Am Mittwoch, 16. April 2008 14:56 schrieb Hans Aberg:
When I load the State module in Hugs, then I can define the function f below, but I do not immediately see exactly what function "return" returns. Explanation welcome.
For example:
f [2..4] [6..9]
[6,7,8,9,6,7,8,9,6,7,8,9] That is, it just repeats the second argument as many times as the length of the second argument.
Hans Aberg
-------- import Control.Monad.State
f :: Monad a => a b -> a c -> a c f x y = x >>= (return y) --------
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 (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 ). 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).