On 16 Apr 2008, at 15:22, Daniel Fischer wrote:
The point is the
instance Monad ((->) a) where
return x = const x
f >>= g = \x -> g (f x) x
which is defined in Control.Monad.Instances...
Thank you. I suspected there was an instance somewhere, and I wanted
to know where it is defined.
(try in GHCI:
Prelude> let f x y = x >>= (return y)
Prelude> :t f
f :: (Monad ((->) a), Monad m) => m a -> m b -> m b
).
It works in Hugs too. If I don't import Control.Monad.State, then
f :: (Monad a, Monad ((->) b)) => a b -> a c -> a c
This is imported into Control.Monad.State and hence the instance is
visible.
By the type of (>>=), (return y) must have type (a -> m b), on the
other hand,
if y has type c, then (return y) has type (m' c) for some monad m'.
Unifying
m' c and a -> m b gives then m' === ((->) a) and c === m b.
Now according to the instance, return y === const y, so f is the
same as
g x y = x >>= (const y).
Good to know the details. Thanks.
Hans
_______________________________________________
Haskell-Cafe mailing list
Haskell-Cafe@haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe
