[Haskell-cafe] Problem using ap -- No instance for (Monad ((->) [[a]]))

[Jim Burton]

If I try  a function to make a point-free version of the function in this
fold --

foldr (\x ys -> ys ++ map (x:) ys) [[]]

:pl gives me

GOA Control.Monad> :pl (\x ys -> ys ++ map (x:) ys)
 ap (++) . map . (:)
GOA Control.Monad> :t  ap (++) . map . (:)
ap (++) . map . (:) :: (Monad ((->) [[a]])) => a -> [[a]] -> [[a]]

but this is what happens if I try to use it:

GOA Control.Monad> let ps = foldr (ap (++) . map . (:)) [[]]

<interactive>:1:16:
    No instance for (Monad ((->) [[a]]))
      arising from use of `ap' at <interactive>:1:16-22
    Possible fix: add an instance declaration for (Monad ((->) [[a]]))
    In the first argument of `(.)', namely `ap (++)'
    In the first argument of `foldr', namely
        `((ap (++)) . (map . (:)))'
    In the expression: foldr ((ap (++)) . (map . (:))) [[]]
(0.00 secs, 0 bytes)

What gives?



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