Hello list,
while trying to learn the secrets of monads, I decided to write a simply
monand for pure educational purpose. But it turned out that it isn't as
easy as I thought... I circumnavigate quite a number of hurdles but now
I reached a point where I'm at a loss. :-(
The source:
#! /usr/bin/env ghc
data Stack a = Stack { run :: [a] -> (a, [a]) }
push :: a -> Stack a
push x = Stack f where f xs = ( x, x:xs )
pop :: Stack a
pop = Stack f where f (x:xs) = ( x, xs )
top :: Stack a
top = Stack f where f (x:xs) = ( x, x:xs )
instance Monad Stack where
return x = Stack f where f xs = ( x, xs )
(>>=) stack g = Stack f where
f s0 = (x2, s2) where
(x1, s1) = run stack s0
(x2, s2) = run (g x1) s1
The errors:
./mymonad.hs:16:24:
Couldn't match expected type `b' (a rigid variable)
against inferred type `a' (a rigid variable)
`b' is bound by the type signature for `>>=' at <no location info>
`a' is bound by the type signature for `>>=' at <no location info>
Expected type: [b] -> (b, [b])
Inferred type: [a] -> (b, [b])
In the first argument of `Stack', namely `f'
In the expression: Stack f
./mymonad.hs:19:28:
Couldn't match expected type `b' (a rigid variable)
against inferred type `a' (a rigid variable)
`b' is bound by the type signature for `>>=' at <no location info>
`a' is bound by the type signature for `>>=' at <no location info>
Expected type: [b]
Inferred type: [a]
In the second argument of `run', namely `s1'
In the expression: run (g x1) s1
I think the problem is that my operator (>>=) is of type:
Stack a -> (a -> Stack a) -> Stack a
but should be:
Stack a -> (a -> Stack b) -> Stack b
But, I have simply no clue how to fix that. :-(
Can anybody give my a hint?
Thank you in advance.
Michael Roth
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