mboehme added inline comments.
================
Comment at: test/clang-tidy/misc-move-const-arg.cpp:75-76
@@ +74,4 @@
+
+class NonMoveable {
+ public:
+ NonMoveable();
----------------
aaron.ballman wrote:
> > Can you expand on this?
> >
> > The standard says: "If the definition of a class X does not explicitly
> > declare a move constructor, one will be implicitly declared as defaulted if
> > and only if - X does not have a user-declared copy constructor [...]"
> > (12.8/9).
> >
> > Because I'm declaring a copy constructor, I would thus have expected not to
> > get an implicitly-declared move constructor. Where am I going wrong here?
>
> Move operations are optimized versions of a copy operation, so a failed move
> operation can fall back to perform a copy operation instead and achieve the
> same operational semantics but with slightly worse performance. Because you
> have a copy constructor, you do not get an implicitly-declared move
> constructor (so that's correct), but that doesn't mean the type cannot be
> used in a context where the user expects a move (i.e., calling `std::move()`
> on it) -- the operation just falls back on the copy constructor.
>
> Basically, there's an operational difference between not having the
> implicitly-declared move constructor and having a deleted move constructor.
> In the former, fallback to copy happens and in the latter you get a
> diagnostic. So when I hear "non-moveable type", my brain assumes you mean
> "get a diagnostic when you try to move."
Ah, now I see what you mean.
I've renamed "NonMoveable" to "NoMoveSemantics" (and "Moveable" to
"MoveSemantics" for consistency). What do you think?
(I don't like NonMoveConstructible because it's prone to the same ambiguity --
or even more so, because std::is_move_constructible<> uses exactly the same
term and the type is move constructible in the sense of that trait.)
http://reviews.llvm.org/D21223
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