今天工作时,我在Java中遇到了volatile关键字。不是很熟悉,我找到了这个解释。

鉴于那篇文章详细解释了所讨论的关键字,您曾经使用过它吗?或者您曾经看到过以正确方式使用该关键字的情况吗?


当前回答

虽然我在这里提到的答案中看到了许多很好的理论解释,但我在这里添加了一个实际的例子来解释:

1.

代码在不使用volatile的情况下运行

public class VisibilityDemonstration {

private static int sCount = 0;

public static void main(String[] args) {
    new Consumer().start();
    try {
        Thread.sleep(100);
    } catch (InterruptedException e) {
        return;
    }
    new Producer().start();
}

static class Consumer extends Thread {
    @Override
    public void run() {
        int localValue = -1;
        while (true) {
            if (localValue != sCount) {
                System.out.println("Consumer: detected count change " + sCount);
                localValue = sCount;
            }
            if (sCount >= 5) {
                break;
            }
        }
        System.out.println("Consumer: terminating");
    }
}

static class Producer extends Thread {
    @Override
    public void run() {
        while (sCount < 5) {
            int localValue = sCount;
            localValue++;
            System.out.println("Producer: incrementing count to " + localValue);
            sCount = localValue;
            try {
                Thread.sleep(1000);
            } catch (InterruptedException e) {
                return;
            }
        }
        System.out.println("Producer: terminating");
    }
}
}

在上面的代码中,有两个线程——生产者和消费者。

生产者线程在循环中迭代5次(睡眠时间为1000毫秒或1秒)。在每次迭代中,生产者线程将sCount变量的值增加1。因此,在所有迭代中,生产者将sCount的值从0更改为5

使用者线程处于一个常量循环中,每当sCount的值发生变化时,它就会打印,直到值达到5为止。

两个循环同时开始。因此,生产者和消费者都应该将sCount的值打印5次。

输出

Consumer: detected count change 0
Producer: incrementing count to 1
Producer: incrementing count to 2
Producer: incrementing count to 3
Producer: incrementing count to 4
Producer: incrementing count to 5
Producer: terminating

分析

In the above program, when the producer thread updates the value of sCount, it does update the value of the variable in the main memory(memory from where every thread is going to initially read the value of variable). But the consumer thread reads the value of sCount only the first time from this main memory and then caches the value of that variable inside its own memory. So, even if the value of original sCount in main memory has been updated by the producer thread, the consumer thread is reading from its cached value which is not updated. This is called VISIBILITY PROBLEM .

2.

代码使用volatile运行

在上面的代码中,用下面的代码替换声明了sCount的代码行:

private volatile  static int sCount = 0;

输出

Consumer: detected count change 0
Producer: incrementing count to 1
Consumer: detected count change 1
Producer: incrementing count to 2
Consumer: detected count change 2
Producer: incrementing count to 3
Consumer: detected count change 3
Producer: incrementing count to 4
Consumer: detected count change 4
Producer: incrementing count to 5
Consumer: detected count change 5
Consumer: terminating
Producer: terminating

分析

当我们声明一个变量为volatile时,这意味着所有对这个变量的读写操作都将直接进入主存。这些变量的值永远不会被缓存。

由于sCount变量的值永远不会被任何线程缓存,消费者总是从主内存中读取sCount的原始值(在那里由生产者线程更新)。因此,在这种情况下,输出是正确的,两个线程都打印了5次不同的sCount值。

通过这种方式,volatile关键字解决了可见性问题。

其他回答

Volatile只保证所有线程,甚至线程本身,都是递增的。例如:计数器在同一时间看到变量的同一面。它不是用来代替同步或原子或其他东西,它完全使读取同步。请不要将其与其他java关键字进行比较。如下例所示,volatile变量操作也是原子性的,它们会立即失败或成功。

package io.netty.example.telnet;

import java.util.ArrayList;
import java.util.List;

public class Main {

    public static volatile  int a = 0;
    public static void main(String args[]) throws InterruptedException{

        List<Thread> list = new  ArrayList<Thread>();
        for(int i = 0 ; i<11 ;i++){
            list.add(new Pojo());
        }

        for (Thread thread : list) {
            thread.start();
        }

        Thread.sleep(20000);
        System.out.println(a);
    }
}
class Pojo extends Thread{
    int a = 10001;
    public void run() {
        while(a-->0){
            try {
                Thread.sleep(1);
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
            Main.a++;
            System.out.println("a = "+Main.a);
        }
    }
}

即使你放不放,结果也总会不一样。但是,如果您像下面那样使用AtomicInteger,结果将始终相同。同步也是如此。

    package io.netty.example.telnet;

    import java.util.ArrayList;
    import java.util.List;
    import java.util.concurrent.atomic.AtomicInteger;

    public class Main {

        public static volatile  AtomicInteger a = new AtomicInteger(0);
        public static void main(String args[]) throws InterruptedException{

            List<Thread> list = new  ArrayList<Thread>();
            for(int i = 0 ; i<11 ;i++){
                list.add(new Pojo());
            }

            for (Thread thread : list) {
                thread.start();
            }

            Thread.sleep(20000);
            System.out.println(a.get());

        }
    }
    class Pojo extends Thread{
        int a = 10001;
        public void run() {
            while(a-->0){
                try {
                    Thread.sleep(1);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                Main.a.incrementAndGet();
                System.out.println("a = "+Main.a);
            }
        }
    }

Volatile具有内存可见性的语义。基本上,volatile字段的值在写操作完成后对所有读取器(特别是其他线程)可见。如果没有volatile,读者可以看到一些未更新的值。

回答您的问题:是的,我使用一个volatile变量来控制某些代码是否继续循环。循环测试易变值,如果为真则继续。可以通过调用“stop”方法将条件设置为false。循环看到false,并在stop方法完成执行后测试该值时终止。

我强烈推荐的《Java并发实践》一书对volatile做了很好的解释。这本书的作者与问题中提到的IBM文章的作者是同一人(事实上,他在那篇文章的末尾引用了他的书)。我对volatile的使用被他的文章称为“模式1状态标志”。

如果您想了解更多关于volatile在底层是如何工作的,请阅读Java内存模型。如果你想超越这个层次,看看像Hennessy & Patterson这样的优秀计算机体系结构书籍,阅读关于缓存一致性和缓存一致性的内容。

Volatile执行以下操作。

不同线程对volatile变量的读写总是从内存,而不是从线程自己的缓存或cpu寄存器。所以每个线程总是处理最新的值。 2>当两个不同的线程在堆中使用相同的实例或静态变量时,其中一个线程可能会认为其他线程的操作是无序的。请看jeremy manson的博客。但不稳定在这里有所帮助。

下面完全运行的代码展示了如何在不使用synchronized关键字的情况下以预定义的顺序执行多个线程并打印输出。

thread 0 prints 0
thread 1 prints 1
thread 2 prints 2
thread 3 prints 3
thread 0 prints 0
thread 1 prints 1
thread 2 prints 2
thread 3 prints 3
thread 0 prints 0
thread 1 prints 1
thread 2 prints 2
thread 3 prints 3

为了实现这一点,我们可以使用以下完整的运行代码。

public class Solution {
    static volatile int counter = 0;
    static int print = 0;
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Thread[] ths = new Thread[4];
        for (int i = 0; i < ths.length; i++) {
            ths[i] = new Thread(new MyRunnable(i, ths.length));
            ths[i].start();
        }
    }
    static class MyRunnable implements Runnable {
        final int thID;
        final int total;
        public MyRunnable(int id, int total) {
            thID = id;
            this.total = total;
        }
        @Override
        public void run() {
            // TODO Auto-generated method stub
            while (true) {
                if (thID == counter) {
                    System.out.println("thread " + thID + " prints " + print);
                    print++;
                    if (print == total)
                        print = 0;
                    counter++;
                    if (counter == total)
                        counter = 0;
                } else {
                    try {
                        Thread.sleep(30);
                    } catch (InterruptedException e) {
                        // log it
                    }
                }
            }
        }
    }
}

下面的github链接有一个自述,它给出了适当的解释。 https://github.com/sankar4git/volatile_thread_ordering

对于长变量和双变量类型的读写操作的处理,目前还没有人提及。读和写对于引用变量和大多数基本变量都是原子操作,长变量和双变量类型除外,它们必须使用volatile关键字作为原子操作。@link

在我看来,除了停止线程之外,使用volatile关键字的两个重要场景是:

Double-checked locking mechanism. Used often in Singleton design pattern. In this the singleton object needs to be declared volatile. Spurious Wakeups. Thread may sometimes wake up from wait call even if no notify call has been issued. This behavior is called spurious wakeup. This can be countered by using a conditional variable (boolean flag). Put the wait() call in a while loop as long as the flag is true. So if thread wakes up from wait call due to any reasons other than Notify/NotifyAll then it encounters flag is still true and hence calls wait again. Prior to calling notify set this flag to true. In this case the boolean flag is declared as volatile.