对可变对象使用 volatile 关键字 [英] Using volatile keyword with mutable object

问题描述

在 Java 中，我知道 volatile 关键字提供了对变量的可见性.问题是，如果变量是对可变对象的引用，volatile 是否也提供对该对象内部成员的可见性?

In Java, I understand that volatile keyword provides visibility to variables. The question is, if a variable is a reference to a mutable object, does volatile also provide visibility to the members inside that object?

在下面的示例中，如果多个线程正在访问 volatile Mutable m 并更改 value，它是否正常工作?

In the example below, does it work correctly if multiple threads are accessing volatile Mutable m and changing the value?

示例

class Mutable {
    private int value;
    public int get()
    {
        return a;
    }
    public int set(int value)
    {
        this.value = value;
    }
}

class Test {
    public volatile Mutable m;
}

推荐答案

这是对 volatile 的一些细节的旁注解释.写在这里是因为评论太多了.我想举一些例子来说明 volatile 如何影响可见性，以及它在 jdk 1.5 中是如何改变的.

This is sort of a side note explanation on some of the details of volatile. Writing this here because it is too much for an comment. I want to give some examples which show how volatile affects visibility, and how that changed in jdk 1.5.

给定以下示例代码:

public class MyClass
{
  private int _n;
  private volatile int _volN;

  public void setN(int i) {
    _n = i;
  }
  public void setVolN(int i) {
    _volN = i;
  }
  public int getN() { 
    return _n; 
  }
  public int getVolN() { 
    return _volN; 
  }

  public static void main() {
    final MyClass mc = new MyClass();

    Thread t1 = new Thread() {
      public void run() {
        mc.setN(5);
        mc.setVolN(5);
      }
    };

    Thread t2 = new Thread() {
      public void run() {
        int volN = mc.getVolN();
        int n = mc.getN();
        System.out.println("Read: " + volN + ", " + n);
      }
    };

    t1.start();
    t2.start();
  }
}

此测试代码的行为在 jdk1.5+ 中定义良好，但在 jdk1.5 之前没有定义良好.

The behavior of this test code is well defined in jdk1.5+, but is not well defined pre-jdk1.5.

在 jdk1.5 之前的世界中，易失性访问和非易失性访问之间没有明确的关系.因此，该程序的输出可能是:

In the pre-jdk1.5 world, there was no defined relationship between volatile accesses and non-volatile accesses. therefore, the output of this program could be:

1. 读取:0, 0
2. 读取:0、5
3. 读取:5, 0
4. 阅读:5、5

在 jdk1.5+ 世界中，volatile 的语义发生了变化，因此 volatile 访问影响非 volatile 访问的方式与同步完全相同.因此，在 jdk1.5+ 世界中只有某些输出是可能的:

In the jdk1.5+ world, the semantics of volatile were changed so that volatile accesses affect non-volatile accesses in exactly the same way as synchronization. therefore, only certain outputs are possible in the jdk1.5+ world:

1. 读取:0, 0
2. 读取:0、5
3. 读取:5, 0 <- 不可能
4. 阅读:5、5

输出 3. 是不可能的，因为从 volatile _volN 读取5"在 2 个线程之间建立了一个同步点，这意味着在分配给 _volN 之前从 t1 采取的所有操作必须对 t2 可见.

Output 3. is not possible because the reading of "5" from the volatile _volN establishes a synchronization point between the 2 threads, which means all actions from t1 taken before the assignment to _volN must be visible to t2.

进一步阅读:

• 修复 Java 内存模型，第 1 部分
• 修复 Java 内存模型，第 2 部分

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