如何避免這種忙碌的等待？

Question

public int getChildrenCount(int groupPosition) { 
    if(children == null){ 
     new SalesRequest().execute(); // runs in other thread which 
             // initialises children with some value. 
     while(children == null){ 
      // I'm doin this to avoid null pointer Exception. 
      // So it comes out of the loop only when childern 
      // gets initialised. 
     } 
    } 
    return children.length; 
} 

但我不滿意我處理這個問題的方式。有一個更好的方法嗎？

Answer 1

您可以使用CountDownLatch等待另一個線程完成。

http://download.oracle.com/javase/1,5.0/docs/api/java/util/concurrent/CountDownLatch.html

Answer 2

有對這個問題的可能解決方案的倍數。最優雅的方式就像Eric上面提到的CountDownLatch。 這裏是你如何才能夠着手：

// Lock to signal Children are created 
CountDownLatch childrenReady = new CountDownLatch(1/*wait for one signal*/); 
public int getChildrenCount(int groupPosition) { 
    if(children == null){ 
     SalesRequest request = new SalesRequest(childrenReady /*pass on this lock to worker thread*/); 
     request().execute(); // runs in other thread which 
             // initialises children with some value. 
     childrenReady.await();  // Wait until salesRequest finishes 
     while(children == null){ 
      // I'm doin this to avoid null pointer Exception. 
      // So it comes out of the loop only when childern 
      // gets initialised. 
     } 
    } 
    return children.length; 
} 

在SalesRequest.execute方法，你可以有以下幾點：

// Populate and create children structure of the calling object 
// When done, signal to callee that we have finished creating children 
childrenReady.countDown(); // This will release the calling thread into the while loop 

此外，還要確保你沒有從UI線程，否則你的應用程序中調用getChildrenCount()將會掛起並且將放棄其響應，直到您從服務器獲得答案。