java - 通过将函数移至其自己的线程来加速 Java 代码

Question

当调用 `Moon m = new Moon(); 时，最底部的以下代码会变慢。为了解决这个问题，我想将以下代码放入线程或其他东西中，以便 MoonCalTimer 更快退出并将值返回给其他函数。知道如何将其放入线程或类似内容中吗？

Moon m = new Moon();
isWaning = m.isWaning();     
fullMoon = MoonPhaseFinder.findFullMoonFollowing(Calendar.getInstance());
newMoon = MoonPhaseFinder.findNewMoonFollowing(Calendar.getInstance());

代码:

Timer moonCalTimer = new Timer();
moonCalTimer.scheduleAtFixedRate(new TimerTask() 
{
@Override
public void run() 
{
try {

//other code here...
//other code here...
//other code here...

Moon m = new Moon();
isWaning = m.isWaning();     
fullMoon = MoonPhaseFinder.findFullMoonFollowing(Calendar.getInstance());
newMoon = MoonPhaseFinder.findNewMoonFollowing(Calendar.getInstance());

} 

catch (Exception ex) 
{
Logger.getLogger(JavaFXApplication4.class.getName()).log(Level.SEVERE, null, ex);
}


}
}, 0, 3600000); 

Answer 1

您可以将计算包装在 Future 中。然后，您将创建 future 并在自己的线程中启动计算，并将 Future 返回给调用者。这将允许调用者在计算完成之前获得其 Future 结果。

private final ExecutorService pool = Executors.newFixedThreadPool(10);

public Future<Boolean> isWaning(final Moon moon) throws IOException {
  return pool.submit(new Callable<Boolean>() {
    @Override
    public Boolean call() throws Exception {
      return moon.isWaning();
    }

  }
  );
}

public Future<Boolean> fullMoonFollowing(final Moon moon) throws IOException {
  return pool.submit(new Callable<Boolean>() {
    @Override
    public Boolean call() throws Exception {
      return MoonPhaseFinder.findFullMoonFollowing(Calendar.getInstance());
    }

  }
  );
}

public Future<Boolean> newMoonFollowing(final Moon moon) throws IOException {
  return pool.submit(new Callable<Boolean>() {
    @Override
    public Boolean call() throws Exception {
      return MoonPhaseFinder.findNewMoonFollowing(Calendar.getInstance());
    }

  }
  );
}