Java：如何在wait（）中区分虚假唤醒和超时

Question

Here is a case where a thread is waiting for notify() or a timeout. Here a while loop is added to handle spurious wake up.

boolean dosleep = true;
while (dosleep){
    try {
        wait(2000);
        /**
         * Write some code here so that
         * if it is spurious wakeup, go back and sleep.
         * or if it is timeout, get out of the loop. 
         */

    } catch (InterruptedException e) {
        e.printStackTrace();
    }
}

In this case how can I distinguish between a spurious wake up and time out? If it is a spurious wake up, i need to go back and wait. And if it is a timeout, i need to get out of the loop.

I can easily identify the case of notify(), because i will be setting the dosleep variable to false while notify() call.

EDIT: i am using 1.4 java version, due to embedded project requirement. I cannot use Condition as it is available only post 1.5.

Thanks in advance.

Answer 1

You could do this:

boolean dosleep = true;
long endTime = System.currentTimeMillis() + 2000;
while (dosleep) {
    try {
        long sleepTime = endTime - System.currentTimeMillis();
        if (sleepTime <= 0) {
            dosleep = false;
            } else {
            wait(sleepTime);
        }
    } catch ...
}

That should work fine in Java 1.4, and it will ensure that your thread sleeps for at least 2000ms.

Answer 2

You need to keep track of your timeout if you want to distinguish the two cases.

long timeout = 2000;
long timeoutExpires = System.currentTimeMillis() + timeout;
while(dosleep) {
  wait(timeout);
  if(System.currentTimeMillis() >= timeoutExpires) {
    // Get out of loop
    break;
  }
}

That said, denis's recommendation of using the Condition class is the better way to do this.

Answer 3

I believe Lock s and Condition will better fit your need in this case. Please check the javadocs for Condition.awaitUntil() - it has an example of usage