downcasting in c++ I couldnt success

[eMRe]

class Animal {


};

class Dog : public Animal {

};

int main ( )
{
stack<Animalmys tack;
Animal *x = new Dog( );
mystack.push(*x );


Animal y = mystack.top( );

Dog z = ( Dog ) y ; // this line doesnt work, how could I fix it?

return 0;
}

[Kai-Uwe Bux]

Re: downcasting in c++ I couldnt success

eMRe wrote:

class Animal {
>
>
};
>
class Dog : public Animal {
>
};
>
int main ( )
{
stack<Animalmys tack;
Animal *x = new Dog( );
mystack.push(*x );

Containers in C++ have value semantics, i.e., the line above only copies the
Animal part of the Dog object. What the stack contains is an honest to God
Animal and no Dog.

>
Animal y = mystack.top( );

That way, you retrieve the Animal you stored.

Dog z = ( Dog ) y ; // this line doesnt work, how could I fix it?

You should get rid of the C-style cast. But anyway, the line above should
not succeed. The Dog part of the object has been lost when you stored it in
the stack.

return 0;
}

If you need a polymorphic stack, you might try

stack< Animal * >


Technically, if D is derived from T, there are two natural maps:

a) a projection from values of type D to values of type T. This map is
called "slicing".

b) an injection from values of type D* to values of type T*. This map is
what people usually call the is-a-relation. It is important to see that it
holds on the level of pointers (or references, but that does not matter in
the context of containers).


Best

Kai-Uwe Bux

[James Kanze]

Re: downcasting in c++ I couldnt success

On Oct 5, 9:36 am, Kai-Uwe Bux <jkherci...@gmx .netwrote:

eMRe wrote:

class Animal {
};

class Dog : public Animal {
};

int main ( )
{
stack<Animalmys tack;
Animal *x = new Dog( );
mystack.push(*x );

Containers in C++ have value semantics, i.e., the line above
only copies the Animal part of the Dog object. What the stack
contains is an honest to God Animal and no Dog.

Animal y = mystack.top( );

That way, you retrieve the Animal you stored.

He retrieves a copy of the Animal he stored. Not the Animal
itself. Even if mystack had type stack<Dog>, all he'd get is a
copy of the Animal part of the Dog in the stack, because his
variable is of type Animal.

--
James Kanze (GABI Software) email:james.kan ze@gmail.com
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