
It sounds a bit like you're under the impression that * is some sort of wildcard for kinds, when you talk about "named kinds vs unnamed kinds". That's not correct; * IS a "named kind", and its name is *. Nat is distinct from * in exactly the same way that Bool is distinct from ().
Kind variables are exactly what you want, if you need to be able to handle both * -> * and Nat -> *.
The reasons for not always using (k -> *) -> Constraint rather than (* -> *) -> Constraint are the same as the reason for not making everything polymorphic at the type level; sometimes you need the assumption that your k really is *, and sometimes you just don't need the polymorphism and prefer to keep it simple (especially as kind variables require extensions).
On 6 May 2015 11:06:24 pm AEST, "Nicholls, Mark"
lets try it the other way around
instance Foo' SNat1 where type Bar' SNat1 = Z1 bar' = SZ1
that works!
(my language will be wrong here...but the jist is)
So in some something of kind "*->*" is substitutable for a parameter of kind "k -> *" "Nat->*" is substitutable for a parameter of kind "k -> *" "*->*" is substitutable for a parameter of kind "* -> *" "Nat->*" is NOT substitutable for a parameter of kind "* -> *"
Hmmmm
So why would I ever want to define a type class of kind (* -> *) -> Constraint Instead of (k -> *) -> Constraint
?
I think I don't completely understand what's going on.
From: Nicholls, Mark Sent: 06 May 2015 1:02 PM To: haskell-cafe@haskell.org Subject: minor confusion over kinds....
And what "*" means as opposed to other things e.g. an explicit name "Nat" or a kind variable "k"
{-# LANGUAGE DataKinds #-} {-# LANGUAGE ExplicitForAll #-} {-# LANGUAGE FlexibleContexts #-} {-# LANGUAGE FlexibleInstances #-} {-# LANGUAGE GADTs #-} {-# LANGUAGE MultiParamTypeClasses #-} {-# LANGUAGE PolyKinds #-} {-# LANGUAGE StandaloneDeriving #-} {-# LANGUAGE TypeFamilies #-} {-# LANGUAGE TypeOperators #-} {-# LANGUAGE UndecidableInstances #-}
what I want is a typeclass with instances of type level Naturals
here's the class, the parameter is of kind (* -> *) i.e. a is some sort of wrapper that maps from types to values
class Foo (a :: * -> *) where type Bar a bar :: a (Bar a)
lets ignore DataKinds define Zero and Succ as distinct types
uninhabited...but of kind "*" and "k -> *"
data Z1 data S1 nat
SNat1 :: * -> *
data SNat1 a where SZ1 :: SNat1 Z1 SS1 :: SNat1 a -> SNat1 (S1 a)
off we go...
instance Foo SNat1 where type Bar SNat1 = Z1 bar = SZ1
now lets do it the datakind route
data Nat where Z :: Nat S :: Nat -> Nat
ahhh...SNat is of kind "Nat -> *"
data SNat a where SZ :: SNat 'Z SS :: SNat a -> SNat ('S a)
so I cant make it an instance of the same type class
"The first argument of 'Foo' should have kind '* -> *', but 'SNat' has kind 'Nat -> *' In the instance declaration for 'Foo SNat'"
--instance Foo SNat where -- type Bar SNat = 'Z -- bar = SZ
unless I redefine my class (which seems a bit silly)
class Foo' (a :: k -> *) where type Bar' a :: k bar' :: a (Bar' a)
now it works.
instance Foo' SNat where type Bar' SNat = 'Z bar' = SZ
clearly named kinds (?) are somehow distinct from unnames ones? CONFIDENTIALITY NOTICE
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