Hi, you can always check the types using GHCi prompt:
*Prelude> :i (,) 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 Functor ((,) a) -- Defined in Control.Monad.Instances 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 that's for a tuple. You can see that tuple has an instance for the Ord class. *Prelude> :i () 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 and that's for a unit type. On 3 March 2011 08:09, Karthick Gururaj <karthick.guru...@gmail.com> wrote: > Hello, > > I'm learning Haskell from the extremely well written (and well > illustrated as well!) tutorial - http://learnyouahaskell.com/chapters. > I have couple of questions from my readings so far. > > In "typeclasses - 101" > (http://learnyouahaskell.com/types-and-typeclasses#typeclasses-101), > there is a paragraph that reads: > Enum members are sequentially ordered types - they can be enumerated. > The main advantage of the Enum typeclass is that we can use its types > in list ranges. They also have defined successors and predecesors, > which you can get with the succ and pred functions. Types in this > class: (), Bool, Char, Ordering, Int, Integer, Float and Double. > > What is the "()" type? Does it refer to a tuple? How can tuple be > ordered, let alone be enum'd? I tried: > Prelude> take 10 [(1,1) ..] > <interactive>:1:8: > No instance for (Enum (t, t1)) > arising from the arithmetic sequence `(1, 1) .. ' > at <interactive>:1:8-17 > Possible fix: add an instance declaration for (Enum (t, t1)) > In the second argument of `take', namely `[(1, 1) .. ]' > In the expression: take 10 [(1, 1) .. ] > In the definition of `it': it = take 10 [(1, 1) .. ] > > This is expected and is logical. > > But, surprise: > Prelude> (1,1) > (1,2) > False > Prelude> (2,2) > (1,1) > True > Prelude> (1,2) > (2,1) > False > Prelude> (1,2) < (2,1) > True > > So tuples are in "Ord" type class atleast. What is the ordering logic? > > Another question, on the curried functions - specifically for infix > functions. Suppose I need a function that takes an argument and adds > five to it. I can do: > Prelude> let addFive = (+) 5 > Prelude> addFive 4 > 9 > > The paragraph: "Infix functions can also be partially applied by using > sections. To section an infix function, simply surround it with > parentheses and only supply a parameter on one side. That creates a > function that takes one parameter and then applies it to the side > that's missing an operand": describes a different syntax. I tried that > as well: > > Prelude> let addFive' = (+5) > Prelude> addFive' 3 > 8 > > Ok. Works. But on a non-commutative operation like division, we get: > Prelude> let x = (/) 20.0 > Prelude> x 10 > 2.0 > Prelude> let y = (/20.0) > Prelude> y 10 > 0.5 > > So a curried infix operator fixes the first argument and a "sectioned > infix" operator fixes the second argument? > > Regards, > Karthick > > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe@haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe > -- Regards, Paul Sujkov
_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
