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# dx = -dx; dy = -dy

Ranch Hand
Posts: 664
Can you explain how variables dx and dy are used to reverse the direction of a bouncing ball?

Source Core Java Vol.I (8th Ed), Ch.14

author
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140

It isn't very hard code to figure out -- in fact, it is incredibly easy. And figuring out another's code is a common task all developers do. Care to take a shot at it, and tell us how you think it works first?

Henry

Jon Camilleri
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Posts: 664

Henry Wong wrote:
It isn't very hard code to figure out -- in fact, it is incredibly easy. And figuring out another's code is a common task all developers do. Care to take a shot at it, and tell us how you think it works first?

Henry

I figured out that by default the ball object moves by (1,1), however, when the ball reaches the enclosing bounds (of type Rectangle2D), the sign is reversed so that the sign of x and y are changed. This is done so when the dx or dy exceeds the bounds the ball switches to the opposing bound. For example, if x is becoming -5 then it becomes +5, so that the ball position starts from (5, y). I just needed a confirmation.

Java Cowboy
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Confirmed...

Ranch Hand
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Jon Camilleri wrote:
I figured out that by default the ball object moves by (1,1), however, when the ball reaches the enclosing bounds (of type Rectangle2D), the sign is reversed so that the sign of x and y are changed. This is done so when the dx or dy exceeds the bounds the ball switches to the opposing bound. For example, if x is becoming -5 then it becomes +5, so that the ball position starts from (5, y). I just needed a confirmation.

No the sign of x and y is never changed, when the ball's position crosses any of the four boundaries, the ball is set to the boundary wall it exceeded and the sign of either dx or dy is changed f.ex when x becomes less than bounds.getMinX() x is set to bounds.getMinX() and dx becomes -dx