c++ - 为什么 fetch_sub 不是释放操作？

Question

引自 C++ 并发实战 $ list 5.9

A fetch_sub operation with memory_order_acquire semantics doesn’t synchronize-with anything, even though it stores a value, because it isn’t a release operation. Likewise, a store can’t synchronize-with a fetch_or with memory_order_release semantics, because the read part of the fetch_or isn’t an acquire operation.

我很难理解上面的段落。如果 fetch_sub memory_order_acquire语义的操作不同步任何东西，为什么fetch_sub接口(interface)给我们留下一个内存顺序参数如下？

T fetch_sub( T arg, std::memory_order order = std::memory_order_seq_cst ) noexcept;

Answer 1

1. “同步于”是单向的，不可交换的。 “A synchronizes with B”并不意味着“B synchronizes with A”(事实上，恰恰相反)，这与人们对英语的期望不同。因此，memory_order_acquire RMW 操作无法与任何内容同步，但是 memory_order_release 存储与 memory_order_acquire RMW 操作同步，后者获取从中读取的值商店。同样，虽然 memory_order_release 存储不与 memory_order_release RMW 同步，但 memory_order_release RMW 可以与 memory_order_acquire 加载。
2. memory_order_acq_rel。