下面是一个非常简单的代码，演示了volatile for variable的需求，它用于从其他线程控制线程执行(这是需要volatile的一个场景)。

// Code to prove importance of 'volatile' when state of one thread is being mutated from another thread.
// Try running this class with and without 'volatile' for 'state' property of Task class.
public class VolatileTest {
    public static void main(String[] a) throws Exception {
        Task task = new Task();
        new Thread(task).start();

        Thread.sleep(500);
        long stoppedOn = System.nanoTime();

        task.stop(); // -----> do this to stop the thread

        System.out.println("Stopping on: " + stoppedOn);
    }
}

class Task implements Runnable {
    // Try running with and without 'volatile' here
    private volatile boolean state = true;
    private int i = 0;

    public void stop() {
        state = false;
    } 

    @Override
    public void run() {
        while(state) {
            i++;
        }
        System.out.println(i + "> Stopped on: " + System.nanoTime());
    }
}

当不使用volatile时:即使在“stop on: xxx”之后，你也永远不会看到“Stopped on: xxx”消息，并且程序继续运行。

Stopping on: 1895303906650500

当使用volatile时:你会立即看到'Stopped on: xxx'。

Stopping on: 1895285647980000
324565439> Stopped on: 1895285648087300

演示:https://repl.it/repls/SilverAgonizingObjectcode

…volatile修饰符保证任何读取字段的线程都能看到最近写入的值。——乔希·布洛赫 如果您正在考虑使用volatile，请仔细阅读java.util.concurrent包，它处理原子行为。 维基百科上关于单例模式的帖子显示了volatile的使用。

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。

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。

挥发性(vɒlətʌɪl):在常温下容易挥发

关于volatile的重要一点:

Synchronization in Java is possible by using Java keywords synchronized and volatile and locks. In Java, we can not have synchronized variable. Using synchronized keyword with a variable is illegal and will result in compilation error. Instead of using the synchronized variable in Java, you can use the java volatile variable, which will instruct JVM threads to read the value of volatile variable from main memory and don’t cache it locally. If a variable is not shared between multiple threads then there is no need to use the volatile keyword.

源

volatile用法示例:

public class Singleton {
    private static volatile Singleton _instance; // volatile variable
    public static Singleton getInstance() {
        if (_instance == null) {
            synchronized (Singleton.class) {
                if (_instance == null)
                    _instance = new Singleton();
            }
        }
        return _instance;
    }
}

我们在第一个请求到来时惰性地创建实例。

如果我们不使_instance变量为volatile，那么创建Singleton实例的线程就不能与其他线程通信。因此，如果线程A正在创建单例实例，在创建后，CPU损坏等，所有其他线程将无法看到_instance的值不为空，他们将认为它仍然被分配为空。

为什么会发生这种情况?因为读线程不做任何锁，直到写线程从同步块中出来，内存不会被同步，_instance的值也不会在主存中更新。使用Java中的Volatile关键字，这是由Java本身处理的，这样的更新将对所有读取线程可见。

结论:volatile关键字也用于线程之间的内存内容通信。

without volatile的用法示例:

public class Singleton {    
    private static Singleton _instance;   //without volatile variable
    public static Singleton getInstance() {   
        if (_instance == null) {  
            synchronized(Singleton.class) {  
                if (_instance == null) 
                    _instance = new Singleton(); 
            } 
        }
        return _instance;  
    }
}

The code above is not thread-safe. Although it checks the value of instance once again within the synchronized block (for performance reasons), the JIT compiler can rearrange the bytecode in a way that the reference to the instance is set before the constructor has finished its execution. This means the method getInstance() returns an object that may not have been initialized completely. To make the code thread-safe, the keyword volatile can be used since Java 5 for the instance variable. Variables that are marked as volatile get only visible to other threads once the constructor of the object has finished its execution completely. Source

Java中不稳定的用法:

快速失败迭代器通常使用list对象上的volatile计数器实现。

当列表更新时，计数器会递增。 创建Iterator时，计数器的当前值嵌入到Iterator对象中。 当执行Iterator操作时，该方法比较两个计数器值，如果不相同则抛出ConcurrentModificationException异常。

故障安全迭代器的实现通常是轻量级的。它们通常依赖于特定列表实现的数据结构的属性。没有一般的模式。

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