我正在创建游戏《蛇与梯子》。我使用的游戏模式可以让您在计算机上玩游戏。
人类玩家掷骰子时，将调用rollTheDiceAndMove方法。我们等待1秒钟（骰子正在滚动...）。然后我们调用move函数，该函数实际上将片段移动到指定的图块。如果碰到了蛇或梯子，我想再等1秒钟，然后再转到最后一块瓷砖。例如。我等待1秒钟，然后滚动5，然后在前面放my_current_pos + dice_roll砖，然后再移动一块。如果蛇或梯子撞到了：另一个延迟，然后递归再次移动。
最后，如果选择了“反对计算机”模式，那么当人类玩家移动时，我想等一秒钟，然后计算机自动掷骰子。如下所示，我正在使用Timer和TimerTask类，但是它很复杂，因为Timer范围内的任何内容都会在一段时间后执行，但是计时器外部的代码会无延迟地执行这导致了我很多错误和异步。
您建议使用什么来创建此延迟？
public void rollTheDiceAndMove() {
int diceRoll = gameBoard.rollDice();
// delay for dice roll.
new Timer().schedule(
new TimerTask() {
@Override
public void run() {
Platform.runLater(new Runnable() {
@Override
public void run() {
////////////////////////////////////////////////////////////////////////////
gameGUI.indicateDiceRoll(diceRoll);
int newIndex = getPlayerIndexAfterRoll(diceRoll);
move(newIndex);
System.out.println("change turns");
swapTurns();
System.out.println(isComputerTurn());
gameGUI.updateCurrentTurnLabel();
if (newIndex == GameBoard.WIN_POINT) {
boolean restartGame = gameBoard.playAgainOrExit();
if (restartGame) {
Player winner = gameBoard.getCurrentPlayer();
gameGUI.updateScore(winner);
gameGUI.playAgain();
} else {
System.exit(0);
}
}
////////////////////////////////////////////////////////////////////////////
}
});
}
}, GameBoard.DICE_ROLL_DELAY
);
// try to recursively call this method again for computer turn.
}
public void move(int currentIndex) {
int[] newCoordinates = gameBoard.getBoardCoordinates(GameBoard.NUMBER_OF_TILES - currentIndex);
gameBoard.getCurrentPlayer().getPlayerPiece().setPosition(currentIndex);
gameGUI.movePieceImages(newCoordinates[0], newCoordinates[1]);
if (gameBoard.getTile(currentIndex).containsLadderOrSnake()) {
// delay for moving to second tile.
new Timer().schedule(
new TimerTask() {
@Override
public void run() {
Platform.runLater(new Runnable() {
@Override
public void run() {
int updatedIndex = gameBoard.getUpdatedPosition(currentIndex);
move(updatedIndex);
}
});
}
}, GameBoard.SECOND_MOVE_DELAY *2
);
return; // we need to return 'cause of recursion. Swap turns will be executed twice, one time for the initial call and the other time on above line.
}
}
答案 0 :(得分：1)
您可以使用java.util.concurrent包中的ScheduledThreadPoolExecutor类的调度方法
进度表（可运行命令，长时间延迟，TimeUnit单位）创建并 执行一次操作，在指定的延迟后启用。
在那之后，您可以检查Future对象是否完成任务。
public void rollTheDiceAndMove() {
int diceRoll = gameBoard.rollDice();
// delay for dice roll.
ScheduledFuture<Void> scheduledFuture = Executors.newScheduledThreadPool(1).schedule(new TimerTask() {
@Override
public void run() {
Platform.runLater(new Runnable() {
@Override
public void run() {
////////////////////////////////////////////////////////////////////////////
gameGUI.indicateDiceRoll(diceRoll);
int newIndex = getPlayerIndexAfterRoll(diceRoll);
move(newIndex);
System.out.println("change turns");
swapTurns();
System.out.println(isComputerTurn());
gameGUI.updateCurrentTurnLabel();
if (newIndex == GameBoard.WIN_POINT) {
boolean restartGame = gameBoard.playAgainOrExit();
if (restartGame) {
Player winner = gameBoard.getCurrentPlayer();
gameGUI.updateScore(winner);
gameGUI.playAgain();
} else {
System.exit(0);
}
}
////////////////////////////////////////////////////////////////////////////
}
});
}
}, GameBoard.DICE_ROLL_DELAY, TimeUnit.SECONDS);
// here we wait and don't go to next actions until run() method of TimerTask is finished.
// we can specify wait timeout value if needed.
scheduledFuture.get();
// try to r
另一个问题是Platform.runLater调用也是异步的，您将无法获得任何结果。在Java FX Task中可以用来实现相同的目的：
Platform.runLater and Task in JavaFX
竞争示例如下：
public void rollTheDiceAndMove() {
int diceRoll = gameBoard.rollDice();
// delay for dice roll.
ScheduledFuture<Void> scheduledFuture = Executors.newScheduledThreadPool(1).schedule(
new Task<Void>() {
@Override
public Void call() {
////////////////////////////////////////////////////////////////////////////
gameGUI.indicateDiceRoll(diceRoll);
int newIndex = getPlayerIndexAfterRoll(diceRoll);
ScheduledFuture<Void> moveScheduledFuture = move(newIndex);
if(moveScheduledFuture != null) {
moveScheduledFuture.get();
}
System.out.println("change turns");
swapTurns();
System.out.println(isComputerTurn());
gameGUI.updateCurrentTurnLabel();
if (newIndex == GameBoard.WIN_POINT) {
boolean restartGame = gameBoard.playAgainOrExit();
if (restartGame) {
Player winner = gameBoard.getCurrentPlayer();
gameGUI.updateScore(winner);
gameGUI.playAgain();
} else {
System.exit(0);
}
}
////////////////////////////////////////////////////////////////////////////
return null;
}
}, GameBoard.DICE_ROLL_DELAY, TimeUnit.SECONDS);
scheduledFuture.get();
// try to recursively call this method again for computer turn.
}
public ScheduledFuture<Void> move(int currentIndex) {
int[] newCoordinates = gameBoard.getBoardCoordinates(GameBoard.NUMBER_OF_TILES - currentIndex);
gameBoard.getCurrentPlayer().getPlayerPiece().setPosition(currentIndex);
gameGUI.movePieceImages(newCoordinates[0], newCoordinates[1]);
if (gameBoard.getTile(currentIndex).containsLadderOrSnake()) {
// delay for moving to second tile.
return Executors.newScheduledThreadPool(1).schedule(
new Task<Void>() {
@Override
public Void call() {
int updatedIndex = gameBoard.getUpdatedPosition(currentIndex);
move(updatedIndex);
}
}, GameBoard.SECOND_MOVE_DELAY *2, TimeUnit.SECONDS);
// we need to return 'cause of recursion. Swap turns will be executed twice, one time for the initial call and the other time on above line.
}
return null;
}
答案 1 :(得分：1)
创建项目的副本，然后尝试使用PauseTransition。
public void rollTheDiceAndMove()
{
int diceRoll = gameBoard.rollDice();
System.out.println("Player: rolling the dice");
PauseTransition pause = new PauseTransition(Duration.seconds(1));
pause.setOnFinished(event ->{
System.out.println("1 second after rolling the dice");
gameGUI.indicateDiceRoll(diceRoll);
int newIndex = getPlayerIndexAfterRoll(diceRoll);
playerMove(newIndex);
if(checkWin(Player))
{
System.out.println("Player won!");
}
else
{
System.out.println("change turns");
swapTurns();
System.out.println(isComputerTurn());
gameGUI.updateCurrentTurnLabel();
computerRollDiceAndMove();
}
});
pause.play();
}
public void computerRollDiceAndMove()
{
int diceRoll = gameBoard.rollDice();
System.out.println("Computer: rolling the dice");
PauseTransition pause = new PauseTransition(Duration.seconds(1));
pause.setOnFinished(event ->{
System.out.println("1 second after rolling the dice");
gameGUI.indicateDiceRoll(diceRoll);
int newIndex = getComputerIndexAfterRoll(diceRoll);
computerMove(newIndex);
if(checkWin(computer))
{
System.out.println("Computer won!");
}
else{
System.out.println(isComputerTurn());
gameGUI.updateCurrentTurnLabel();
}
});
pause.play();
}