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comp.lang.ruby

Add Array#first= and Array#last= to std lib

Robert Klemme

12/27/2007 9:59:00 AM

Hi,

just today I came across a situation where I needed Array#last=
because I wanted to do

an_array.last += 1

Does anybody else see this as useful? Any issues with this? If not
I'll open an RCR.

Kind regards

robert

--
use.inject do |as, often| as.you_can - without end

35 Answers

Joe

12/27/2007 2:39:00 PM

0

Is there a problem with doing something like this:

an_array[-1] += 1

Joe

On Dec 27, 2007 4:59 AM, Robert Klemme <shortcutter@googlemail.com> wrote:
> Hi,
>
> just today I came across a situation where I needed Array#last=
> because I wanted to do
>
> an_array.last += 1
>
> Does anybody else see this as useful? Any issues with this? If not
> I'll open an RCR.
>
> Kind regards
>
> robert
>
> --
> use.inject do |as, often| as.you_can - without end
>
>

George

12/27/2007 3:45:00 PM

0

On Dec 27, 2007 8:59 PM, Robert Klemme <shortcutter@googlemail.com> wrote:
> Hi,
>
> just today I came across a situation where I needed Array#last=
> because I wanted to do
>
> an_array.last += 1
>
> Does anybody else see this as useful? Any issues with this? If not
> I'll open an RCR.

I'll bite... ;-)

What would you expect this to do?

[].last = 1

Regards,
George.

Sebastian Hungerecker

12/27/2007 3:55:00 PM

0

George wrote:
> What would you expect this to do?
>
> =A0 [].last =3D 1

The same thing as "[][-1] =3D 1", I'd imagine.
The problem I'm seeing would be this: If you allow arr.last =3D x, you'd al=
so=20
have to allow arr.last(n) =3D x if you want to be consistent, but that's no=
t=20
syntactically possible.


=2D-=20
NP: Katatonia - Endtime
Jabber: sepp2k@jabber.org
ICQ: 205544826

MonkeeSage

12/27/2007 5:53:00 PM

0

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.

Regards,
Jordan

Trans

12/27/2007 6:53:00 PM

0



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.

MonkeeSage

12/27/2007 7:15:00 PM

0

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

Sebastian Hungerecker

12/27/2007 7:38:00 PM

0

MonkeeSage wrote:
> IOW, #pop returns a value, and this is just what #last does.

How is [] different in that regard? That only returns a value, too.


--
Jabber: sepp2k@jabber.org
ICQ: 205544826

MonkeeSage

12/27/2007 7:41:00 PM

0

On Dec 27, 1:38 pm, Sebastian Hungerecker <sep...@googlemail.com>
wrote:
> MonkeeSage wrote:
> > IOW, #pop returns a value, and this is just what #last does.
>
> How is [] different in that regard? That only returns a value, too.
>
> --
> Jabber: sep...@jabber.org
> ICQ: 205544826

Exactly. It's not... [] = 3 => syntax error...

Regards,
Jordan

Sebastian Hungerecker

12/27/2007 7:58:00 PM

0

MonkeeSage wrote:
> On Dec 27, 1:38 pm, Sebastian Hungerecker wrote:
> > MonkeeSage wrote:
> > > IOW, #pop returns a value, and this is just what #last does.
> >
> > How is [] different in that regard? That only returns a value, too.
>
> Exactly. It's not... [] = 3 => syntax error...

That [] there is an emtpy array. That's not the [] I was talking about.
I'm sorry if I wasn't clear, let me rephrase:
You seemed to say that while there is a method Array#[]= (which I assume,
you're ok with, since you haven't stated otherwise), there shouldn't be a
method Array#last= since Array#last only returns a value like Array#pop does.
Now my question to you is: How is Array#[] different in that regard than
Array#last? I mean some_array[-1] also only returns a value. But you don't
have a problem with people being able to write some_array[-1] = some_value
do you?


--
Jabber: sepp2k@jabber.org
ICQ: 205544826

Sebastian Hungerecker

12/27/2007 8:06:00 PM

0

MonkeeSage wrote:
> It's tempting to treat #first / #last as 0 / -1, but in
> actuality they are method calls and simply return a value

Ok, I'm confused. You seem to be saying that first and last are different than
using [] because they are methods and return a value, right? Does that mean
that you mean to imply that Array#[] is not a method or that it doesn't return
a value?
I'm asking because it is starting to look to me like that is what you're
saying, but Array#[] most certainly is a method and it most certainly does
return a value.


--
NP: Jon Oliva's Pain - Walk Alone
Jabber: sepp2k@jabber.org
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