Michael Roth wrote:
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]) }
You want: data Stack a b = Stack { run :: [a] -> (b, [a]) } All monads have the property that they can represent calculations of values of any type. Without that, they're not so useful. So your "Stack" needs two type variables : one for the kind of thing stacked, and one for what we're actually calculating at this instant. The correct types for the other functions are:
push :: a -> Stack a
push :: a -> Stack a () [you could have a -> Stack a a, if you want to echo back the pushed value, but why bother? ]
pop :: Stack a
pop :: Stack a a
top :: Stack a
top :: Stack a a With those clues I think you will be able to write >>= and return more successfully! There are some other interesting combinators to consider, like: isolate :: Stack b x -> Stack a x -- runs the computation with an empty stack, therefore -- guaranteeing it does more pushes than pops or isolate :: Stack b x -> Stack a ([b],x) -- if you want to collect the pushes. and so on. Jules