問題描述

問題解答

如果你了解信號量的實現機制，那么這道題目也是一個意思。

public class Test { private final Integer maxCounter = 3; private Integer current = 0; public void call1() {//在這里補充代碼synchronized (this) { try {while (current.equals(maxCounter)) { // 請求 到達上限 wait();} } catch (InterruptedException ex) { } current++; notifyAll();}call2(current);synchronized (this) { try {while (current == 0) { wait();} } catch (InterruptedException ex) { } current--; notifyAll();} } private void call2(Integer current) {System.out.println(Thread.currentThread().getName() + ': I’m called ' + current);// 下面的休眠 2 秒鐘用于測試try { Thread.sleep(2000);} catch (InterruptedException ex) { ex.printStackTrace(System.err);} } static class TestThread implements Runnable {private Test t;public TestThread(Test t) { this.t = t;}@Overridepublic void run() { t.call1();} } public static void main(String[] args) {Test t1 = new Test();TestThread tt = new TestThread(t1);for (int i = 0; i < 10; i++) { Thread t = new Thread(tt, 'Thread-' + i); t.start();} }}

運行這段代碼，你可以發現每 2 秒內，最多只有 3 （maxCounter）個線程在運行。

用CountDownLatch。。。