我尝试过延迟(或休眠)我的Java程序,但是出现了一个错误。
我无法使用Thread.sleep(x)或wait()。同样的错误信息出现:
interruptedexception;必须被捕获或宣布被丢弃。
在使用Thread.sleep()或wait()方法之前,是否需要任何步骤?
我尝试过延迟(或休眠)我的Java程序,但是出现了一个错误。
我无法使用Thread.sleep(x)或wait()。同样的错误信息出现:
interruptedexception;必须被捕获或宣布被丢弃。
在使用Thread.sleep()或wait()方法之前,是否需要任何步骤?
当前回答
public static void main(String[] args) throws InterruptedException {
//type code
short z=1000;
Thread.sleep(z);/*will provide 1 second delay. alter data type of z or value of z for longer delays required */
//type code
}
eg:-
class TypeCasting {
public static void main(String[] args) throws InterruptedException {
short f = 1;
int a = 123687889;
short b = 2;
long c = 4567;
long d=45;
short z=1000;
System.out.println("Value of a,b and c are\n" + a + "\n" + b + "\n" + c + "respectively");
c = a;
b = (short) c;
System.out.println("Typecasting...........");
Thread.sleep(z);
System.out.println("Value of B after Typecasting" + b);
System.out.println("Value of A is" + a);
}
}
其他回答
当使用Android(我唯一一次使用Java)时,我会建议使用处理程序而不是让线程进入睡眠状态。
final Handler handler = new Handler();
handler.postDelayed(new Runnable() {
@Override
public void run() {
Log.i(TAG, "I've waited for two hole seconds to show this!");
}
}, 2000);
参考:http://developer.android.com/reference/android/os/Handler.html
public static void main(String[] args) throws InterruptedException {
//type code
short z=1000;
Thread.sleep(z);/*will provide 1 second delay. alter data type of z or value of z for longer delays required */
//type code
}
eg:-
class TypeCasting {
public static void main(String[] args) throws InterruptedException {
short f = 1;
int a = 123687889;
short b = 2;
long c = 4567;
long d=45;
short z=1000;
System.out.println("Value of a,b and c are\n" + a + "\n" + b + "\n" + c + "respectively");
c = a;
b = (short) c;
System.out.println("Typecasting...........");
Thread.sleep(z);
System.out.println("Value of B after Typecasting" + b);
System.out.println("Value of A is" + a);
}
}
一种更简单的等待方法是使用System.currentTimeMillis(),它返回自UTC 1970年1月1日午夜以来的毫秒数。例如,等待5秒:
public static void main(String[] args) {
//some code
long original = System.currentTimeMillis();
while (true) {
if (System.currentTimeMillis - original >= 5000) {
break;
}
}
//more code after waiting
}
这样,您就不必处理线程和异常。 希望这能有所帮助!
看看这篇关于如何正确做到这一点的简短文章。
本质上:捕获InterruptedException。记住,您必须添加这个捕获块。这篇文章对此作了进一步的解释。
Thread.sleep()对于初学者来说很简单,可能适用于单元测试和概念证明。
但是请不要在产品代码中使用sleep()。最终sleep()可能会狠狠地咬你一口。
多线程/多核java应用程序使用“线程等待”概念的最佳实践。Wait释放线程持有的所有锁和监控器,这允许其他线程获得这些监控器,并在您的线程平静睡眠时继续执行。
下面的代码演示了该技术:
import java.util.concurrent.TimeUnit;
public class DelaySample {
public static void main(String[] args) {
DelayUtil d = new DelayUtil();
System.out.println("started:"+ new Date());
d.delay(500);
System.out.println("half second after:"+ new Date());
d.delay(1, TimeUnit.MINUTES);
System.out.println("1 minute after:"+ new Date());
}
}
DelayUtil实现:
import java.util.concurrent.TimeUnit;
import java.util.concurrent.locks.Condition;
import java.util.concurrent.locks.ReentrantLock;
public class DelayUtil {
/**
* Delays the current thread execution.
* The thread loses ownership of any monitors.
* Quits immediately if the thread is interrupted
*
* @param durationInMillis the time duration in milliseconds
*/
public void delay(final long durationInMillis) {
delay(durationInMillis, TimeUnit.MILLISECONDS);
}
/**
* @param duration the time duration in the given {@code sourceUnit}
* @param unit
*/
public void delay(final long duration, final TimeUnit unit) {
long currentTime = System.currentTimeMillis();
long deadline = currentTime+unit.toMillis(duration);
ReentrantLock lock = new ReentrantLock();
Condition waitCondition = lock.newCondition();
while ((deadline-currentTime)>0) {
try {
lock.lockInterruptibly();
waitCondition.await(deadline-currentTime, TimeUnit.MILLISECONDS);
} catch (InterruptedException e) {
Thread.currentThread().interrupt();
return;
} finally {
lock.unlock();
}
currentTime = System.currentTimeMillis();
}
}
}