“M&M 规则”是否适用于 std::atomic 数据成员？
Does “M&M rule” applies to std::atomic data-member?
”mutable用于指定成员不影响class的外部可见状态（经常用于mutexes，备忘录缓存、惰性评估和访问检测）。" [参考：cv (const and volatile) type qualifiers, mutable specifier]
这句话让我纳闷：
”指南：记住“M&M 规则”：对于成员变量，可变和互斥（或原子）一起使用. [参考：GotW #6a Solution: Const-Correctness, Part 1 (updated for C ++11/14)]
我理解为什么“M&M 规则”适用于 std::mutex 数据成员：允许 const 函数是线程安全的，尽管它们 lock/unlock互斥体数据成员，但“M&M 规则”是否也适用于 std::atomic 数据成员？
你把它弄反了。本文不建议让所有原子成员可变。相反，它说：
(1) For a member variable, mutable implies mutex (or equivalent): A
mutable member variable is presumed to be a mutable shared variable
and so must be synchronized internally—protected with a mutex, made
atomic, or similar.
(2) For a member variable, mutex (or similar synchronization type)
implies mutable: A member variable that is itself of a synchronization
type, such as a mutex or a condition variable, naturally wants to be
mutable, because you will want to use it in a non-const way (e.g.,
take a std::lock_guard) inside concurrent const member
functions.
(2) 表示您希望互斥锁成员可变。因为通常您还想在 const 方法中锁定互斥锁。 (2) 没有提到原子成员。
(1) 另一方面说，如果一个成员是可变的，那么你需要在内部处理同步，无论是通过互斥体还是通过使成员成为 atomic。那是因为之前文章提到的子弹：
If you are implementing a type, unless you know objects of the type can never be shared (which is generally impossible), this means that each of your const member functions must be either:
- truly physically/bitwise const with respect to this object, meaning that they perform no writes to the object’s data; or else
- internally synchronized so that if it does perform any actual writes to the object’s data, that data is correctly protected with a mutex or equivalent (or if appropriate are atomic<>) so that any possible concurrent const accesses by multiple callers can’t tell the difference.
可变成员不是“真正的常量”，因此您需要在内部处理同步（通过互斥锁或使成员成为原子成员）。
TL;DR：这篇文章并不建议让所有原子成员都可变。它更建议使互斥锁成员可变并对所有可变成员使用内部同步。
”mutable用于指定成员不影响class的外部可见状态（经常用于mutexes，备忘录缓存、惰性评估和访问检测）。" [参考：cv (const and volatile) type qualifiers, mutable specifier]
这句话让我纳闷：
”指南：记住“M&M 规则”：对于成员变量，可变和互斥（或原子）一起使用. [参考：GotW #6a Solution: Const-Correctness, Part 1 (updated for C ++11/14)]
我理解为什么“M&M 规则”适用于 std::mutex 数据成员：允许 const 函数是线程安全的，尽管它们 lock/unlock互斥体数据成员，但“M&M 规则”是否也适用于 std::atomic 数据成员？
你把它弄反了。本文不建议让所有原子成员可变。相反，它说：
(1) For a member variable, mutable implies mutex (or equivalent): A mutable member variable is presumed to be a mutable shared variable and so must be synchronized internally—protected with a mutex, made atomic, or similar.
(2) For a member variable, mutex (or similar synchronization type) implies mutable: A member variable that is itself of a synchronization type, such as a mutex or a condition variable, naturally wants to be mutable, because you will want to use it in a non-const way (e.g., take a std::lock_guard) inside concurrent const member functions.
(2) 表示您希望互斥锁成员可变。因为通常您还想在 const 方法中锁定互斥锁。 (2) 没有提到原子成员。
(1) 另一方面说，如果一个成员是可变的，那么你需要在内部处理同步，无论是通过互斥体还是通过使成员成为 atomic。那是因为之前文章提到的子弹：
If you are implementing a type, unless you know objects of the type can never be shared (which is generally impossible), this means that each of your const member functions must be either:
- truly physically/bitwise const with respect to this object, meaning that they perform no writes to the object’s data; or else
- internally synchronized so that if it does perform any actual writes to the object’s data, that data is correctly protected with a mutex or equivalent (or if appropriate are atomic<>) so that any possible concurrent const accesses by multiple callers can’t tell the difference.
可变成员不是“真正的常量”，因此您需要在内部处理同步（通过互斥锁或使成员成为原子成员）。
TL;DR：这篇文章并不建议让所有原子成员都可变。它更建议使互斥锁成员可变并对所有可变成员使用内部同步。