为什么 fetch_sub 不是释放操作？
Why isn't fetch_sub a release operation?
引自 C++ 并发实战 $清单 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
使用memory_order_acquire语义的操作不与任何东西同步，为什么fetch_sub的接口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;
- "synchronize with" 是单向的，不可交换。 "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 存储与从存储中读取值的 memory_order_acquire RMW 操作同步。同样，虽然 memory_order_release 商店不与 memory_order_release RMW 同步，但 memory_order_release RMW 可以与 memory_order_acquire 负载同步。
memory_order_acq_rel.
引自 C++ 并发实战 $清单 5.9
A
fetch_suboperation withmemory_order_acquiresemantics 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 afetch_orwithmemory_order_releasesemantics, because the read part of thefetch_orisn’t an acquire operation.
我很难理解上面的段落。如果 fetch_sub
使用memory_order_acquire语义的操作不与任何东西同步，为什么fetch_sub的接口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;
- "synchronize with" 是单向的，不可交换。 "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_acquireRMW 操作无法与任何内容同步，但是memory_order_release存储与从存储中读取值的memory_order_acquireRMW 操作同步。同样，虽然memory_order_release商店不与memory_order_releaseRMW 同步，但memory_order_releaseRMW 可以与memory_order_acquire负载同步。 memory_order_acq_rel.