我很好的JavaScript承諾。所以我想試用Java Futures(仍然停留在Java 7中)。但是這些Java期貨對我來說沒有意義。下面是一個修整和改進的journaldev版本:Java未來:有人可以解釋我的程序流程
import java.util.concurrent.*;
public class FutureTest {
static class MyCallable implements Callable<String> {
private long waitTime;
MyCallable(int timeInMillis){
this.waitTime=timeInMillis;
}
@Override
public String call() throws Exception {
Thread.sleep(waitTime);
return Thread.currentThread().getName();
}
}
public static void main(String[] args) throws Exception {
MyCallable callable1 = new MyCallable(500);
MyCallable callable2 = new MyCallable(1000);
FutureTask<String> futureTask1 = new FutureTask<String>(callable1);
FutureTask<String> futureTask2 = new FutureTask<String>(callable2);
ExecutorService executor = Executors.newFixedThreadPool(2);
executor.execute(futureTask2);
executor.execute(futureTask1);
while (true) {
try {
boolean done1 = futureTask1.isDone();
boolean done2 = futureTask2.isDone();
if(futureTask1.isDone() && futureTask2.isDone()){
System.out.println("Done");
executor.shutdown();
return;
}
System.out.println("Done1:" + done1 + " - 2:" + done2);
String x = futureTask1.get(100L, TimeUnit.MILLISECONDS);
if (x != null)
System.out.println("FutureTask1 output="+x);
else
System.out.println("Waiting for FutureTask1");
String s = futureTask2.get(200L, TimeUnit.MILLISECONDS);
if(s != null)
System.out.println("FutureTask2 output="+s);
else
System.out.println("Waiting for FutureTask2");
Thread.sleep(100);
} catch(TimeoutException e) {}
}
}
}
它的輸出是:
Done1:false - 2:false
Done1:false - 2:false
Done1:false - 2:false
Done1:false - 2:false
Done1:false - 2:false
FutureTask1 output=pool-1-thread-2
Done1:true - 2:false
FutureTask1 output=pool-1-thread-2
Done1:true - 2:false
FutureTask1 output=pool-1-thread-2
FutureTask2 output=pool-1-thread-1
Done
爲什麼在Waiting for FutureTaskX
系統超時不執行?我希望主線程能夠循環和系統輸出Waiting for ...
,直到期貨得到解決。
我對解決這個問題的不同方式不感興趣,只是在這段代碼的程序流程中。謝謝。
*請*獲得使用括號甚至一個行塊的習慣。它會爲你節省很多錯誤,對於我們其他人來說更容易閱讀。 – slim