Question

我试图创建一个接口，在实现时，方法内的任何内容都将在一个单独的线程上连续执行。

希望这个描述有意义，这就是想法。

接口：

public interface Painting {

void onPaint();
}

主题：

public class Paintable implements Runnable, Painting {

@Override
public void run() {
    onPaint();
}

@Override
public void onPaint() {
}
}

最后，目标是你实现绘画，这需要你实现和覆盖onPaint方法。然后，当执行该线程时，无论您决定放入该方法，都将在Paintable Thread上运行。

public class Main implements Painting {

private static final ScheduledThreadPoolExecutor THREAD_POOL_EXECUTOR = new ScheduledThreadPoolExecutor(3);

public static void main(String[] args) {
    THREAD_POOL_EXECUTOR.scheduleAtFixedRate(new Paintable(), 0, 100, TimeUnit.MILLISECONDS);

@Override
public void onPaint() {
System.out.println("This text should be continously printed, but is not.");
}
}

Answer 1

好的，所以你有一个绘图界面，并想把它转换成一个调用onPaint（）的Runnable。

你可以这样做：

public class PaintingRunnable implements Runnable {
    private final Painting painting;

    public PaintingRunnable(Painting painting) {
        this.painting = painting;
    }

    @Override
    public void run() {
        this.painting.onPaint();
    }
}

因此，您可以使用

从绘画中制作出Runnable

Runnable runnableThatCallsOnPaint = new PaintingRunnable(somePainting);

但这有点矫枉过正，因为你可以在没有任何使用lambda的类的情况下做同样的事情：

Runnable runnableThatCallsOnPaint = () -> somePaining.onPaint()

您甚至可以使用方法参考：

Runnable runnableThatCallsOnPaint = somePaining::onPaint

Answer 2

正如上面的回答中所提到的，我同意这一点。你必须创建一个模板类来定义你的步骤，并且在每个步骤中执行什么，你必须由子类或类的委托定义，这将执行精确的逻辑。

主要课程：

 public class Main implements Painting {

        private static final ScheduledThreadPoolExecutor THREAD_POOL_EXECUTOR = new ScheduledThreadPoolExecutor(3);

        public static void main(String[] args) {
            THREAD_POOL_EXECUTOR.scheduleAtFixedRate(new Paintable(new Main()), 0, 100, TimeUnit.MILLISECONDS);

        }

        @Override
        public void onPaint() {
            System.out.println("This text should be continously printed, but is not.");
        }
    }

可上漆的司机班

public class Paintable implements Runnable {

    private final Painting painting;

    public Paintable(Painting painting) {
        this.painting = painting;
    }

    @Override
    public void run() {
        painting.onPaint();
    }

}

如何使接口实现runnable？

2 个答案: