[solved]getBounds()

Newbie & Debugging Questions
#1

I tried to search for the exact explanation of this method and how to get the (x, y) of the rectangle.

and I found out getX() and getY() as double.

But the one I want to get is the Y-axis of the hitted blocked when the BALL hitted it.

i tried to get the TOP and BOTTOM side of the BLOCK.

public boolean TOP() {
        boolean top = true;
        if (game.blocks.getTopY() == getTopY())
            top = true;
        return top;
    }
    
    public boolean BTM() {
        boolean btm = true;
        if (game.blocks.getLowY() == getLowY())
            btm = true;
        return btm;
    }

    public int getLowY() {
        int LowY = 0;
        if (hit())
            LowY = y + DIAMETER;
        return LowY;
    }

    public int getTopY() {
        int TopY = 0;
        if (hit())
            TopY = y;
        return TopY;
    }

If my information is not enough, please inform me so that I can add more.

Thanks.

#2

These methods always return true, so… as for the ‘exact explanation’ of these methods… they are pretty much useless :slight_smile:

#3

LOL. did not notice it.

Thanks.

#4

Hello guys, I am trying to tell to the ball if what side of the block it hitted.

 public boolean TOP() {
        boolean top = false;
        if (hit() && y <= game.blocks.getTopY())
            top = true;
        return top;
    }
    
    public boolean BTM() {
       boolean btm = false;
       if (hit() && y <= game.blocks.getBtmY())
           btm = true;
       return btm;
    }

    public boolean LEFT() {
        boolean left = false;
        if (y < game.blocks.getBtmY() && y > game.blocks.getTopY()
                && x == game.blocks.getLeftX() - DIAMETER)
            left = true;
        return left;
    }

    public boolean RIGHT() {
        boolean right = false;
        if (y < game.blocks.getBtmY() && y > game.blocks.getTopY()
                && x == game.blocks.getRightX())
            right = true;
        return right;
    }

If there`s anything that I should do so that I can improve it, please tell me.
I am willing to learn more.

If there`s anything wrong with my code please tell me.

Thanks.

#5

Minor, but


 public boolean TOP() {

        if (hit() && y <= game.blocks.getTopY())
            top = true;
        else
            top = false;
        return top;  
    }

is better, I think.

#6

In this case,

I don`t need the BTM() anymore, right?

Making the code shorter, is better.

Thanks dude.

#7

If you still need to check for bottom collision then keep it. I’m just saying you don’t need to initialize a boolean every time you call TOP() BTM() LEFT() or RIGHT().

#8

This will allow you to get all collisions in one method call:


static final int TOP = 1, BOTTOM = 2, LEFT = 4, RIGHT = 8;

public int collisionResult() {
	int ret = 0;
	if (hit() && y <= game.blocks.getTopY()) ret |= TOP;
	if (hit() && y <= game.blocks.getBtmY()) ret |= BOTTOM;
	if (y < game.blocks.getBtmY() && y > game.blocks.getTopY()
			&& x == game.blocks.getLeftX() - DIAMETER) ret |= LEFT;
	if (y < game.blocks.getBtmY() && y > game.blocks.getTopY()
			&& x == game.blocks.getRightX()) ret |= RIGHT;
			
	return ret;
}
#9

First time to see this

ret |= TOP;

What is the explanation of it?

Thanks.

#10 

static final int TOP = 1, BOTTOM = 2, LEFT = 4, RIGHT = 8;

public int collisionResult() {
	int ret = 0;
	if (hit() && y <= game.blocks.getTopY()) ret |= TOP;
	if (hit() && y <= game.blocks.getBtmY()) ret |= BOTTOM;
	if (y < game.blocks.getBtmY() && y > game.blocks.getTopY()
			&& x == game.blocks.getLeftX() - DIAMETER) ret |= LEFT;
	if (y < game.blocks.getBtmY() && y > game.blocks.getTopY()
			&& x == game.blocks.getRightX()) ret |= RIGHT;
			
	return ret;
}

It made my code shorter.
Thanks for the idea.

This is my new code:

  
        private static final TOP = 1, BTM = 2, LEFT = 3; RIGHT = 4;
        public int Side() {
        int side = 0;
        if (hit() && y <= game.blocks.getTopY())
            side = TOP;
        else if (hit() && y >= game.blocks.getBtmY() - DIAMETER)
            side = BTM;
        else if (y < game.blocks.getBtmY() && y > game.blocks.getTopY()
                && x == game.blocks.getLeftX() - DIAMETER)
            side = LEFT;
        else if (y < game.blocks.getBtmY() && y > game.blocks.getTopY()
                && x == game.blocks.getRightX())
            side = RIGHT;
        return side;
    }

Movement of the ball

 public void moveBall() {
        if (x + xa < 0)
            xa = game.speed;
        else if (x + xa > game.getWidth() - DIAMETER)
            xa = -game.speed;
        else if (y + ya < 0)
            ya = game.speed;
        else if (y + ya > game.getHeight() - DIAMETER)
            game.gameOver();
        else if (collision()) {
            ya = -game.speed;
            y = game.racquet.getTopY() - DIAMETER;
        }
        if (Side() == TOP) {
            ya = -game.speed;
            y = game.blocks.getTopY() - DIAMETER;
            System.out.println("ytop: "+y);
        }
        else if (Side() == BTM) {
            ya = game.speed;
            y = game.blocks.getBtmY();
            System.out.println("ybtm: "+y);
        }
        else if (Side() == LEFT) {
            xa = -game.speed;
            x = game.blocks.getLeftX() - DIAMETER;
            System.out.println("yleft: "+y);
        }
        else if (Side() == RIGHT) {
            xa = game.speed;
            x = game.blocks.getRightX();
            System.out.println("yryt: "+y);
        }
        x = x + xa;
        y = y + ya;        
    }

*Dont mind the System.out.print().
I put it to monitor the (x, y).

#11

Basically playing around with bits like legos.

Looking at the 4 least significant bits it becomes quite clear:

static final int TOP = 1, BOTTOM = 2, LEFT = 4, RIGHT = 8;


0 = 0....0000 // 0x0 (HEX)
1 = 0....0001 // 0x1
2 = 0....0010 // 0x2
4 = 0....0100 // 0x4
8 = 0....1000 // 0x8

Bitwise OR is like stacking the bits on top of each other and adding up all the 1’s.

Therefore

ret |= TOP;

is


 0....0000 // ret
+0....0001 // TOP
=0....0001

and if you add BOTTOM or LEFT or RIGHT you’ll just be adding that single 1 bit to a specific position.

#12

OMG! It is too complicated. LOL

I can`t use this thing, I think.

Ok, can you explain the difference of

ret = TOP to ret |= TOP?

Thanks.

#13

the diffrence is this:

when ret = 2:
ret = TOP: ret = 1
ret |= TOP: ret = 3

when ret = 1:
ret = TOP: ret = 1
ret |= TOP: ret = 1

Just think in bits, an int is in people language 1, 2, 3, 4, 5, 6, 7 etc.
A computer can only be on or off, so an 2 does not exist in memory.
It counts like 0001, 0010, 0011, 0100, 0101, 0110, 0111, etc

With bit shifting you just paste these numbers on top of each other.

00 | 00 = 00
00 | 01 = 01
00 | 10 = 10
00 | 11 = 11
01 | 00 = 01
01 | 01 = 01
01 | 10 = 11
01 | 11 = 11
10 | 00 = 10
10 | 01 = 11
10 | 10 = 10
10 | 11 = 11
11 | 00 = 11
11 | 01 = 11
11 | 10 = 11
11 | 11 = 11
#14

Oh, I see.

That is why the value of sides is 1, 2, 4, 8 respectively.

Thanks for the explanation sir.

#15

To add on to what everyone else has said, the bitwise OR operation allows you to create bit flags. Since each bit has it’s own position, and a boolean value is simply a single bit, you can think of it as creating several booleans within a byte.

One byte can contain eight boolean values:
0000 0001 = 1
0000 0010 = 2
0000 0100 = 4
0000 1000 = 8
0001 0000 = 16
0010 0000 = 32
0100 0000 = 64
1000 0000 = 128

The bitwise OR operator has the following truth table:
0 | 0 = 0
0 | 1 = 1
1 | 1 = 1

This means that:…
0000 0000 | 0000 0000 = 0000 0000
0000 0001 | 0000 0000 = 0000 0001
0000 0100 | 0000 0010 = 0000 0110
0000 0110 | 0000 0100 = 0000 0110

So, to sum it all up, bit flags allow you to place several booleans within a single byte. You can read the binary value of the byte, with a 0 meaning false and a 1 meaning true.
I really recommend this article.

Edit: It might be easier to read if you specify the values as binary. You can do this like so:


byte a = 0b00000000; // 0
byte b = 0b00000001; // 1
byte c = 0b00000010; // 2
byte d = 0b00000100; // 4
byte e = 0b00001000; // 8

byte x = a | c | d; // 0110, 6
#16

Thank you for the informations guys.

#17 
[quote="Troubleshoots,post:15,topic:46549"]
-snip-

[/quote]
You made a pretty fatal mistake there.
A byte has 8 bits, or ‘booleans’ in this matter, not just 4!

Have a nice day.

  • Longor1996
#18

Oh wow, I can’t have been thinking consciously. I edited the post. :expressionless: