[Haskell-beginners] type constructors

[Daniel Fischer]

Am Samstag 01 August 2009 19:44:49 schrieb Michael P Mossey:
> I was playing around with type constructors and I wrote this:
>
> data Foo a b = Foo1 [a]
>
>               | Foo2 (a -> b)
>
> t3 = Foo1 [1, 2, 3]
>
> I wanted to see what ghci thought the type of t3 was. Essentially, it's
> data that doesn't use all of the type variables. So this ran fine, and
>
> *Main> :t t3
> t3 :: Foo Integer b

Actually, the type of t3 could be
t3 :: Num a => Foo a b

but without a type signature the monomorphism restriction applies and a is defaulted to 
Integer.

*MFoo> :t Foo2 even
Foo2 even :: (Integral a) => Foo a Bool

>
> Wow! The data exists but it doesn't have a "complete type" so to speak.

It does, it has a polymorphic type, or one could say it has many types:

*MFoo> :t [t3,Foo2 even]
[t3,Foo2 even] :: [Foo Integer Bool]
*MFoo> :t [t3,Foo2 id]
[t3,Foo2 id] :: [Foo Integer Integer]
*MFoo> :t (t3 :: Foo Integer Bool)
(t3 :: Foo Integer Bool) :: Foo Integer Bool
*MFoo> :t (t3 :: Foo Integer Char)
(t3 :: Foo Integer Char) :: Foo Integer Char
*MFoo> :t (t3 `asTypeOf` (Foo2 even))
(t3 `asTypeOf` (Foo2 even)) :: Foo Integer Bool
*MFoo> :t (t3 `asTypeOf` (Foo2 id))
(t3 `asTypeOf` (Foo2 id)) :: Foo Integer Integer

> This worked, too:
>
> f3 (Foo1 xs) = length xs
>
> *Main> f3 t3
> 3

*MFoo> :t (\(Foo1 xs) -> length xs)
(\(Foo1 xs) -> length xs) :: Foo t1 t -> Int

>
> This is surprising to a conventional programmer. But does this naturally
> relate to other features of Haskell. Perhaps laziness? (I.e. data of type
> Foo doesn't always need a type b so it just doesn't have one until it needs
> one.)

Polymorphism and type inference, not laziness.

>
> Thanks,
> Mike