Are volatile data members trivially copyable?

问题

Whilst writing this answer I realised that I'm not as confident about my conclusions as I usually would ensure before hitting Post Your Answer.

I can find a couple of reasonably convincing citations for the argument that the trivial-copyability of volatile data members is either implementation-defined or flat-out disallowed:

• https://groups.google.com/forum/?fromgroups=#!topic/comp.std.c++/5cWxmw71ktI
• http://gcc.gnu.org/bugzilla/show_bug.cgi?id=48118
• http://www.open-std.org/jtc1/sc22/wg21/docs/cwg_active.html#496

But I haven't been able to back this up in the standard¹ itself. Particularly "worrying" is that there's no sign of the proposed wording change from that n3159 issues list in the actual standard's final wording.

So, what gives? Are volatile data members trivially copyable, or not?

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¹   C++11

回答1:

I'm seeing the following definition for "trivially copyable" (C++11 §3.9, paragraph 9):

...Scalar types, trivially copyable class types, arrays of such types, and cv-qualified versions of these types are collectively called trivially copyable types....

cv-qualified by definition includes const and/or volatile (§3.9.3). It would therefore appear that volatile values are explicitly trivially copyable, if the unqualified type would be trivially copyable (a scalar or trivially copyable class type, or array thereof).

来源：https://stackoverflow.com/questions/13407737/are-volatile-data-members-trivially-copyable