public class Test{
private final Integer maxCounter = 3;
private Integer current = 0;
public void call1(){
//在这里补充代码
}
private void call2(Integer current){
System.out.println("I'm called"+current);
}
static class TestThread implements Runnable{
private Test t;
public TestThread(Test t){
this.t = t;
}
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);
t.start();
}
}
}
忘了,大概是这样;
要求并发调用call2的线程数小于maxCounter(也即3),请问如何添加代码(在call1中)?
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。
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。