在 volatile 变量存在的情况下会发生重新排序吗? [英] Can a reordering happen in the presence of a volatile variable?

问题描述

我的做法如下，我的程序中只有 2 个线程.

I am doing as follows, there's only 2 threads in my program.

// Thread 1
write a = 0
write a = 1
write volatile b = 1

// Thread 2
read volatile b // this I always do after write volatile b in the 1st thread
read a

我读过 Java 内存模型，根据我在线程 2 中的理解 read a 总是给我 1.

I've read on Java Memory Model and from what I understand in thread 2 read a will ALWAYS give me 1.

我想知道我的理解是否正确.

I would like to know if this my understanding is correct or not.

特别是重新排序是否仍然发生，所以我在第二个线程中看到 a = 0?

In particular CAN A REORDERING still HAPPEN so I see a = 0 in the 2nd thread?

推荐答案

您的假设基本正确.但是，我会稍微重申一下以符合 JMM 的保证.

Your assumptions are mostly correct. However, i would restate it slightly to match what the JMM guarantees.

如果线程 2 读取 b 并看到值 1，那么后续读取的 a 将为 1.就像你说的，如果线程2 总是在线程 1 完成写入后在"读取 b，然后线程 2 将看到值 1，并且读取 a 将如您所料.

If Thread 2 reads b and sees value 1, then the subsequent read of a will be 1. Like you said, if Thread 2 always reads b "after" Thread 1 finishes writing it, then Thread 2 will see the value 1 and the read of a will be as you expect.

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