Difference between revisions of "No kind signatures"
From HaskellWiki
(How to live without KindSignatures) |
m (parens) |
||
Line 5: | Line 5: | ||
E.g. I want to define | E.g. I want to define | ||
<haskell> | <haskell> | ||
− | newtype MonadTransformer t m a = MonadTransformer t m a | + | newtype MonadTransformer t m a = MonadTransformer (t m a) |
</haskell> | </haskell> | ||
with <hask>a :: *</hask>, <hask>m :: * -> *</hask>, <hask>t :: (* -> *) -> (* -> *)</hask>, | with <hask>a :: *</hask>, <hask>m :: * -> *</hask>, <hask>t :: (* -> *) -> (* -> *)</hask>, |
Revision as of 18:21, 25 March 2013
Question
I have a datatype that needs other kinds than the compiler infers
but my compiler does not support KindSignatures
.
E.g. I want to define
newtype MonadTransformer t m a = MonadTransformer (t m a)
with a :: *
, m :: * -> *
, t :: (* -> *) -> (* -> *)
,
but the compiler infers
a :: *
, m :: *
, t :: * -> * -> *
!
Answer
You can achieve this using the phantom type of the Const
functor.
import Control.Applicative (Const(Const))
newtype MonadTransformer t m a = MonadTransformer (Const (t m a) (m a))
monadTransformer :: t m a -> MonadTransformer t m a
monadTransformer = MonadTransformer . Const
runMonadTransformer :: MonadTransformer t m a -> t m a
runMonadTransformer (MonadTransformer (Const m)) = m
For more than one kind specification you still need only one Const
,
since you can use a tuple type like so Const (t m a) (t m a, m a, a)
.
So to speak, the second type argument of Const
allows you to specify examples of how to use the type parameters
of MonadTransformer
.