[Haskell-beginners] List.sort

[Geoffrey Marchant]

Try
Prelude> :info (,)
data (,) a b = (,) a b -- Defined in GHC.Tuple
instance (Bounded a, Bounded b) => Bounded (a, b)
  -- Defined in GHC.Enum
instance (Eq a, Eq b) => Eq (a, b) -- Defined in Data.Tuple
instance (Ord a, Ord b) => Ord (a, b) -- Defined in Data.Tuple
instance (Read a, Read b) => Read (a, b) -- Defined in GHC.Read
instance (Show a, Show b) => Show (a, b) -- Defined in GHC.Show


On Sat, Jul 11, 2009 at 12:25 PM, Patrick LeBoutillier <
patrick.leboutillier at gmail.com> wrote:

> On Sat, Jul 11, 2009 at 1:13 PM, Chaddaï Fouché<chaddai.fouche at gmail.com>
> wrote:
> > On Sat, Jul 11, 2009 at 5:57 PM, Patrick
> > LeBoutillier<patrick.leboutillier at gmail.com> wrote:
> >>> Does this make sense?
> >>
> >> Yes, perfectly. I guess you just have to know about it... :)
> >>
> >
> > Well, it's the lexicographic order, the most "natural", in any case
> > the most used order for tuples, so it would be strange to have other
> > expectations by default
> >
> > Of course, the mechanism by which this work is interesting by itself.
>
> Really interesting in fact. It shows how generic these concepts really
> are. But it wasn't obvious to me that a pair could be an instance of
> Ord.
>
> This kind of gives you a clue:
>
> Prelude> :info ()
> data () = ()    -- Defined in GHC.Unit
> instance Bounded () -- Defined in GHC.Enum
> instance Enum () -- Defined in GHC.Enum
> instance Eq () -- Defined in Data.Tuple
> instance Ord () -- Defined in Data.Tuple
> instance Read () -- Defined in GHC.Read
> instance Show () -- Defined in GHC.Show
>
> but is there a way to do the equivalent of:
>
> Prelude> :info (a,b)
> or
> Prelude> :info (Int,String)
>
> ?
>
> Patrick
>
> >
> > --
> > Jedaï
> >
>
>
>
> --
> =====================
> Patrick LeBoutillier
> Rosemère, Québec, Canada
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> http://www.haskell.org/mailman/listinfo/beginners
>
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