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
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