混合原子和非原子变量以及缓存 [英] mix atomic and non atomic variables and caches

问题描述

假设我们有一段正确的代码(至少希望如此):

Let's say we have this piece of code that it is correct (I hope at least) :

std::atomic<int> a;
std::atomic<bool> ready{false};
void threadA() {
    a.store(666, std::memory_order_relaxed);
    ready.store(true, std::memory_order_release);
}

void threadB() {
    while(!ready.load(std::memory_order_acquire));
    process(a.load(std::memory_order_relaxed));
}

我的问题是:如果您使用的是int a;而不是std::atomic<int> a;，那么它也是正确的吗?还是存在缓存刷新/失效的问题?

My question is : In the case you are using a int a; instead of std::atomic<int> a;, it is correct as well? Or is there a problem of cache flushing / invalidation?

推荐答案

这是否一个好主意，例如，您的代码就可以了.

Whether or not this is a good idea, as an example, your code is fine..

您可以将a的原子类型替换为常规的int(或与此相关的任何类型).
C ++标准使用以下短语(第1.10.1-6节)来支持您的情况:

You may replace the atomic type of a with a regular int (or any type for that matter).
The C++ standard supports your case with the following phrase (§ 1.10.1-6):

某些库调用与另一个线程执行的其他库调用同步.例如，原子存储发行版与从其存储中获取其值的负载获取同步

Certain library calls synchronize with other library calls performed by another thread. For example, an atomic store-release synchronizes with a load-acquire that takes its value from the store

由于threadB加载了threadA存储的ready的值(正在循环等待)，所以建立了 synchronizes-with 关系. 因此，a.load()观察a.store()的记忆效应.另一种说法是a.store() 发生之前 a.load()

Since threadB loads the value of ready stored by threadA (it is waiting for it in a loop), the synchronizes-with relationship is established. Therefore, a.load() observes the memory effects of a.store(). Another way to say this is that a.store() happens-before a.load()

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