From: MonkeeSage <MonkeeSage@...>
Date: 2007-12-28T04:19:58+09:00
Subject: Re: Add Array#first= and Array#last= to std lib

On Dec 27, 12:52 pm, Trans <transf...@gmail.com> wrote:
> On Dec 27, 12:54 pm, MonkeeSage <MonkeeS...@gmail.com> wrote:
>
>
>
> > On Dec 27, 9:55 am, Sebastian Hungerecker <sep...@googlemail.com>
> > wrote:
>
> > > George wrote:
> > > > What would you expect this to do?
>
> > > >   [].last = 1
>
> > > The same thing as "[][-1] = 1", I'd imagine.
> > > The problem I'm seeing would be this: If you allow arr.last = x, you'd also
> > > have to allow arr.last(n) = x if you want to be consistent, but that's not
> > > syntactically possible.
>
> > > --
> > > NP: Katatonia - Endtime
> > > Jabber: sep...@jabber.org
> > > ICQ: 205544826
>
> > Agree. It's tempting to treat #first / #last as 0 / -1, but in
> > actuality they are method calls and simply return a value; they don't
> > subscript an array. Setting #last is not semantically different than
> > [1,2,3].pop = 4, it's just that #last is just a bit more subtle.
>
> I don't see what you are getting at here. #pop is destructive, #last
> is not. What does #last return when it is called? It returns a
> reference to the last element. So why would #last= do anything other
> then set the reference of the last element? Seems obvious to me. So we
> can't do last(n) = x, due to syntax constraints, oh well. It would
> still be convenient to have the obvious n=1, no arg case. I find that
> my programs are usually easier to read when I can use words rather non-
> alphabetic symbols.
>
> T.

IOW, #pop returns a value, and this is just what #last does. One could
argue that #last and #[-1] *should be* synonymous (which may be the
point of the proposal); but as it currently stands, #last means the
same thing as `def l(a); a[-1]; end` so it makes no sense to have a
setter for it. Unless the semantics change, (to me at least) it is non-
sense to have #last=. It's the same as a.pop = 3.

Regards,
Jordan