Java - Most optimal way to poll for timeout

Question

I have code such that:

while(isResponseArrived)
   Thread.yield();

But what I'd really like to do is something like this:

long startTime = System.currentTimeInMilliseconds();

while(isResponseArrived)
{
   if(isTimeoutReached(startTime))
       throw new TimeOutExcepton(); 
   Thread.yield();
}

I'm not yet sure about throwing an exception or not (it's not important for this question), but what I'd like to know is how to make it as performant as possible, so I'm not chugging away on the processor. In other words how can I make isTimeoutReached(long startTime) as performance friendly as possible.

I tested:

for(int x=0; x<99999999; x++)
    System.nanoTime();

versus

for(int x=0; x<99999999; x++)
    System.currentTimeInMilliseconds();

And the difference was minimal, less than 10% in terms of time to complete

I also look at using Thread.sleep(), but I really want the user to be notified as quickly as possible if there's an update and the processor is just waiting. Thread.yield() doesn't get the processor churning, it's just a NOP, giving anyone else processor priority, until it's good to go.

Anyways, what's the best way to test for a timeout without throttling the CPU? Is this the right method?

Answer 1

I think it would be more efficient to use wait / notify

boolean arrived;

public synchronized void waitForResponse(long timeout) throws InterruptedException, TimeoutException {
    long t0 = System.currentTimeMillis() + timeout;
    while (!arrived) {
        long delay = System.currentTimeMillis() - t0;
        if (delay < 0) {
            throw new TimeoutException();
        }
        wait(delay);
    }
}

public synchronized void responseArrived() {
    arrived = true;
    notifyAll();
}