I get exception when using Thread.sleep(x) or wait()

user_200876 0
Viewed 892595

I have tried to delay - or put to sleep - my Java program, but an error occurs.

I'm unable to use Thread.sleep(x) or wait(). The same error message appears:

unreported exception java.lang.InterruptedException; must be caught or declared to be thrown.

Is there any step required before using the Thread.sleep() or wait() methods?

sleep java
13 Answers

A simpler way to wait is to use System.currentTimeMillis(), which returns the number of milliseconds since midnight on January 1, 1970 UTC. For example, to wait 5 seconds:

public static void main(String[] args) {
    //some code
    long original = System.currentTimeMillis();
    while (true) {
        if (System.currentTimeMillis - original >= 5000) {
            break;
        }
    }
    //more code after waiting
}

This way, you don't have to muck about with threads and exceptions. Hope this helps!

user_83875160

Use java.util.concurrent.TimeUnit:

TimeUnit.SECONDS.sleep(1);

Sleep for one second or

TimeUnit.MINUTES.sleep(1);

Sleep for a minute.

As this is a loop, this presents an inherent problem - drift. Every time you run code and then sleep you will be drifting a little bit from running, say, every second. If this is an issue then don't use sleep.

Further, sleep isn't very flexible when it comes to control.

For running a task every second or at a one second delay I would strongly recommend a [ScheduledExecutorService][1] and either [scheduleAtFixedRate][2] or [scheduleWithFixedDelay][3].

To run the method myTask every second (Java 8):

public static void main(String[] args) {
    final ScheduledExecutorService executorService = Executors.newSingleThreadScheduledExecutor();
    executorService.scheduleAtFixedRate(App::myTask, 0, 1, TimeUnit.SECONDS);
}

private static void myTask() {
    System.out.println("Running");
}
user_57700040
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