我在模拟两个竞争对手之间的比赛时遇到了麻烦。这是您典型的比赛计划，您可以使用随机数生成器来确定竞争对手使用的“移动”。如下面的代码所示，轨道由50个矩形组成，填充的矩形显示每个竞争者在轨道上的位置。例如，一些“移动”使竞争者向右跳9个方格，或向后跳2个方格。当我运行applet时，只显示初始起始位置;小程序不起作用。我意识到这是很多代码，但是我需要做些什么来解决我的问题呢？我真的陷入了困境。任何帮助表示赞赏。我只能使用AWT，而不是挥杆。这是类的赋值：/这是代码：
import java.awt.*;
import java.applet.*;
public class Example extends Applet
{
Image tortoise, hare;
int tortX = 250, hareX = 250;
final int tortY = 100, hareY = 300, WIDTH = 15, HEIGHT = 50;
int turn; String turnNum;
int move; String tMove, hMove;
public void init()
{
tortoise = getImage( getDocumentBase(), "images/tortoise.gif" );
hare = getImage( getDocumentBase(), "images/hare.gif" );
move = 0; turn = 0;
}
public void control()
{
while (( tortX < 985 ) || ( hareX < 985 ))
{
move = (int)(10 * Math.random());
switch (move)
{
case 1:
case 2:
tortX += (3 * WIDTH);
hareX += (9 * WIDTH);
tMove = "Fast Plod"; hMove = "Big Hop";
break;
case 3:
case 4:
case 5:
tortX += (3 * WIDTH);
hareX += WIDTH;
tMove = "Fast Plod"; hMove = "Small Hop";
break;
case 6:
tortX += WIDTH;
if (hareX == 250) {} // DO NOTHING
else if (hareX <= (250 + (11 * WIDTH)))
hareX = 250;
else
hareX -= (12 * WIDTH);
tMove = "Slow Plod"; hMove = "Big Slip";
break;
case 7:
case 8:
tortX += (1 * WIDTH);
if (hareX == 250) {} // DO NOTHING
else if (hareX <= (250 + (WIDTH)))
hareX = 250;
else
hareX -= (2 * WIDTH);
tMove = "Slow Plod"; hMove = "Small Slip";
break;
case 9:
case 10:
if (tortX == 250) {} // DO NOTHING
else if (tortX <= (250 + (5 * WIDTH)))
tortX = 250;
else
tortX -= (6 * WIDTH);
tMove = "Slip"; hMove = "Fall Asleep. Zzz...";
break;
// Hare falls asleep. No action.
}
turn++; turnNum = (turn + "");
repaint();
for (int i = 1; i <= 10; i++)
{
delay();
}
}
tortX = 985; hareX = 985;
repaint();
}
public void paint( Graphics screen )
{
drawRace(screen);
if (tortX >= 985)
{
screen.setFont(new Font("Times New Roman", Font.ITALIC, 48));
screen.drawString("Tortoise Wins", 650, 240);
clearCurrent(screen);
fillNext(screen);
}
else if (hareX >= 985)
{
screen.setFont(new Font("Times New Roman", Font.ITALIC, 48));
screen.drawString("Tortoise Wins", 650, 240);
clearCurrent(screen);
fillNext(screen);
}
else
{
screen.drawString(("Turn " + turnNum), 621, 55);
screen.setFont(new Font("Times New Roman", Font.ITALIC, 12));
screen.drawString(tMove, 59, 65); screen.drawString(hMove, 66, 255);
clearCurrent(screen);
fillNext(screen);
}
stop();
}
public void clearCurrent( Graphics s )
{
s.clearRect(tortX+1, tortY+1, WIDTH-1, HEIGHT-1);
s.clearRect(hareX+1, hareY+1, WIDTH-1, HEIGHT-1);
}
public void fillNext( Graphics s )
{
s.fillRect(tortX+1, tortY+1, WIDTH-1, HEIGHT-1);
s.fillRect(hareX+1, hareY+1, WIDTH-1, HEIGHT-1);
}
public void drawRace( Graphics s )
{
// GENERATES INITIAL GRAPHICS FOR RACE
s.drawRect(250, 100, 750, 50);
s.drawRect(250, 300, 750, 50);
int lineX = 265, lineYi = 100, lineYf = 150;
for (int i = 1; i <= 98; i++)
{
if (lineX == 1000)
{
lineX = 265; lineYi = 300; lineYf = 350;
}
s.drawLine(lineX, lineYi, lineX, lineYf);
lineX += 15;
}
s.fillRect(tortX+1, tortY+1, WIDTH-1, HEIGHT-1);
s.fillRect(hareX+1, hareY+1, WIDTH-1, HEIGHT-1);
s.drawImage(tortoise, 59, 80, this);
s.drawImage(hare, 66, 271, this);
s.setFont(new Font("Times New Roman", Font.BOLD, 24));
s.drawString("Race", 250, 55);
}
public void delay()
{
for (int i = 0; i < 90000000; i++)
{
}
}
public void stop()
{
}
}
答案 0 :(得分：3)
你的第一个问题是你实际上从未“开始”比赛......确定你的init小程序，但是没有......
你的第二个问题是control方法是阻止事件调度线程，这意味着，当你在这个方法中时，NOTHING将被绘制到屏幕上。这是因为Event Dispatching Thread还负责调度重绘请求。
第三个问题是您违反了paint联系人。你有责任调用super.paint(screen) - paint是一种复杂的方法，除非你有充分的理由这样做，否则不应该忽略它。
您的第四个问题是，您使用的是Applet而不是JApplet。最好忽略AWT控件，转而使用Swing控件。 Swing更灵活，更容易扩展。
你的第五个问题是你画在顶层容器上，这是绝对不推荐的。您最好使用JPanel之类的内容并覆盖它的paintComponent方法（不要忘记调用super.paintComponent）。除了其他一切，它是双缓冲的，并且会在屏幕更新时减少轻弹。
看看......