Hi Nicolas,
On 12/11/06, Nicolas Frisby <[EMAIL PROTECTED]> wrote:
The interpreter infers that m = (e ->) because of the types of snd and fst.
When snd and fst are considered as monadic computations in the (e ->)
monad, there types are:
Prelude> :t fst
fst :: (a, b) -> a
Prelude> :t snd
snd ::
Hi All, Hi Cale,
Can you tell me if I understood things right ? Please see below ...
On 12/11/06, Cale Gibbard <[EMAIL PROTECTED]> wrote:
The monad instance which is being used here is the instance for ((->)
e) -- that is, functions from a fixed type e form a monad.
So in this case:
liftM2 ::
Hi Cale !
On 12/11/06, Cale Gibbard <[EMAIL PROTECTED]> wrote:
The monad instance which is being used here is the instance for ((->)
e) -- that is, functions from a fixed type e form a monad.
So in this case:
liftM2 :: (a1 -> a2 -> r) -> (e -> a1) -> (e -> a2) -> (e -> r)
I bet you can guess wh
Hi All,
I'm loving learning Haskell quite a bit.
It is stretching my brain but in a delightfull way.
I've googled, I've hoogled but I haven't found a clear explanation for
what exactly liftM2 does in the context below.
Using the cool lambdabot "pointless" utility I found out that:
\x -> snd(x
Hi Spencer,
On 11/30/06, Spencer Janssen <[EMAIL PROTECTED]> wrote:
I believe you're talking about the `pl' plugin for lambdabot.
Lambdabot has an offline mode, visit the homepage for the source:
http://www.cse.unsw.edu.au/~dons/lambdabot.html
That's exactly what I was looking for!
Tha
Hi All!
Haskell newbie here with a very simple question because google and
hoogle are of no help.
On the IRC channel #haskell (which I cannot access now from work) I
saw somebody using a tool which automatically simplifies
expressions,composition of multiple functions to the bare minimum. It
was
