Why can't we define re-assignable const reference variable?

[Sergey Gromov]

Steven Schveighoffer <schveiguy at yahoo.com> wrote:
> "Sergey Gromov" wrote
> > If you create a generic algorithm which works with T, and declare
> > something as
> >
> > const(T) t;
> >
> > then you effectively assert that you're going to replace t with
> > different other Ts, but promise to keep integrity of a referenced T
> > intact, if anything is referenced at all.  This means that for int,
> > which is not referencing anything,
> >
> > const(int) t;
> >
> > is completely and safely equivalent to
> >
> > int t;
> >
> > without breaking any contracts of your algorithm.  Please provide an
> > example where "const(int) x;" absolutely must mean "const int x;" for a
> > generic algorithm to work properly.
> 
> Sergey,  I am not arguing that the functionality that you wish to have is 
> invalid.  In fact, I would have been ok with that syntax, and understood it 
> once it was explained thoroughly.  But I have to say that I agree with 
> Janice and Walter on this one, it was very confusing.
> 
> It just goes against the traditional usage of parentheses.  The idea is 
> correct, the syntax is confusing.  Find another syntax.

The traditional usage of parentheses is this: foo(x).  This means that a 
function foo is applied to its argument x, and not applied to everything 
else.  You are probably talking about some different tradition.  Please 
clarify.

-- 
SnakE