On 12/5/10 12:06 AM, Steven Schveighoffer wrote:
On Sun, 05 Dec 2010 00:42:20 -0500, Andrei Alexandrescu
<seewebsiteforem...@erdani.org> wrote:
Yah, still not getting the original point there.
I responded better in another part of this thread.
Basically, inout doesn't mean what you think it means.
Oh, I see. I thought its behavior can be equivalent to producing three
functions, each replacing inout with (a) nothing, (b) const, (c)
immutable. This approximation works only if all methods actually end up
having the same code.
Andrei
