Question

求一段J2ME的代码？

我在J2ME模拟器上做了一个程序，一个方块从左上角向右下角移动，但是会超出屏幕，超出去后程序继续就是没有方块了，只剩背景，我想让方块碰到屏幕边上反弹这样无限反弹，应该加什么代码（我需要具体代码），是不是应该加在方块移动的代码后面？

Answer 1

给你一段参考吧 只不过是圆形的物体在屏幕上碰到墙后反弹 原理都一样 而且是用swt做的,不是j2me 你参考一下吧
public class Ball {
/** 球的半径 */
int diameter;
/** 最大半径 */
private final int maxR = 60;
/** 最大半径 */
private final int minR = 2;
/** 球的大小 */
public Rectangle bounds; // Bounding rectangle for the ball
/** 球的xy速度 */
Point vector; // The direction the ball is traveling

/** 球的颜色 */
Color c;

// 速度
private final int maxSpeed = 20;

private MyCanvas canvas;

/** 向量 */
VectorQuantity quantity;

private int timer;
private int timerMax = 6;

public void checkWallCollision() {
// Get the Client Rectangle
Rectangle cr;
cr = canvas.getBounds();
// Check for collision with Top Wall
if (bounds.y <= cr.y) {
// Reverse the Y vector
vector.y = -vector.y;

// Reset the bounding rectangle so the ball never appears outside of
// the screen
bounds.y = cr.y + 1;
bounds.height = bounds.y + diameter;
}
// Check for collision with Bottom Wall
else if (bounds.height >= cr.height) {
// Reverse the Y vector
vector.y = -vector.y;

// Reset the bounding rectangle so the ball never appears outside of
// the screen
bounds.height = cr.height - 1;
bounds.y = bounds.height - diameter;
}
// Check for collision with Left Wall
else if (bounds.x <= cr.x) {
// Reverse the X vector
vector.x = -vector.x;

// Reset the bounding rectangle so the ball never appears outside of
// the screen
bounds.x = cr.x + 1;
bounds.width = bounds.x + diameter;
}
// Check for collision with Right Wall
else if (bounds.width >= cr.width) {
// Reverse the X vector
vector.x = -vector.x;

// Reset the bounding rectangle so the ball never appears outside of
// the screen
bounds.width = cr.width - 1;
bounds.x = bounds.width - diameter;
}
}

Answer 2

可以用if。代码如下：
//x，y为方块坐标，w为反弹时x坐标。等于屏宽减方形边长
while(true){
Boolean ft=false;
if(ft){
x--;
y--;
}else{
x++;
y++;
}
if(x>w || x<0){
ft=!ft;
//当满足反弹条件时,变换ft的值
}
repaint();
}

追问

public void run() {
while (true) {
try {
Thread.sleep(5);
} catch (Exception ex) {

}
repaint();
x++;
y++;
}
}
这是我的代码，应该加在哪里怎么加

Answer 3

学习是要用脑子的，
我在表扬你好问的同事，再批评你，凡事用心想想，
你可以实现向右下移动，就不能让他返回来？你懒的可以，或都说*的可以