pointer variable.)
> 
> What do you try to achieve by having myfruit3 point to a different object?

I think the OP is seeking for something like a clone method.


Marcel
0
Reply news.5.maazl2 (236) 5/18/2008 9:29:28 AM

On 17 mai, 23:11, timid <visicalcena...@googlemail.com> wrote:
> I'm attempting to learn how to use inheritance in C++, the
> following program should print the word 'Orange' on the screen
> but it prints 'Apple' instead. Can someone explain what is
> wrong?

> #include <iostream>

> class Fruit
> {
> public:
>     virtual const char* get_name() {return "(nothing)";}
> };

> class Apple: public Fruit
> {
> public:
>     const char* get_name() {return "Apple";}
> };

> class Orange: public Fruit
> {
>     const char* get_name() {return "Orange";}
> };

> int main()
> {
>     std::cout << "Test inheritation" << std::endl;

>     Fruit *myfruit1 =3D new Apple();
>     Fruit *myfruit2 =3D new Orange();
>     Fruit *myfruit3 =3D new Apple();

>     std::cout << "myfruit1 =3D " << myfruit1->get_name() <<
> std::endl; // Prints "Apple"
>     std::cout << "myfruit2 =3D " << myfruit2->get_name() <<
> std::endl; // Prints "Orange"
>     std::cout << "myfruit3 =3D " << myfruit3->get_name() <<
> std::endl; // Prints "Apple"

>     // The following does not work
>     std::cout << "Now change myfruit3 from Apple to Orange..." <<
> std::endl;
>     *myfruit3 =3D *myfruit2;

And what do you expect this to do?  You cannot change the type
of an existing object.  In general, when inheritence is
involved, you don't support assignment and copy.

>     // Prints "Apple" but should print "Orange"
>     std::cout << "myfruit3 =3D " << myfruit3->get_name() << std::endl;

No.  Should print "Apple".  But a lot of typical implementations
of the assignment operator (for more complicated types) will
result in undefined behavior.  The base class for an inheritence
hierarchy should normally forbid assignment, so that the
situation can't arrive, even accidentally.

--
James Kanze (GABI Software)             email:james.kanze@gmail.com
Conseils en informatique orient=E9e objet/
                   Beratung in objektorientierter Datenverarbeitung
9 place S=E9mard, 78210 St.-Cyr-l'=C9cole, France, +33 (0)1 30 23 00 34
0
Reply james.kanze (9590) 5/18/2008 1:17:08 PM

On 18 mai, 11:29, Marcel M=FCller <news.5.ma...@spamgourmet.com> wrote:
> I think the OP is seeking for something like a clone method.

Or the letter/envelope idiom.  Or perhaps just simply some
information on OO design; he hasn't really expressed a concrete
need for assignment.

--
James Kanze (GABI Software)             email:james.kanze@gmail.com
Conseils en informatique orient=E9e objet/
                   Beratung in objektorientierter Datenverarbeitung
9 place S=E9mard, 78210 St.-Cyr-l'=C9cole, France, +33 (0)1 30 23 00 34
0
Reply james.kanze (9590) 5/18/2008 1:19:05 PM

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