使用volatile而不同步 [英] Uses of volatile without synchronization

问题描述

知道

读取和写入是 atomic

问题1：这可以理解为

private volatile int x = 0;

x ++; 操作是原子的？

那

标记变量volatile并不能消除所有需要同步
原子操作，因为仍然可以内存一致性错误。 / p>

Marking variable volatile does not eliminate all need to synchronize atomic actions, because memory consistency errors are still possible.

问题2：我想知道在什么情况下（如果有的话）可以看到标记为<$的变量c $ c> volatile 并且看不到标记为同步的块的任何方法（试图访问/修改变量）？

Question2: I wonder under what circumstances (if any) it is possible to see a variable marked volatile and not see any methods of blocks marked synchronized (that attempt to access/ modify the variable)?

换句话说，如果所有需要保护免于并发修改的变量都标记为 volatile ？

In other words, should all variables that need to be protected from concurrent modification be marked volatile?

推荐答案

volatile只为你提供额外的可见性保证，长/双的原子写/读（否则JLS无法保证，是的）和一些内存顺序保证。 没有同步（虽然可以构建以volatile为开头的同步块，但是可以实现 - Dekker的算法）
所以不，它对 x ++ 没有帮助 - 这仍然是一个读取，包含和写入，需要某种形式的同步。

The volatile only gives you additional visibility guarantees, atomic writes/reads for longs/doubles (otherwise not guaranteed by the JLS, yes) and some memory order guarantees. No synchronization (it is possible though to build synchronization blocks starting with just volatile - Dekker's algorithm ) So no, it does not help you with x++ - that's still a read, inc and write and needs some form of synchronization.

volatile的一个例子是着名的双重检查锁定，我们大部分时间都避免同步，因为排序保证是我们所需要的：

One example of volatile is the famous double-checked locking, where we avoid synchronization most of the time because the ordering guarantees are all we need:

private volatile Helper helper = null;
public Helper getHelper() {
    if (helper == null) {
        synchronized(this) {
            if (helper == null) {
                helper = new Helper();
            }
        }
    }
    return helper;
}

一个完全没有涉及同步的例子，是一个简单的退出标志，这里它不是关于订购保证而是关于保证的可见性

An example where there's absolutely no synchronization involved, is a simple exit flag, here it's not about ordering guarantees but only about the guaranteed visibility

public volatile boolean exit = false;
public void run() {
   while (!exit) doStuff();
   // exit when exit set to true
}

如果另一个线程设置 exit = true 执行while循环的其他线程可以保证看到更新 - 没有volatile，它可能不会。

If another thread sets exit = true the other thread doing the while loop is guaranteed to see the update - without volatile it may not.

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