Question

编辑：找到答案！尽管CAG确实让我走上正轨，但我会奖励他。我提供了正确的答案。

我正在使用Canvas在JavaFX中制作Snake游戏。

我让游戏在一个循环中运行：

通过设置打印网格的可视化表示正确的VBox细胞的背景颜色。
等待输入（Thread.sleep（1000）。
生成下一个视觉效果。

问题是，如果我使用Thread.sleep（），我的画布根本不会加载。然而，在幕后，游戏仍在运行，直到我撞墙而死。

我在这里做错了吗？ thread.sleep（）是否暂停了加载和显示JavaFX节点的能力？

    Thread gameThread = new Thread() {

        @Override
        public synchronized void start() {
            super.start();
            printGridToGUI();
            while (KEEP_PLAYING) {
                generateNextGrid();
                try {
                    Thread.sleep(1000);
                } catch (InterruptedException ex) {
                    Logger.getLogger(SnakeGUIController.class.getName()).log(Level.SEVERE, null, ex);
                }
                Platform.runLater(() -> {
                    printGridToGUI();
                });
            }
            /*Stop continuing to play. You either won or lost.*/
            if (WON_GAME) {
                System.out.println("Congratulations!");
            } else {
                System.out.println("You lose.");
            }
        }
    };
    gameThread.start();

其中printGrid（）是：

/**
 * Prints the grid, with chars in place of on and off areas.
 */
public void printGridToGUI() {
    resetCanvas();
    for (Coordinate c : coordinates) {
        drawCell(c.row, c.col, true);
    }
    drawCell(food.row, food.col, true);
}

和resetCanvas是：

/**
 * Clears the boolean array, setting all values to false. A quick way to
 * wipe the grid.
 */
public final void resetCanvas() {
    /*Lay out the grid on the canvas.*/
    GraphicsContext gc = canvas.getGraphicsContext2D();
    for (int row = 0; row < GRID_SIZE; row++) {
        for (int col = 0; col < GRID_SIZE; col++) {
            drawCell(row, col, false);
        }
    }
}

和drawCell是：

/**
 * Draws a cell on the canvas at the specified row and col. The row, col
 * coordinates are translated into x,y coordinates for the graphics context.
 *
 * @param row The row of the cell to paint.
 * @param col The col of the cell to paint.     
 * @param cellON The state of the cell, if it is on or off.
 */
private void drawCell(int row, int col, boolean cellON) {
    /*Translate the row, col value into an x-y cartesian coordinate.*/
    int xCoord = 0 + col * CELL_SIZE;
    int yCoord = 0 + row * CELL_SIZE;
    /*Draw on the canvas.*/
    GraphicsContext gc = canvas.getGraphicsContext2D();
    gc.setFill(Color.BLACK);
    gc.fillRect(xCoord, yCoord, CELL_SIZE, CELL_SIZE);
    if (!cellON) {
        gc.setFill(Color.WHITE);
        int BORDER = 1;
        gc.fillRect(xCoord + BORDER, yCoord + BORDER, CELL_SIZE - BORDER, CELL_SIZE - BORDER);
    }
}

Answer 1

我的猜测是你在FX应用程序线程上调用Thread.sleep()。该线程负责保持您的UI响应，因此使其处于睡眠状态会冻结您的UI，同时让您的游戏机制处于响应状态（假设这些是从FX应用程序线程执行的）。

解决方案是在新线程中执行游戏循环，如下所示：

Thread gameLoop = new Thread(() ->
{
     while (KEEP_PLAYING)
     {
         printGrid(); //<- I assume this prints the state of the grid to the console, and so is safe to execute off of the FX Application Thread
         try
         {
             Thread.sleep(1000);
         }
         catch (InterruptedException ex) {}
         Platform.runLater(() ->
         {
             generateNextGrid(); //<- execute this on the FX Application Thread as it modifies your UI
         });
     }
     if (WON_GAME)
     {
         ...
     }
     else
     {
         ...
     }
});
gameLoop.start();

这可以防止任何冻结，只要您没有执行任何长时间运行的任务或在FX应用程序线程上调用睡眠。

Answer 2

想出来！我不得不使用JavaFX的并发包并利用Task而不是简单的Thread。

    Task<Boolean> gameTask = new Task() {

        @Override
        protected Object call() throws Exception {
            while (KEEP_PLAYING) {
                generateNextGrid();
                try {
                    Thread.sleep(GAME_SPEED_DELAY);
                } catch (InterruptedException ex) {
                    Logger.getLogger(SnakeGUIController.class.getName()).log(Level.SEVERE, null, ex);
                }
                Platform.runLater(() -> printGridToGUI());
            }
            /*Stop continuing to play. You either won or lost.*/
            if (WON_GAME) {
                System.out.println("Congratulations!");
            } else {
                System.out.println("You lose.");
            }
            return true;
        }
    };
    Thread gameThread = new Thread(gameTask);
    gameThread.start();

JavaFX Snake Thread.Sleep（）不加载FXML

2 个答案: