Prelude Control.Monad> liftM2 (\a b -> a : b : []) "abc" "123" ["a1","a2","a3","b1","b2","b3","c1","c2","c3"] Prelude Control.Monad> Got it! Thanks to all. Michael --- On Sat, 7/24/10, aditya siram wrote: From: aditya siram Subject: Re: [Haskell-cafe] Heavy lift-ing To: "Max Rabkin" Cc: "haskell-cafe@haskell.org" , "Lennart Augustsson" Date: Saturday, July 24, 2010, 10:35 AM Perhaps I'm being unclear again. All I was trying to say was that: liftM2 (-) [0,1] [2,3] /= liftM2 (-) [2,3] [0,1] -deech On Sat, Jul 24, 2010 at 9:30 AM, Max Rabkin wrote:

On Sat, Jul 24, 2010 at 4:08 PM, aditya siram wrote:

...

I wouldn't-it was a bad example. My only point was that because of the way (>>=) is implemented for lists the order of the arguments 'a' and 'b' in 'liftM2 f a b' matters.

-deech

No, it's not. The type of liftM2 makes this clear:

liftM2 :: (Monad m) => (a -> b -> r) -> m a -> m b -> m r

The arguments to the function *must* come in the right order, because there is no way to match (a) with (m b) or (b) with (m a). Since liftM2 is parametrically polymorphic in (a) and (b), it can't behave differently in the case where (a = b).

--Max

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On 17/08/10 17:13, Tilo Wiklund wrote:

On 24/07/2010, aditya siram wrote:

...

Perhaps I'm being unclear again. All I was trying to say was that: liftM2 (-) [0,1] [2,3] /= liftM2 (-) [2,3] [0,1]

-deech

I'm sorry if I'm bumping an old thread, but why should "liftM2 f" be commutative when "f" isn't?

(I hope I'm not responding incorrectly)

I think the point that was being made is that: liftM2 (flip f) /= flip (liftM2 f) This is because the former (well: liftM2 (flip f) a b) effectively does: do {x <- a; y <- b; return (f y x)} Whereas the latter (flip (liftM2 f) a b) effectively does: do {y <- b; x <- a; return (f y x)} That is, the order of the arguments to liftM2 matters because they are executed in that order. So lifting the flipped function has a different effect to flipping the lifted function. Thanks, Neil.