2017-07-28 45 views
3

我有一個如下所示的API,其中Baz是工作者的實施。 這個Bar需要是線程安全的,當與Baz的回調交互時,這會變得棘手。維護線程安全,同時防止死鎖可能同步回調

當前baz實例需要在回調中引用(可以在工作線程或同步調用)。 的意見應該顯示問題:

final class Bar { 
    final Lock lock = new ReentrantLock(); 
    Baz baz; // Guarded by lock. 

    void run() { // Called by any thread. 
    lock.lock(); 
    if (baz.isRunning()) { 
     lock.unlock(); 
     return; 
    } 
    baz = new Baz(); 
    // If it unlocks here, the next line may execute on the wrong Baz. 
    // If it doesn't unlock here, there will be a deadlock when done() is called synchronously. 
    // lock.unlock(); 
    baz.run(new Baz.Callback() { // May be called synchronously or by Baz worker thread. 
     @Override 
     public void done() { 
     lock.lock(); 
     baz = new Baz(); 
     lock.unlock(); 
     } 
    }); 
    } 
} 

有沒有正確地做這項工作,同時也不會導致死鎖的好方法?

編輯:更簡潔:

final class Foo { 
    final Lock lock = new ReentrantLock(); 

    void run() { 
    lock.lock(); 
    worker.enqueue(new Callback() { 
     @Override void complete() { 
     lock.lock(); // Could cause deadlock. 
     } 
    }); 
    lock.unlock(); 
    } 
} 

回答

0

不知道該completelly得到你所要實現的目標,但也許這是你在找什麼?

final class Bar { 
    final Lock lock = new ReentrantLock(); 
    Baz baz = new Baz(); 

    void run() { 
     if (!lock.tryLock()) { 
      return; 
     } 
     try { 
      CountdownLatch callbackFlag = new CountdownLatch(1); 
      baz.run(new Baz.Callback() { 
       @Override 
       public void done() { 
        callbackFlag.countDown(); 
       } 
      }); 
      try { 
       callbackFlag.await(); // better use overloaded method with max timeout waiting. you don't probably want to wait forever 
       baz = new Baz(); // do you really want to reinit Baz on each execution? 
      } catch (InterruptedException e) { 
       // decide what you want to happen here 
      } 
     } finally { 
      lock.unlock(); 
     } 
    } 
} 
+0

巴茲指派需要在回調,但。 –

+0

編輯澄清。 –

+0

爲什麼它應該在回調中被引用?我發佈的代碼確保baz不會重新啓動直到回調完成 –