使用挥发性只要原子 [英] Using volatile long as an atomic

问题描述

如果我有这样的事情...

If I have something like this...

volatile long something_global = 0;

long some_public_func()
{
    return something_global++;
}

难道是合理的当多个线程访问，以期望这code不破（竞争条件）？如果它不是标准的，可以将它仍然可以为关于现代编译一个合理的假设做了什么？

Would it be reasonable to expect this code to not break (race condition) when accessed with multiple threads? If it's not standard, could it still be done as a reasonable assumption about modern compilers?

请注意：所有我使用这个的，是原子递增和递减 - 没有什么票友

NOTE: ALL I'm using this for is atomic increment and decrement - nothing fancier.

推荐答案

没有 - 挥发性并不意味着同步。它只是意味着，每次访问会返回最先进的最新值（而不是在线程本地缓存副本）。

No - volatile does not mean synchronized. It just means that every access will return the most up-to-date value (as opposed to a copy cached locally in the thread).

后递增不是一个原子操作，则存储器访问后跟一个存储器写。交错二者可以意味着该值实际递增一次。

Post-increment is not an atomic operation, it is a memory access followed by a memory write. Interleaving two can mean that the value is actually incremented just once.

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