How and why is ap defined as liftM2 id in Haskell

Posted by luke_randall on Stack Overflow See other posts from Stack Overflow or by luke_randall
Published on 2011-03-18T23:40:26Z Indexed on 2011/03/19 0:10 UTC
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Whilst trying to better understand Applicative, I looked at the definition of <*>, which tends to be defined as ap, which in turn is defined as:

ap                :: (Monad m) => m (a -> b) -> m a -> m b
ap                =  liftM2 id

Looking at the type signatures for liftM2 and id, namely:

liftM2  :: (Monad m) => (a1 -> a2 -> r) -> m a1 -> m a2 -> m r
id                      :: a -> a

I fail to understand how just by passing in id, the relevant part of the type signature seems to transform from (a1 -> a2 -> r) -> m a1 to m (a -> b). What am I missing here?

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