Reputation: 302
This is The Code :
public void paint(Graphics g) {
g.fillOval(x, y, 25, 25);
repaint();
}
This will create an output like This where if i move the pointer it doesn't clear the previous Graphics and creates a Path .
And by adding the super statement like this it Doesn't show the Path but only the Current location of the Oval Graphic.
public void paint(Graphics g) {
super.paint(g);
g.fillOval(x, y, 25, 25);
repaint();
}
I just want to the Mechanism and The Reason Behind This.
Upvotes: 1
Views: 110
Reputation: 285403
Sorry, but there is so much wrong with your code, that we have to fix it:
super.paintComponent(g)
as this will continue to propagate the painting chain, whereas by not calling this you break this chain. Calling the super's method will call the component's housekeeping painting including the overpainting of "dirty" pixels which is why the trails are removed. This is likely the answer that you're looking for.repaint()
from within any painting method. Doing this results in a poor-man's animation that is completely uncontrollable, and puts too much responsibility on the painting method which should focus on its sole job -- to paint the component. Use a game loop such as a Swing Timer instead.Most important -- read the tutorials as most of this is already explained there:
Upvotes: 3
Reputation: 4135
Straight from the docs
public void paint(Graphics g)
Paints the container. This forwards the paint to any lightweight components that are children of this container. If this method is reimplemented, super.paint(g) should be called so that lightweight components are properly rendered. If a child component is entirely clipped by the current clipping setting in g, paint() will not be forwarded to that child.
As he said that super.paint() cleans the dirty pixels. This says it all .!! :)
Upvotes: 2