Heim > Fragen und Antworten > Hauptteil
怪我咯2017-04-18 10:48:36
如果你了解信号量的实现机制,那么这道题目也是一个意思。
public class Test {
private final Integer maxCounter = 3;
private Integer current = 0;
public void call1() {
//在这里补充代码
synchronized (this) {
try {
while (current.equals(maxCounter)) { // 请求 到达上限
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 秒钟用于测试
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;
}
@Override
public 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();
}
}
}
运行这段代码,你可以发现每 2 秒内,最多只有 3 (maxCounter)个线程在运行。