#include "StdAfx.h"
int main()
{
//Creating Objects
Actor *pActorArray[2];
Player *pPlayer = new Player();
Enemy *pEnemy = new Enemy();
//Setting up Array...
pActorArray[0] = pPlayer;
pActorArray[1] = pEnemy;
cout << pActorArray[0]->getAnimationName(2) << endl;
cout << pActorArray[1]->getAnimationName(4) << endl;
for( int Seek = 0; Seek < 2; Seek++ )
{
//This works fine
pActorArray[Seek]->getAnimationName(2);
//This doesn't----------------
pActorArray[Seek]->Modify(5);
}
getch();
return 0;
}
Inherited Classes in Class Array?
Author
Here is my main() so far-
the error I'm receiving is-
main.cpp(24) : error C2039: 'Modify' : is not a member of 'Actor'
Which I know, its a member of Player and Enemy (It does different things in both classes though)
Is what I'm trying to do even possible? Or would I have to create 2 separate arrays, one to store Players and the other to store Enemy's (even though they are both derived from Class 'Actor'?)
Please, tell me if I need to elaborate :)
if I understand you, Player is derived from Actor.. player has a member 'Modify'
Actor *a is assigned to players memory
you want a->Modify() to access players Modify..
just put a 'virtual' in front of Actors 'Modify' declaration
[edit]
(just to explain a bit)
if you declare a virtual function in a base class, then reference that
function using a base class pointer to a derived class the base class will
'lookup' the function in the derived class, and if present, will execute it..
if its not present, the base class virtual function will be called..
to assure a derived class provides this function you can put
'virtual derived_must_provide() = 0;' in the base class.. This just tells the
compiler that all derived classes must provide a 'derived_must_provide()'
member.. also if you do the '= 0' thing, you cannot use an instance of the
base class but must use an instance of the derived class (this will result in
a compiler error if you try to 'base something;'
Actor *a is assigned to players memory
you want a->Modify() to access players Modify..
just put a 'virtual' in front of Actors 'Modify' declaration
[edit]
(just to explain a bit)
if you declare a virtual function in a base class, then reference that
function using a base class pointer to a derived class the base class will
'lookup' the function in the derived class, and if present, will execute it..
if its not present, the base class virtual function will be called..
to assure a derived class provides this function you can put
'virtual derived_must_provide() = 0;' in the base class.. This just tells the
compiler that all derived classes must provide a 'derived_must_provide()'
member.. also if you do the '= 0' thing, you cannot use an instance of the
base class but must use an instance of the derived class (this will result in
a compiler error if you try to 'base something;'
Sounds like, while Modify is in both Player and Enemy, it's not in Actor (the common base class). You'll want to make sure Actor has a Modify as well, and make it virtual. Failing that, post the class definitions.
(I should note, however, that making it virtual is not required for the code to compile; however without it you won't get the runtime behavior you probably want).
(I should note, however, that making it virtual is not required for the code to compile; however without it you won't get the runtime behavior you probably want).
You said you have the Modify function in both Player and Enemy but you didn't mention it being in the Actor base class. If it's not in the base class, what you're doing won't work. The fact that they have the same function and inherit the same base is not enough. You should have something like this in your base.
The virtual keyword tells it to use a derived class's version of the function.
That should let you do what you're trying to do.
Hope that helps.
[Edit] I was second to respond when I started this ... you people type like mad-men.
class Actor{ public: virtual void Modify(int);};The virtual keyword tells it to use a derived class's version of the function.
That should let you do what you're trying to do.
Hope that helps.
[Edit] I was second to respond when I started this ... you people type like mad-men.
This topic is closed to new replies.
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