public class Test extends Thread { static int x=0; public static void a(){ x++; System.out.println(x); } public void run(){ a(); } public static void main(String[] args) throws InterruptedException { Thread t1=new Thread(()->Test.a()); Thread t2=new Thread(()->Test.a()); Thread t3=new Thread(()->Test.a()); Thread t4=new Thread(()->Test.a()); Thread t5=new Thread(()->Test.a()); t1.start(); t1.join(); t2.start(); t2.join(); t3.start(); t3.join(); t4.start(); t4.join(); t5.start(); t5.join(); } }用信号量实现输出总为5

可以使用信号量来限制同时执行 `a()` 方法的线程数量，从而保证输出总为5。代码如下：

import java.util.concurrent.Semaphore;

public class Test extends Thread {
    static int x = 0;
    static Semaphore semaphore = new Semaphore(1);

    public static void a() {
        try {
            semaphore.acquire();
            x++;
            System.out.println(x);
        } catch (InterruptedException e) {
            e.printStackTrace();
        } finally {
            semaphore.release();
        }
    }

    public void run() {
        a();
    }

    public static void main(String[] args) throws InterruptedException {
        Thread t1 = new Thread(() -> Test.a());
        Thread t2 = new Thread(() -> Test.a());
        Thread t3 = new Thread(() -> Test.a());
        Thread t4 = new Thread(() -> Test.a());
        Thread t5 = new Thread(() -> Test.a());
        t1.start();
        t2.start();
        t3.start();
        t4.start();
        t5.start();
        t1.join();
        t2.join();
        t3.join();
        t4.join();
        t5.join();
    }
}

在 `a()` 方法中，使用 `semaphore.acquire()` 申请信号量，如果已经有5个线程在执行，则会被阻塞，直到有一个线程执行完毕后释放信号量。同时在 `finally` 中释放信号量。这样就可以保证同时只有最多5个线程在执行 `a()` 方法，从而保证输出总为5。