java 多线程面试题4个线程按顺序打印ABCD

4个线程,一个打印A,一个打印B,一个打印C,一个打印D,要求按 ABCDABCD 的顺序打印

多线程同步,首先想到的就是 synchronized 和 wait notify 机制。

import java.util.concurrent.CountDownLatch;

public class Main {

    final static int N = 4;
    static int nextNum = 1;

    public static void main(String[] args) {

        Object o1 = new Object();
        Object o2 = new Object();
        Object o3 = new Object();
        Object o4 = new Object();

        CountDownLatch countDownLatch = new CountDownLatch(4);

        new Thread(() -> {
            for (int i = 0; i < N; i++) {
                synchronized (o1) {
                    while (nextNum != 1) {
                        try {
//                            System.out.println("o1 wait");
                            o1.wait();
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }
                    }
                }
                System.out.println("A");
                synchronized (o2) {
                    nextNum = 2;
//                    System.out.println("o2 notifyAll");
                    o2.notifyAll();
                }
            }
            countDownLatch.countDown();
        }).start();

        new Thread(() -> {
            for (int i = 0; i < N; i++) {
                synchronized (o2) {
                    while (nextNum != 2) {
                        try {
//                            System.out.println("o2 wait");
                            o2.wait();
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }
                    }
                }
                System.out.println("B");
                synchronized (o3) {
                    nextNum = 3;
//                    System.out.println("o3 notifyAll");
                    o3.notifyAll();
                }
            }
            countDownLatch.countDown();
        }).start();

        new Thread(() -> {
            for (int i = 0; i < N; i++) {
                synchronized (o3) {
                    while (nextNum != 3) {
                        try {
//                            System.out.println("o3 wait");
                            o3.wait();
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }
                    }
                }
                synchronized (o4) {
                    System.out.println("C");
                    nextNum = 4;
//                    System.out.println("o4 notifyAll");
                    o4.notifyAll();
                }
            }
            countDownLatch.countDown();
        }).start();

        new Thread(() -> {
            for (int i = 0; i < N; i++) {
                synchronized (o4) {
                    while (nextNum != 4) {
                        try {
//                            System.out.println("o4 wait");
                            o4.wait();
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }
                    }
                }
                synchronized (o1) {
                    System.out.println("D");
                    nextNum = 1;
//                    System.out.println("o1 notifyAll");
                    o1.notifyAll();
                }
            }
            countDownLatch.countDown();
        }).start();

        try {
            countDownLatch.await();
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
    }
}

4 个线程,根据条件分别阻塞在不同的 object 锁上

我只用一个 Object 锁可以吗???当然可以只用一个锁,4个线程同时竞争一个锁,然后同时唤醒再竞争,再附加一个条件变量就行了。代码如下:

import java.util.concurrent.CountDownLatch;

public class Main {

    final static int N = 4;
    static int nextNum = 1;

    public static void main(String[] args) {

        Object o1 = new Object();

        CountDownLatch countDownLatch = new CountDownLatch(4);

        new Thread(() -> {
            for (int i = 0; i < N; i++) {
                synchronized (o1) {
                    while (nextNum != 1) {
                        try {
                            o1.wait();
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }
                    }
                    System.out.println("A");
                    nextNum = 2;
                    o1.notifyAll();
                }
            }
            countDownLatch.countDown();
        }).start();

        new Thread(() -> {
            for (int i = 0; i < N; i++) {
                synchronized (o1) {
                    while (nextNum != 2) {
                        try {
                            o1.wait();
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }
                    }
                    System.out.println("B");
                    nextNum = 3;
                    o1.notifyAll();
                }
            }
            countDownLatch.countDown();
        }).start();

        new Thread(() -> {
            for (int i = 0; i < N; i++) {
                synchronized (o1) {
                    while (nextNum != 3) {
                        try {
//                            System.out.println("o3 wait");
                            o1.wait();
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }
                    }
                    System.out.println("C");
                    nextNum = 4;
                    o1.notifyAll();
                }
            }
            countDownLatch.countDown();
        }).start();

        new Thread(() -> {
            for (int i = 0; i < N; i++) {
                synchronized (o1) {
                    while (nextNum != 4) {
                        try {
                            o1.wait();
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }
                    }
                    System.out.println("D");
                    nextNum = 1;
                    o1.notifyAll();
                }
            }
            countDownLatch.countDown();
        }).start();

        try {
            countDownLatch.await();
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
    }
}

notifyAll 会唤醒所有等待线程,所有线程再同时竞争锁。notify 会随机唤醒一个等待线程。随机唤醒一个线程,被唤醒的线程条件不一定满足,又会进入等待,在高并发场景下性能不好。这就是为什么又出现了 Condition 机制。

我用 synchronized 方法可以吗???当然可以,synchronized 方法默认会锁住当前对象,就不用 new 一个对象锁了。

下面看看 Condition 实现

import java.util.concurrent.CountDownLatch;
import java.util.concurrent.locks.Condition;
import java.util.concurrent.locks.ReentrantLock;

public class Main {

    final static int N = 4;
    static int nextNum = 1;

    public static void main(String[] args) {

        ReentrantLock lock = new ReentrantLock();
        Condition condition1 = lock.newCondition();
        Condition condition2 = lock.newCondition();
        Condition condition3 = lock.newCondition();
        Condition condition4 = lock.newCondition();

        CountDownLatch countDownLatch = new CountDownLatch(4);

        new Thread(() -> {
            for (int i = 0; i < N; i++) {
                lock.lock();
                try {
                    while (nextNum != 1) {
                        try {
//                            System.out.println("condition1 await");
                            condition1.await();
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }
                    }
                    System.out.println("A");
                    nextNum = 2;
//                    System.out.println("condition2 signalAll");
                    condition2.signalAll();
                } finally {
                    lock.unlock();
                }
            }
            countDownLatch.countDown();
        }).start();

        new Thread(() -> {
            for (int i = 0; i < N; i++) {
                lock.lock();
                try {
                    while (nextNum != 2) {
                        try {
//                            System.out.println("condition2 await");
                            condition2.await();
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }
                    }
                    System.out.println("B");
                    nextNum = 3;
//                    System.out.println("condition3 signalAll");
                    condition3.signalAll();
                } finally {
                    lock.unlock();
                }
            }
            countDownLatch.countDown();
        }).start();

        new Thread(() -> {
            for (int i = 0; i < N; i++) {
                lock.lock();
                try {
                    while (nextNum != 3) {
                        try {
//                            System.out.println("condition3 await");
                            condition3.await();
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }
                    }
                    System.out.println("C");
                    nextNum = 4;
//                    System.out.println("condition4 signalAll");
                    condition4.signalAll();
                } finally {
                    lock.unlock();
                }
            }
            countDownLatch.countDown();
        }).start();

        new Thread(() -> {
            for (int i = 0; i < N; i++) {
                lock.lock();
                try {
                    while (nextNum != 4) {
                        try {
//                            System.out.println("condition4 await");
                            condition4.await();
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }
                    }
                    System.out.println("D");
                    nextNum = 1;
//                    System.out.println("condition1 signalAll");
                    condition1.signalAll();
                } finally {
                    lock.unlock();
                }
            }
            countDownLatch.countDown();
        }).start();

        try {
            countDownLatch.await();
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
    }
}

ReentrantLock 就等价于 Object 锁。condition.await() 等价于 object.wait(), condition.signal() 等价于 object.notify()。

我可以只用一个条件变量吗???当然可以,那样和上面只用一个对象锁没什么区别。

一个ReentrantLock 绑定多个 Condition 对象意味着什么???为什么用 Condition 写生产者消费者效率高???

synchronized方法 或 Object锁 都只有一个等待队列,所有阻塞的线程都挂在当前对象 或一个 object 实例上。 notify 会随机唤醒队列中的一个线程,而唤醒的线程不一定满足条件,所以又进入等待,这样效率低了一点。

而 ReentrantLock 绑定多个 Condition 对象, 每个 Condition 对象上都有一个等待队列,我们可以只唤醒一个Condition 对象上等待的线程 。也就是 Condition 能实现粒度更细的锁。


看别人写的信号量机制

import java.util.concurrent.CountDownLatch;
import java.util.concurrent.Semaphore;

public class Main {

    final static int N = 4;

    public static void main(String[] args) {

        Semaphore semaphore1 = new Semaphore(1);
        Semaphore semaphore2 = new Semaphore(0);
        Semaphore semaphore3 = new Semaphore(0);
        Semaphore semaphore4 = new Semaphore(0);

        CountDownLatch countDownLatch = new CountDownLatch(4);

        new Thread(() -> {
            for (int i = 0; i < N; i++) {
                try {
                    semaphore1.acquire();
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                System.out.println("A");
                semaphore2.release();
            }
            countDownLatch.countDown();
        }).start();

        new Thread(() -> {
            for (int i = 0; i < N; i++) {
                try {
                    semaphore2.acquire();
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                System.out.println("B");
                semaphore3.release();
            }
            countDownLatch.countDown();
        }).start();

        new Thread(() -> {
            for (int i = 0; i < N; i++) {
                try {
                    semaphore3.acquire();
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                System.out.println("C");
                semaphore4.release();
            }
            countDownLatch.countDown();
        }).start();

        new Thread(() -> {
            for (int i = 0; i < N; i++) {
                try {
                    semaphore4.acquire();
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                System.out.println("D");
                semaphore1.release();
            }
            countDownLatch.countDown();
        }).start();

        try {
            countDownLatch.await();
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
    }
}

我可以只用一个信号量吗???当然可以,那样还得附加一个条件变量,跟只用一个 object 锁也差不多。

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