Hi Nicola, Am Montag, den 11.12.2006, 17:15 +0100 schrieb Nicola Paolucci:
Using the cool lambdabot "pointless" utility I found out that:
\x -> snd(x) - fst(x)
is the same as:
liftM2 (-) snd fst
I like the elegance of this but I cannot reconcile it with its type. I can't understand it. I check the signature of liftM2 and I get:
Prelude> :t liftM2 Prelude> liftM2 :: (Monad m) => (a1 -> a2 -> r) -> m a1 -> m a2 -> m r
Can someone help me understand what's happening here ? What does a Monad have to do with a simple subtraction ? What is actually the "m" of my example ?
You came across a very nice Monad: “((->) a)”, or the Monad of all functions that take a parameter of type a. Do not be confused by the arrow _before_ the a, it is actually behind the a: A function of type “a -> b” has type “(->) a b”. The same syntax as for infix operators applies, and can be curried to “((->) a)”. So in your example, snd and fst are computations in the ((->) (a,b)) Monad, and liftM2 (-) get’s the type: Num a => ((->) (a,a) a) -> ((->) (a,a) a) -> ((->) (a,a) a) or Num a => ((a,a) -> a) -> ((a,a) -> a) -> ((a,a) -> a) (at least I think so...) So if you ever feel like squaring something: square = join (*) Greetings, Joachim
I am sure if I get this I'll be another step closer to illumination ...
Thanks, Nick _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
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