Question

我有这个方法：

public class NFeRoboServiceImpl implements NFeRoboService {

 private static volatile ThreadPoolExecutor executor;

    private ThreadPoolExecutor getExecutor() {
        if (executor == null) {
            synchronized (NFeRoboServiceImpl.class) {
                if (executor == null) {
                    executor = new ThreadPoolExecutor(2, 10, 10, TimeUnit.SECONDS,
                            new ArrayBlockingQueue<Runnable>(100, true), new RetryRejectedExecutionHandler());
                    Monitor monitor = new Monitor(nFePersistenceService, executor, 3);
                    Thread monitorThread = new Thread(monitor);
                    monitorThread.start();
                }
            }
        }
        return executor;
    }
}

问题很奇怪，当我把光标放在新的ThreadPoolExecutor之后的'executor'之上时，它说对象是null，当我选择执行器并且 ctrl + shift + i ，它显示了我创建的对象。

我在线程执行器上有空：

getExecutor().execute(worker);

Answer 1

我认为，执行者是静态的。在这种情况下，您应该将其定义为volatile。请参阅＆＃34;仔细检查＆＃34;回答Should I synchronize a static volatile variable?

见：

http://www.cs.umd.edu/~pugh/java/memoryModel/DoubleCheckedLocking.html

由于Bean-Convention，当eclipse试图确定执行者的值时，可能会使用getExecutor。如果将隐式初始化函数重命名为与fieldname无关的内容会发生什么？

ThreadPoolExecutor返回null

1 个答案: