Acquire/Release 语义
Acquire/Release semantics
在回答中
@Peter Cordes 写过
For Acquire/Release semantics to give you the ordering you want, the
last store has to be the release-store, and the acquire-load has to be
the first load. That's why I made y a std::atomic, even though you're
setting x to 0 or 1 more like a flag.
我想问一些问题，以便更好地理解它。
我已阅读http://preshing.com/20120913/acquire-and-release-semantics as well. And this article contains:
而且写的是保证r2 == 42。我不明白为什么。在我看来，这是可能的：
1、Thread2执行了第一行。它是原子的并且是 memory_order_acquire 所以它必须在后续内存操作之前执行。
现在，Thread2 执行第二行：int r2 = A 并且 r2 等于 0。
然后，Thread1会执行他的代码。
为什么我错了？
完整的报价是：
If we let both threads run and find that r1 == 1, that serves as
confirmation that the value of A assigned in Thread 1 was passed
successfully to Thread 2. As such, we are guaranteed that r2 == 42.
获取-释放语义只保证
A = 42 不会在线程 1Ready = 1 之后发生 r2 = A 不会在线程 2r1 = Ready 之前发生
所以必须在线程2中检查r1的值，以确保A已经被线程1写入。问题中的场景确实可以发生，但是r1 在这种情况下将是 0。
在回答中
For Acquire/Release semantics to give you the ordering you want, the last store has to be the release-store, and the acquire-load has to be the first load. That's why I made y a std::atomic, even though you're setting x to 0 or 1 more like a flag.
我想问一些问题，以便更好地理解它。
我已阅读http://preshing.com/20120913/acquire-and-release-semantics as well. And this article contains:
而且写的是保证r2 == 42。我不明白为什么。在我看来，这是可能的：
1、Thread2执行了第一行。它是原子的并且是 memory_order_acquire 所以它必须在后续内存操作之前执行。
现在，Thread2 执行第二行：
int r2 = A并且r2等于0。然后，Thread1会执行他的代码。
为什么我错了？
完整的报价是：
If we let both threads run and find that r1 == 1, that serves as confirmation that the value of A assigned in Thread 1 was passed successfully to Thread 2. As such, we are guaranteed that r2 == 42.
获取-释放语义只保证
A = 42不会在线程 1Ready = 1之后发生r2 = A不会在线程 2r1 = Ready之前发生
所以必须在线程2中检查r1的值，以确保A已经被线程1写入。问题中的场景确实可以发生，但是r1 在这种情况下将是 0。