# [Haskell-beginners] liftM2, how does it work?

Kyle Murphy orclev at gmail.com
Fri Feb 24 21:11:42 CET 2012

```It might be helpful to fully replace the >>= operators with their
definitions:

m1 >>= \x1 -> m2 >>= \x2 -> return (f x1 x2)
foldr ((++) . (\x1 -> m2 >>= \x2 -> return (f x1 x2))) [] m1
foldr ((++) . (\x1 -> foldr ((++) . (\x2 -> return (f x1 x2))) [] m2)) [] m1
foldr ((++) . (\x1 -> foldr ((++) . (\x2 -> [f x1 x2])) [] m2)) [] m1

foldr ((++) . (\x1 -> foldr ((++) . (\x2 -> [x1 + x2])) [] [3,4])) [] [1,2]

Does that make a bit more sense?

-R. Kyle Murphy
--
Curiosity was framed, Ignorance killed the cat.

On Fri, Feb 24, 2012 at 11:42, <franco00 at gmx.com> wrote:

>  I am a bit puzzled by liftM2. I know what it does, but let's take a look
> at
> the implementation.
>
> liftM2  :: (Monad m) => (a1 -> a2 -> r) -> m a1 -> m a2 -> m r
> liftM2 f m1 m2          = do { x1 <- m1; x2 <- m2; return (f x1 x2) }
>
> Which could be rewritten (I am a bit uncomfortable with the do notation):
>
> m1               >>= \x1 ->
> m2               >>= \x2 ->
> return (f x1 x2)
>
> so: liftM2 (+) [1,2] [3,4] = [4,5,5,6]
>
> The instance of the List monad tells me that:
>
>     m >>= k   = foldr ((++) . k) [] m
>     m >> k    = foldr ((++) . (\ _ -> k)) [] m
>     return x  = [x]
>     fail _    = []
>
> That foldr is just doing a concatMap, it seems to me.
> But I still don't get the | return (f x1 x2) | part.
> The first thing I thought was:
>
> return ( (+) [1,2] [3,4] )
>
> but of course this is not the IO monad, the bind operator does not just
> fetch stuff from outer space.
>
> What is there really 'inside' x1 and x2 when the return statement is
> evaluated, though?
>
> I think I intuitively got it (x1 and x2 are bound to m1 and m2, every
> f x1 x2 expression is evaluated and then put together in a list), but
> it would be helpful to see what really happens to interiorise it.
>
> I started from the first step
>
> (>>=) m1 = foldr. ((++) . k) [] m1 -- partially applying m1
>
> but cannot really proceed from there. Any help will be appreciated,
> if you think the question is not clear enough, do not hesitate to ask.
>
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