On 23 August 2011 13:13, Julian Gonggrijp <j.gonggrijp@gmail.com> wrote:
Nevin Liber wrote:
In the C++11 move case, the moved-from object itself doesn't know if it is about to be destroyed or reused, so it has to be put in a state where either can happen.
Excuse me, but that doesn't seem logical. You are saying that since the object doesn't know what is going to happen, the programmer has to prepare it for both scenarios.
C++11 move semantics have a requirement that the source object is left in a valid but unspecified state. Given that T is a MoveConstructible and MoveAssignable type: T a(T_Factory()); T b(T_Factory()); // ... a = std::move(b); 1. At this point, what operations can you perform on a? 2. At this point, what operations can you perform on b? a still has to be destructible and assignable, as a cannot possibly know how it is going to be used past this point. Any other operations on a would have to be documented (for instance, there are stronger requirements on move semantics for standard library types). b has no restrictions on the operations that can be called. My question is, given: T c(T_Factory()); T d(T_Factory()); // ... move_raw(c, d); 3. At this point, what operations can you perform on c? 4. At this point, what operations can you perform on d? -- Nevin ":-)" Liber <mailto:nevin@eviloverlord.com> (847) 691-1404