On Tue, Nov 30, 2010 at 12:08 PM, Larry Evans <cppljevans@suddenlink.net> wrote:

so now I must "manually" figure out what the a and b in
the ap declaration correspond to in the return(:) type:

       m    ( a             ->           b           )
       __     _                          _
    1: []     Char          ->     [Char]->[Char]
    2: []     Char->[Char]  ->          [Char]

A type a -> b -> c is always equivalent to the type a -> (b->c), not (a->b) -> c.

In particular, breaking down sequence (c:cs) = return (:) `ap` c `ap` sequence cs

 return (:) :: m (a -> [a] ->[a])
 (\c -> return (:) `ap` c) :: m a -> m ([a] -> [a])
 (\c cs' -> return (:) `ap` c `ap` cs') :: m a -> m [a] -> m [a]

therefore

 sequence :: [ m a ] -> m [a]

Perhaps a special tutorial interpreter would be of use, but I've had some success simply passing in anonymous functions to ghci's :t operator, since that lets me simplify a program one bit at a time, inferring the types that might confuse me.