On 2011-06-21 12:13:32 -0400, so <[email protected]> said:
On Tue, 21 Jun 2011 18:18:26 +0300, Michel Fortin
<[email protected]> wrote:
Actually, no copy is needed. Move takes the argument by ref so it can
obliterates it. Obliteration consists of replacing its bytes with those
in S.init. That way if you have a smart pointer, it gets returned
without having to update the reference count (since the source's
content has been destroyed). It was effectively be moved, not copied.
T move(ref T a) {
T b;
move(a, b);
return b;
}
T a;
whatever = move(a);
If T is a struct, i don't see how a copy is not needed looking at the
current state of move.
Actually, that depends on how you look at this.
The essence of a move operation is that you just copy the bits and then
obliterate the old ones. So yes, there's indeed a copy to do, but
there's no need to call a copy constructor or a destructor because no
new instance has been created, it has just been moved. If you don't
call the copy constructor (postblit) then it's a move operation, not a
copy operation, even though there's still a bitwise copy inside the
move operation.
In the return statement above, 'b' gets copied to 'whatever', then
disappears along with the stack frame belonging to the function. So it
becomes a move operation. (And it's even more direct than that with the
named-value optimization.)
--
Michel Fortin
[email protected]
http://michelf.com/
