Java wait() notify()循环打印A和B

Question

public class Test {
    
    public static Object object = new Object();
    public static boolean printA = true;
    
    public static void main(String[] args) {
        ThreadA threadA = new ThreadA();
        threadA.start();
        
        ThreadB threadB = new ThreadB();
        threadB.start();
    }
}
 
class ThreadA extends Thread {
    
    @Override
    public void run() {
        for(int i = 0; i < 10; i++) {
            synchronized (Test.object) {
                if(!Test.printA) {
                    try {
                        Test.object.wait();
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                } else {
                    System.out.println("A");
                    Test.printA = false;
                    Test.object.notify();
                }
            }
        }
    }
}

class ThreadB extends Thread {
    
    @Override
    public void run() {
        for(int i = 0; i < 10; i++) {
            synchronized (Test.object) {
                if(Test.printA) {
                    try {
                        Test.object.wait();
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                } else {
                    System.out.println("B");
                    Test.printA = true;
                    Test.object.notify();
                }
            }
        }
    }
}

运行以上代码，为什么会一直不结束无限wait下去？

------------------ 分割线，以下是正确的写法 -------------------

实际上线程中不应该有else，把else注释掉就正确了。 如果加上else后，该线程会在wait后不再notify，导致另一个线程无限wait。

public class Test {
    
    public static Object object = new Object();
    public static boolean printA = true;
    
    public static void main(String[] args) {
        ThreadA threadA = new ThreadA();
        threadA.start();
        
        ThreadB threadB = new ThreadB();
        threadB.start();
    }
}
 
class ThreadA extends Thread {
    
    @Override
    public void run() {
        for(int i = 0; i < 10; i++) {
            synchronized (Test.object) {
                if(!Test.printA) {
                    try {
                        Test.object.wait();
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                }/* else {*/
                    System.out.println("A");
                    Test.printA = false;
                    Test.object.notify();
                /*}*/
            }
        }
    }
}

class ThreadB extends Thread {
    
    @Override
    public void run() {
        for(int i = 0; i < 10; i++) {
            synchronized (Test.object) {
                if(Test.printA) {
                    try {
                        Test.object.wait();
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                }/* else { */
                    System.out.println("B");
                    Test.printA = true;
                    Test.object.notify();
                /*}*/
            }
        }
    }
}

Answer 1

public class Test {

    public static Object object = new Object();
    public static boolean printA = true;

    public static void main(String[] args) {
        ThreadA threadA = new ThreadA();
        threadA.setName("Thread-A");
        threadA.start();

        ThreadB threadB = new ThreadB();
        threadB.setName("Thread-B");
        threadB.start();
    }
}

class ThreadA extends Thread {
    @Override
    public void run() {
        System.out.println(this.getName() + ":start...");
        for (int i = 0; i < 4; i++) {
            System.out.println(this.getName() + ":i=" + i + "...");
            synchronized (Test.object) {
                System.out.println(this.getName() + ":i=" + i + " got lock...");
                if (!Test.printA) {
                    try {
                        System.out.println(this.getName() + ":i=" + i + " prepare to wait...");
                        Test.object.wait();
                        System.out.println(this.getName() + ":i=" + i + " wait over...");
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                } else {
                    System.out.println(this.getName() + ":i=" + i + " print A...");
                    System.out.println("A");
                    Test.printA = false;
                    System.out.println(this.getName() + ":i=" + i + " prepare to notify...");
                    Test.object.notify();
                    System.out.println(this.getName() + ":i=" + i + " notify over...");
                }
            }
        }
    }
}

class ThreadB extends Thread {
    @Override
    public void run() {
        System.out.println(this.getName() + ":start...");
        for (int i = 0; i < 4; i++) {
            System.out.println(this.getName() + ":i=" + i + "...");
            synchronized (Test.object) {
                System.out.println(this.getName() + ":i=" + i + " got lock...");
                if (Test.printA) {
                    try {
                        System.out.println(this.getName() + ":i=" + i + " prepare to wait...");
                        Test.object.wait();
                        System.out.println(this.getName() + ":i=" + i + " wait over...");
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                } else {
                    System.out.println(this.getName() + ":i=" + i + " print B...");
                    System.out.println("B");
                    Test.printA = true;
                    System.out.println(this.getName() + ":i=" + i + " prepare to notify...");
                    Test.object.notify();
                    System.out.println(this.getName() + ":i=" + i + " notify over...");
                }
            }
        }
    }
}

帮你的代码加了点注释，同时把i缩小到4，这样好分析点。你执行一遍然后看看输出的日志你就知道问题在哪了。
其实问题就是A执行完了，B正好处于wait状态，以至于没有人来notify他，你可以直接给A多循环一次即可。

Answer 2

这个和生产者消费者的问题很像。你题目中无限wait下去唯一的情况是：当ThreadA或ThreadB有一方已执行完，未执行完的线程wait之后，另一线程并不能唤醒它，因为那个线程已跑完了。

Answer 3

A线程先跑，如果B还没有wait的时候，A已经notify了，这时这个notify就没有效果，等B开始wait了，A的notify已经跑完了，根本永远得不到通知，你这么写逻辑是有问题的

Answer 4

很明显ThreadB先wait，但代码却ThreadA先调用.... 只要让ThreadB先执行就可以了

ThreadB threadB = new ThreadB();
threadB.start();
        
Thread.sleep(10);
        
ThreadA threadA = new ThreadA();
threadA.start();

注意：两个Thread同时执行，java并不能能保证先调用的Thread先执行。