Re: C++11 move semantics

Hans Åberg wrote: > > On 5 Mar 2018, at 18:15, Frank Heckenbach wrote: > > > > Let's see if I can add "$$ = std::move ($1)" as a default action in > > my patches. (Not as a pre-action as in C, that would be wrong with > > move, only as a default action when no user

Re: C++11 move semantics

> On 5 Mar 2018, at 18:15, Frank Heckenbach wrote: > > Let's see if I can add "$$ = std::move ($1)" as a default action in > my patches. (Not as a pre-action as in C, that would be wrong with > move, only as a default action when no user action is supplied.) That might

Re: C++11 move semantics

> On 5 Mar 2018, at 18:15, Frank Heckenbach wrote: > > Let's see if I can add "$$ = std::move ($1)" as a default action in > my patches. (Not as a pre-action as in C, that would be wrong with > move, only as a default action when no user action is supplied.) That might

Re: C++11 move semantics

Hans Åberg wrote: > >>> That's what I thought too, but it doesn't seem to (and it wouldn't > >>> work with move-only types without std::move). It just seems to leave > >>> $$ unset (i.e. default-constructed) without an explicit action. > >> > >> If you don't have an {...}, one should get $$ =

Re: C++11 move semantics

> On 5 Mar 2018, at 00:36, Frank Heckenbach wrote: > >>> It would work in C, but users were advised not to rely on it. >> >> No problem in C, as there are no copy constructors, just an overhead. > > I think users were advised against it because it was just an >