[Haskell-beginners] liftM2, how does it work?
franco00 at gmx.com
franco00 at gmx.com
Fri Feb 24 17:42:51 CET 2012
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:
instance Monad [] where
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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