Volatile变量是轻量级同步。当所有线程之间的最新数据可见性是必需的，并且原子性可能会受到损害时，在这种情况下，Volatile变量必须是首选。对volatile变量的读取总是返回任何线程最近完成的写操作，因为它们既不缓存在寄存器中，也不缓存在其他处理器看不到的缓存中。Volatile是无锁的。当场景满足上面提到的条件时，我使用volatile。

如果您正在开发多线程应用程序，您将需要使用'volatile'关键字或'synchronized'以及您可能拥有的任何其他并发控制工具和技术。桌面应用就是这样一个例子。

If you are developing an application that would be deployed to application server (Tomcat, JBoss AS, Glassfish, etc) you don't have to handle concurrency control yourself as it already addressed by the application server. In fact, if I remembered correctly the Java EE standard prohibit any concurrency control in servlets and EJBs, since it is part of the 'infrastructure' layer which you supposed to be freed from handling it. You only do concurrency control in such app if you're implementing singleton objects. This even already addressed if you knit your components using frameworkd like Spring.

因此，在Java开发的大多数情况下，应用程序是一个web应用程序，并使用IoC框架，如Spring或EJB，你不需要使用'volatile'。

使用volatile的一个常见示例是使用volatile布尔变量作为终止线程的标志。如果您已经启动了一个线程，并且希望能够安全地从另一个线程中断它，您可以让线程定期检查标志。要阻止它，将标志设置为true。通过将标志设置为易失性，可以确保正在检查它的线程在下次检查时看到它已经设置好，甚至不必使用同步块。

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

是的，我经常使用它——它对多线程代码非常有用。你指的那篇文章很好。不过有两件重要的事情要记住:

You should only use volatile if you completely understand what it does and how it differs to synchronized. In many situations volatile appears, on the surface, to be a simpler more performant alternative to synchronized, when often a better understanding of volatile would make clear that synchronized is the only option that would work. volatile doesn't actually work in a lot of older JVMs, although synchronized does. I remember seeing a document that referenced the various levels of support in different JVMs but unfortunately I can't find it now. Definitely look into it if you're using Java pre 1.5 or if you don't have control over the JVMs that your program will be running on.

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。