Ok, now that I'm at work I took another look at this thread and it seems that what rypyr mentioned is what I need. I'll give an update when I can try it out tonight.

In response to alnite:
I was just using "AddControl()" as an example that I need to include "Ship" in "Player" so I can run "Ship" member functions - More specifically, "AddControl()" would be used like "myShip->AddControl(THRUST);" after "Player" gets the inputs from the input handlers.

The other questions you asked go hand in hand - When I forward declare "Player" in "Ship", I can then give "Ship" a pointer to a Player as a member. It doesnt, however, let me use functions of that member. I can't say "myPlayer->anything();", and I assume it is because "Player" is only forward declared, not actually "truely included".

Thanks for the help so far guys

*EDIT* Response to Ishan:
I said before that I understand why I can't do it literally, but that I want achieve the same effect doing so would have. Simple form of what I want to do in case you don't want to read the description I just put up:

I have classes Player and Ship. Not only do they need to know OF eachother, they need to be able to access the data members of eachother (something forward declaration doesn't let you do). For example, Player needs to know of and directly effect Ship it owns (control it) and Ship needs to know the information thier Player holds to determine how it moves (each Player has unique statistics that control thier ships: movement speed, turn rate, etc).
What I need is a method that will allow me to access the data equally back and forth without making accessor class "C" which seems very ugly and cumbersome (to me anyway). But as I said, rypyr seems to have given me the answer I was looking for, which I will try tonight when I get home .

[edited by - neonfaktory on May 1, 2003 2:33:09 PM]

neonfaktory: you still must make the data members accessible "outside" the class. Hopefully you know how to expose data to the outside world (i.e. you shouldn''t just blindly make all player and ship attributes public).

Hopefully you know this, but if you don''t here is an example.

in Player.h:

class Ship;class Player {  Ship* m_pShip;  int m_Intelligence;public:  int getIntelligence() const { return m_Intelligence; }  void flyShip();}; 

in Ship.h:

class Player;class Ship {  Player* m_pPlayer;  int m_ShieldStrength;public:  int getShieldStrength() const { return m_ShieldStrength; }  void fly();} 

in Player.cpp:

#include "Player.h"#include "Ship.h"void Player::flyShip() {  // do something with m_pShip->getShieldStrength() here...  m_pShip->fly();} 

in Ship.cpp:

#include "Ship.h"#include "Player.h"void Ship::fly() {  // do something with m_pPlayer->getIntelligence() here...} 

Regards,
Jeff

Hahaha, ISOPimp, your description sounds funny, but I'll explain it.

Basically, each Player is just an "entity" of someone in the game - I guess kind of a way of leaving the door open for multiplayer. When you respawn, the Ship you will control is then tied to your Player object. Each Player object holds thier own set of attributes, like thrust speed, fire rate, etc., so each Player can create thier own play-style. These attributes determine how the Ship they are controlling performs.

To do this, I have 'Player' accept the input from the input handler and translate it to simplified actions that 'Ship' uses. So when the game loop handles the 'Player's, 'Player" is where it says "if(key == up) myShip->AddControl(THRUST);" When the game loop gets around to handling all the 'Ship's in the universe, it runs through a small queue of controls, acting accordingly.

So like in the example I gave, the Ship you are controlling sees a "THRUST" command, which then should access a "myPlayer" member to do something along the lines of "XVEL += myPlayer->thrust * cosf(angle);" (from the top of my head). Soo, to make this run smoothly I need to have access back and forth between these 2 classes, hence the "Include Eachother (or achieve the same effect)".

Thats just to clear everything up though, as it seems that I already have the answer to my solution (thanks rypyr), but thanks to everybody that helped and gave input. If it doesn't work out right, I'm sure to be back again .

[edited by - neonfaktory on May 2, 2003 12:39:27 PM]