网格计时问题

我正在沿着“网格”移动形状。我似乎终于弄清楚了，但我有一个小问题;无论我如何接近每个动作之间的特定时间间隔，似乎仍然会脱离电网。我想使用我正在使用的当前方法，因为我知道thigs如何以这种方式工作。我知道它有时会出现一个随机运动“小故障”，因为如果我将它移回第四个位置，无论如何它都应该锁定在电网上。这里是我的代码（对不起，我缺乏的意见和东西，我只是测试代码怪异的位置）：网格计时问题

主类：

import java.awt.Graphics; 
import java.awt.Graphics2D; 
import java.awt.Rectangle; 
import javax.swing.JFrame; 

public class WhyOhWhy { 
public int x; 
public int y; 

public static void main(String[] args) { 

    JFrame f = new JFrame(); 
    InputHandler input = new InputHandler(); 
    f.add(input); 
    f.setVisible(true); 
    f.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); 
    f.setSize(800,600); 
    input.doStuff(); 
    } 
}

InputHandler类：

import javax.swing.*; 
import java.awt.*; 
import java.awt.event.*; 
import java.awt.geom.*; 
import java.util.logging.Level; 
import java.util.logging.Logger; 

public class InputHandler extends JPanel implements ActionListener, KeyListener { 

Timer t = new Timer(5, this); 
int x = 0, y = 0, velX = 0, velY = 0; 
int i = 0, j = 0; 
TimeKeeper timeStart; 


public void doStuff(){ 
    velX = 0; 
    velY = 0; 
} 
public InputHandler() { 
    t.start(); 
    addKeyListener(this); 
    setFocusable(true); 
    setFocusTraversalKeysEnabled(false); 
} 


public void paintComponent(Graphics g) { 
    super.paintComponent(g); 
    Graphics2D g2 = (Graphics2D) g; 
    g2.fill(new Ellipse2D.Double(x, y, 32, 32)); 
    for(int i = 0;i <500;i+=32){ 
     g2.drawRect(i, j, 32, 32); 
     for(int j=0;j<500;j+=32){ 
      g2.drawRect(i, j, 32, 32); 
     } 
    } 
} 

public void actionPerformed(ActionEvent e) { 

    repaint(); 
    x += velX; 
    y += velY; 

    if(TimeKeeper.isFinished() == true){ 
     System.out.println("DONE"); 
     TimeKeeper.resetTimer(false); 
     velX = 0; 
     velY = 0; 
     setEnabled(true); 
    } 
} 

public void up() { 
    //System.out.println("Moving up"); 

    timeStart = new TimeKeeper(185); 
    velY = -1; 
    velX = 0; 
    setEnabled(false); 
} 

public void down() { 
    //System.out.println("Moving down"); 
    setEnabled(false); 
    timeStart = new TimeKeeper(185); 
    velY = 1; 
    velX = 0; 
} 

public void left() { 
    //System.out.println("Moving left"); 
    setEnabled(false); 
    timeStart = new TimeKeeper(185); 
    velY = 0; 
    velX = -1; 
} 

public void right() { 
    //System.out.println("Moving right"); 
    setEnabled(false); 
    timeStart = new TimeKeeper(185); 
    velY = 0; 
    velX = 1; 
} 

public void keyPressed(KeyEvent e) { 

    int code = e.getKeyCode(); 
    if (code == KeyEvent.VK_W) { 
     up(); 
     } 
    if (code == KeyEvent.VK_S) { 
     down(); 
     } 
    if (code == KeyEvent.VK_A) { 
     left(); 
     } 
    if (code == KeyEvent.VK_D) { 
     right(); 
     } 
    } 

public void keyTyped(KeyEvent e) { 
    } 

public void keyReleased(KeyEvent e) { 
    velX = 0; 
    velY = 0; 
    } 
}

TimeKeeper的类：

import java.util.Timer; 
import java.util.TimerTask; 

public class TimeKeeper { 

Timer timer; 
public static boolean isDone = false; 

public TimeKeeper(int seconds) { 

    timer = new Timer(); 
    isDone = false; 
    timer.schedule(new RemindTask(), seconds); 
} 

class RemindTask extends TimerTask { 

    public void run() { 
     //System.out.println("Time's up!"); 
     isDone = true; 
     timer.cancel(); //Terminate the timer thread 
    } 
} 

public static boolean isFinished() { 
    return isDone; 
} 

public static void resetTimer(boolean done) { 
    isDone = done; 
} 
}

谢谢！

来源

2012-11-21 Dillon Burton

没关系，想通了。需要使用％运算符来查看形状移动的数字是否可以在x和y两个方向上被32整除。

来源

2012-11-22 02:05:31

网格计时问题

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