精灵的对角运动

Question

如何实现对精灵的对角线运动？我创建了一个可移动的精灵（一个矩形），它可以在四个方向上移动。

为了给矩形设置动画，使用了一个计时器对象和执行动作的方法。我在keyPressed和keyReleased方法中实现了以下代码，以使其在四个方向上移动。

public void keyPressed(KeyEvent arg0){
    int c=arg0.getKeyCode();
    if(c==KeyEvent.VK_LEFT){
        velx=-4;
        vely=0;
    }
    else if(c==KeyEvent.VK_RIGHT){
        velx=4;
        vely=0;
    }
    else if(c==KeyEvent.VK_UP){
        velx=0;
        vely=-4;
    }
    else if(c==KeyEvent.VK_DOWN){
        velx=0;
        vely=4;
    }
}

public void keyReleased(KeyEvent arg0){
    velx=0;
    vely=0;
}

Answer 1

这是基于您的最后评论的代码：

// Set of currently pressed keys
private final Set<Integer> pressed = new TreeSet<Integer>();

@Override
public void keyPressed(KeyEvent arg0) {
    int c = arg0.getKeyCode();
    pressed.add(c);
    if (pressed.size() > 1) {
        Integer[] array = pressed.toArray(new Integer[] {});
        if (array[0] == KeyEvent.VK_LEFT && array[1] == KeyEvent.VK_UP) {
            velx = -4;
            vely = -4;
        } else if (array[0] == KeyEvent.VK_UP && array[1] == KeyEvent.VK_RIGHT) {
            velx = 4;
            vely = 4;
        } else if (array[0] == KeyEvent.VK_RIGHT && array[1] == KeyEvent.VK_DOWN) {
            velx = 4;
            vely = -4;
        } else if (array[0] == KeyEvent.VK_LEFT && array[1] == KeyEvent.VK_DOWN) {
            velx = -4;
            vely = 4;
        }
    } else {
        if (c == KeyEvent.VK_LEFT) {
            velx = -4;
            vely = 0;
        } else if (c == KeyEvent.VK_RIGHT) {
            velx = 4;
            vely = 0;
        } else if (c == KeyEvent.VK_UP) {
            velx = 0;
            vely = -4;
        } else if (c == KeyEvent.VK_DOWN) {
            velx = 0;
            vely = 4;
        }
    }

}

@Override
public void keyReleased(KeyEvent arg0) {
    velx = 0;
    vely = 0;

    pressed.remove(Integer.valueOf(arg0.getKeyCode()));
}