线程是否在 wait() 之后唤醒并再次获得监视器从 wait() 之后的下一条指令继续或从头开始继续？

Question

假设 thread_1 这样做：

synchronized(o) {
    dosomthing1;
    o.wait();
    dosomthing2;
}

假设 thread_1 被唤醒，在获取 o 监视器时被阻止，明白了。下一个 thread_1 指令是什么？ dosomthing1 或 dosomthing2?

Answer 1

假设我们有 2 种方法printFirst() 将打印从 0 到 5，printSecond() 将打印从 0 到 10

void printFirst() {
    for (int i = 0; i <= 5; i++) {
        System.out.print(i + " ");
    }
}

void printSecond() {
    for (int i = 0; i <= 10; i++) {
        System.out.print(i + " ");
    }
}

这两个方法将由 Thread-0 使用，正如您首先在 try catch 中看到的那样，首先它将在方法完成其工作后调用 printFirst 方法 wait() 方法会将当前放在 WAIT STATE 上， wait() 下面的任何代码都不会执行，直到 notify 被调用。

synchronized void testThreadWait(){
    System.out.println(Thread.currentThread().getName() +" starting to work");
    try {
        printFirst();
        wait();
        System.out.println(Thread.currentThread().getName() + "starts");
        printSecond();
    } catch (InterruptedException e) {
        e.printStackTrace();
    }
}

另一个方法将从 Thread-1 执行，它会打印一些东西，它会 notify() 处于当前 wait 状态的线程

synchronized void testThreadNotify(){
    System.out.println("\n" + Thread.currentThread().getName() + "starting to work");
    System.out.println("Thread one is waiting");
    System.out.println("Preparing to notify Thread One");
    notify();
}

当 wait() 方法被调用时 Thread-0 将继续执行他的工作，在这种情况下将调用 printSecond() wait() 方法下面的方法，并且它将打印来自0-10

完整代码

class TestThread{

    void printFirst() {
        for (int i = 0; i <= 5; i++) {
            System.out.print(i + " ");
        }
    }

    void printSecond() {
        for (int i = 0; i <= 10; i++) {
            System.out.print(i + " ");
        }
    }

    synchronized void testThreadWait(){
        System.out.println(Thread.currentThread().getName() +" starting to work");
        try {
            printFirst();
            wait();
            System.out.println(Thread.currentThread().getName() + "starts");
            printSecond();
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        System.out.println("This will be printed after wait");
    }

    synchronized void testThreadNotify(){
        System.out.println(Thread.currentThread().getName() + "starting to work");
        System.out.println("Thread one is waiting");
        System.out.println("Preparing to notify Thread One");
        notify();
    }
}

public class Test{
    public static void main(String args[]){
        final TestThread c=new TestThread();
        new Thread(){
            public void run(){c.testThreadWait();}
        }.start();
        new Thread(){
            public void run(){c.testThreadNotify();}
        }.start();
    }
}

输出

Thread-0 starting to work
0 1 2 3 4 5 
Thread-1starting to work
Thread one is waiting
Preparing to notify Thread One
Thread-0starts
0 1 2 3 4 5 6 7 8 9 10 <--- prints only printSecond()

在这里你可以看到，当 Thread-0 被通知时，它只打印 wait() 方法下面的第二个方法

线程是否在 wait() 之后唤醒并再次获得监视器从 wait() 之后的下一条指令继续或从头开始继续？

Does a thread waked after wait() and got the monitor again continues from the next instruction after wait() or from the beginning?

java

multithreading

synchronized

notify

wait