我该如何避免这种忙碌的等待？

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;
}

但我对我处理这个问题的方式并不满意。 有一个更好的方法吗？

Answer 1

您可以使用CountDownLatch等待其他线程完成。

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

Answer 2

这个问题有多种可能的解决方案。 最优雅的方式是Eric上面提到的CountDownLatch。 以下是您可以继续的方式：

// 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方法中，您可以具有以下内容：

// 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，请确保您没有从UI线程调用getChildrenCount() ，否则您的应用程序将挂起并将失去其响应性，直到您从服务器获得答案。