在易失性场周围重新排列法线场 [英] Reorder normal field around volatile field

问题描述

基于

volatile有什么作用? http://www.cs.umd.edu/~pugh/java/memoryModel/jsr-133-faq.html#incorrectlySync

和

易失性的新保证 http://www. ibm.com/developerworks/library/j-jtp03304/

class VolatileExample {
  int x = 0;
  volatile boolean v = false;
  public void writer() {
    x = 42;
    v = true;
  }

  public void reader() {
    if (v == true) {
      //uses x - guaranteed to see 42.
    }
  }
}

看来.

1a) write to non-volatile variable x 
1b) write to volatile variable v

1a永远不会移动通行证1b

我想知道是否将源代码修改为以下内容

class VolatileExample {
  int x = 42;
  volatile boolean v = true;
  public void writer() {
    v = false;
    x = 0;
  }

  public void reader() {
    if (v == true) {
      //uses x - guaranteed to see 42?????
    }
  }
}

下面的序列可以排列吗?

2a) write to volatile variable v
2b) write to non-volatile variable x

我想知道2b能否在2a之前移动?这是因为如果2b能够在2a之前移动，那么读者将无法再保证在if块内看到42.

根据以下信息，我认为2b可以在2a之前移动.

---

http://www.cs .umd.edu/〜pugh/java/memoryModel/jsr-133-faq.html#reordering

写入易失性字段具有 与监视器具有相同的记忆效果 释放，并从易失性中读取 栏位的记忆效果与 监控获取.

---

这意味着任何内存操作 之前对线程可见的 退出同步块 进入后对任何线程可见 同步块受 同一台显示器，因为所有内存 操作发生在发布之前， 而发布发生在 获取.

---

和蟑螂汽车旅馆和Java内存模型

volatile_v = true;      <-- monitor release
non_volatile_x = 42;
(volatile_v will act as a roach motels, and it is fine for non_volatile_x to move into roach motels)


non_volatile_x = 42;
volatile_v = true;      <-- monitor release
(volatile_v will act as a roach motels, and it is not OK for non_volatile_x to move out from roach motels)

解决方案

我们知道，对volatile变量的写入不能相对于以前的任何读取或写入(从Java5开始)进行重新排序，但这不是反之亦然.因此，将程序重新排序为x = 0; v =假；据我所知是正确的.

写入v后，可以保证读取v时可以看到写入v之前发生的每个动作，但是它并没有说明写入v之后的动作-这些可能发生或未发生甚至可以重新排序以将其写入v.

Based on

What does volatile do? http://www.cs.umd.edu/~pugh/java/memoryModel/jsr-133-faq.html#incorrectlySync

and

New guarantees for volatile http://www.ibm.com/developerworks/library/j-jtp03304/

class VolatileExample {
  int x = 0;
  volatile boolean v = false;
  public void writer() {
    x = 42;
    v = true;
  }

  public void reader() {
    if (v == true) {
      //uses x - guaranteed to see 42.
    }
  }
}

It seems that.

1a) write to non-volatile variable x 
1b) write to volatile variable v

1a can never moved pass 1b

I was wondering, if I modify the source code to the following

class VolatileExample {
  int x = 42;
  volatile boolean v = true;
  public void writer() {
    v = false;
    x = 0;
  }

  public void reader() {
    if (v == true) {
      //uses x - guaranteed to see 42?????
    }
  }
}

Can the following sequence be permuted?

2a) write to volatile variable v
2b) write to non-volatile variable x

I was wondering, can 2b ever move before 2a? This is because if 2b able to move before 2a, reader can no longer guaranteed to see 42 within if block.

I feel 2b can be move before 2a based on the following information.

---

http://www.cs.umd.edu/~pugh/java/memoryModel/jsr-133-faq.html#reordering

Writing to a volatile field has the same memory effect as a monitor release, and reading from a volatile field has the same memory effect as a monitor acquire.

---

This means that any memory operations which were visible to a thread before exiting a synchronized block are visible to any thread after it enters a synchronized block protected by the same monitor, since all the memory operations happen before the release, and the release happens before the acquire.

---

and Roach Motels and The Java Memory Model

volatile_v = true;      <-- monitor release
non_volatile_x = 42;
(volatile_v will act as a roach motels, and it is fine for non_volatile_x to move into roach motels)


non_volatile_x = 42;
volatile_v = true;      <-- monitor release
(volatile_v will act as a roach motels, and it is not OK for non_volatile_x to move out from roach motels)

解决方案

Well we know that a write to a volatile variable can't be reordered with respect to any previous read or write (as of Java5), but this isn't true the other way around. So reordering the program to x = 0; v = false; would be correct as I understand it.

After we write to v we're guaranteed that when reading v that every action that happened before writing to v will be visible, but it doesn't say anything about actions AFTER writing to v - those may or may not have happened and can even be reordered to happen before v is written to.

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