[Haskell-cafe] Gentle introduction questions / comments

[Cristiano Paris]

2009/1/27 Matthijs Kooijman <matthijs at stdin.nl>
>
> Hi all,
>
> I've been reading the gentle introduction to Haskell a bit more closely today
> and there a few things which I can't quite understand (presumably because they
> are typo's). I've found two issues with the "Using monads" section [1]. Not
> sure if this is the right place to report this, but there's probably somewhere
> here who can fix it :-)
>
> The first one is in the state monad data type. It reads:
>
>        data SM a = SM (S -> (a,S))  -- The monadic type
>
> I  can't seem to find out what "S" is exactly here. It seems to be a
> universally quantified type variable, but then it should be lowercase.
> Considering that it is referred to later on as "s", I suspect this is a type?

I guess it's wrong.

S should be the type of the state but then it should've been declared
as a type variable. I think a more readable declaration could be:

data SM s a =  SM { runState :: s -> (a,s) }

> Further on, a resource monad is described. Here, a lift1 function is given
> to allow one to lift a normal computation into a resource aware computation,
> with the following type signature:
>
>        lift1                   :: (a -> b) -> (R a -> R b)

Mmmmhhh... this seems the signature of the liftM function, whose
purpose is to make a function operate on monadic values instead of
pure values. Notice that this is different from the "lift" function
you described above. A computation is a monadic value (i.e. an object
of the type "Monad m => m a") and the "lift" function wraps that value
into a monad transformer, making it a new monadic value (i.e. an
object of the type "MonadTrans t, Monad m => t m a").

> A few lines down, the inc operation is defined as follows:
>
>        inc i                   =  lift1 (i+1)

This is clearly wrong, as the expression "i+1" is not of the function
type, as required by the signature of lift1 (i.e. a -> b).

> To me, this seems wrong. Even without going into the semantics, it seems that
> the (i+1) expression does not satisfy the type of the first argument to lift1.
> I suspect this should have been:
>
>        inc                     = lift1 (+1)

You're right.

Cristiano