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Re: On "transparently replaceable" in std::vector operations

From: Giuseppe D'Angelo <giuseppe.dangelo_at_[hidden]>
Date: Wed, 28 Apr 2021 10:53:25 +0200
Il 28/04/21 10:40, Edward Catmur ha scritto:
> Consider a T that is (CopyInsertable,CopyAssignable, and) MoveInsertable
> but not MoveAssignable. An implementation may choose to shuffle the
> elements along by destruct-and-move rather than copy assign.
> Invalidating everything after the insertion point gives greater
> implementation freedom.

Sorry, could you please elaborate on this a bit more? A type can't be
CopyAssignable without also being MoveAssignable:


Thank you,

Giuseppe D'Angelo | giuseppe.dangelo_at_[hidden] | Senior Software Engineer
KDAB (France) S.A.S., a KDAB Group company
Tel. France +33 (0)4 90 84 08 53, http://www.kdab.com
KDAB - The Qt, C++ and OpenGL Experts

Received on 2021-04-28 03:53:33