How do I write properly this function?

[Shishka]

Hello everyone, first of all I want to thank you for trying to help me.
I was working on a simple project, trying to program something similar to the game Pong.
I defined these two classes (you don't need to pay too much attention to the constructor):

const int HEIGHT= 128, 
             WIDTH= 160;
class Settings
{
  friend class Game;
  int id,
      x,
      y,
      width,
      height;
  public:
  int sp;
}batl, batr, top, bot, ball;
class Game
{
  public:
  Game()
  {
   batl.height=HEIGHT/4;
   batl.width=WIDTH/32;
   batl.x=0;
   batl.y=HEIGHT/4;
   batl.id=3;
   
   batr=batl;
   batr.x=WIDTH-batr.width;
   batr.id=4;
   
   top.height=WIDTH/64;
   top.width=WIDTH;
   top.x=0;
   top.y=0;
   top.id=1
   
   bot=top;
   bot.y=HEIGHT-bot.height;
   bot.id=2
   
   ball.x=WIDTH/2;
   ball.y=HEIGHT/2;
   ball.width=WIDTH/10;
   ball.id=5;
  }
};

I have created the parameter ID for each element because I was going to create a bool function to understand when two different objects are touching or not.
Now that I have to write it I have a problem: knowing the ID of an object, can I know what object it belongs to and make the function work for it?
What I mean is: do I have to write this function like

bool contact (int id1, int id2)
{
 if (id1==batl.id && id2== top.id)
 {
   //some code
 }
else if (id1==batl.id && id2==bot.id)
 {
   // some code
 }
 //and so on
}

?
Or is there a better way to do it? Thank you for your attention!

[AbstractionAnon]

Your question is not clear. What do you mean by "find the right variable"?

[Shishka]

Sorry, my question should have been "How do I write properly this function?"

[vlkon]

For your code:
What is top and bot?
Why is x and y set to 0 for top?
I also see some undefined values - for example where is batr.y?

You should make some fixed vector position (x,y) = (0,0) for example left bottom corner of your window and make all "pixel" coordinates (that you will ever define) relative to this position.


I would do something like this: (not actual code - just a hint)
Considering that everithing is some sort of rectangle (even the ball).

class Obstacle
{
public:
   Obstacle(int x, int y, int width, int height); //Constructor
   //For comparison of collision - overload operator==
   //If colision occurs then True else False
   bool operator==(const Obstacle& secondObstacle);
private:
   Obstacle(); //Block default constructor from public access - not entirely needed
   int x; //you can measure distance in pixels so int is better than double - my opinion
   int y;
   int width;
   int height;
};

int main()
{
   Obstacle leftBar(int numbers for x, y, width, height);
   Obstacle rightBar(again 4 int numbers...);
   Obstacle ball(4 int numbers...);

   while(true){
      if(leftBar == ball) //Left bar is colliding with the ball
         //... Do stuff
      else if(rightBar == ball) //Right bar is colliding with the ball
         //... Do some other stuff
      else      //No collision
         //... Continue game or check for game exit and break infinite while loop
   }

    //You could also check (leftBar == rightBar) but that case should never happen
    //At least in classic Pong :-)
}

Then content of overloaded operator== would be something like this:
(different bounds may be set by different <, >, <=, >=; it's all up to your preferences)

bool Obstacle::operator==(const Obstacle& secondObstacle)
{
//"this" means left hand side of operator ==. So "a==b" meand that "this" is "a"
//Probably you don't even need to write words "this" in the first time
   if (this.x              < secondObstacle.x &&
      (this.x+this.width)  > secondObstacle.x && //X coordinate checked
      this.y               < secondObstacle.y &&
      (this.y+this.height) > secondObstacle.y) //Y coordinate checked
         return true; //We have collision
//Now we checked secondObstacle corned at coordinates (X,Y)

   else if(this.x < (secondObstacle.x + secondObstacle.width) &&
             //... Other checks for other corners of secondObstacle
             //... I don't really want to write it all
            //... We have corners secondObject-> (x,y), (x+width,y), (x, y+height), (x+w, y+h)

    else
          return false; //No collision
}

You may also include several checks if for example is one object Obstacle inside the other one. "secondObstacle" can be bigger that "this" and the code I posted only checks if some corner of "secondObstacle" is inside of "this" object (left hand side argument of ==).

I hope I didn't make it too confusing. If you didn't get some part from this code snippet just ask.

[Shishka]

Okay, I think I got it even though it don't know very well how to use the

this

pointer and operators, but I will try!
By the way I just realized I haven't said it before, but I had to say that I was making this program for an Arduino connected to a small display (160x128) so yes, there actually is a fixed vector and measures are in pixel! Thanks anyway for your help!

[giblit]

this isn't even necessary in that bit of code. The only reason I think think to use it is if you wish to return a pointer to the current object or you have two variables with the same name.

Examples

class SomeClass
{
    int x;
    public:
        SomeClass returnObject(void);
        void setX(int x); //same name as private
};

SomeClass SomeClass::returnObject(void)
{
    return *this;
}

void SomeClass::setX(int x)
{
    this->x = x;
}

BTW vlkon this is a pointer to the current object so you either need *this.something or this->something Though in your case there is no point.

[vlkon]

@giblit: Thank you for a correction. I didn't pay too much attention when I wrote this code - it was only for a demonstration and I wanted to somehow distinguish secondObstacle (right side of ==) from the left side (this).
Other way would be to overload it as a friend function with Obstacle& lhs and Obstacle& rhs as arguments.

@Shishka: If you remove all this from my example you should still be fine as giblit said.