This gets me pretty close. I now have to figure out how to return a PU Props instead of a PU [Prop] which is what it's returning. Still, it's a huge step forward. props :: Props -> PU Props props m = props' $ sort $ M.toList m where props' [] = lift [] props' ((_, (_, x)):xs) = wrap (\(a, b) -> a : b, \(a : b) -> (a, b)) (pair (prop x) (props' xs)) ./Script/Prop.hs:82:10: Couldn't match `Props' against `[Prop]' Expected type: PU Props Inferred type: PU [Prop] In the expression: props' $ (sort $ (Data.Map.toList m)) In the definition of `props': props m = props' $ (sort $ (Data.Map.toList m)) where props' [] = lift [] props' ((_, (_, x)) : xs) = wrap (\ (a, b) -> a : b, \ (a : b) -> (a, b)) (pair (prop x) (props' xs)) Big thanks to Andrew and Malcolm on #haskell for pointing me in the right direction! Thanks, Joel On Nov 8, 2005, at 5:23 PM, Andrew Pimlott wrote:

You might have better luck getting if you send a complete example, since it's hard for me to figure out exactly what this code is supposed to do. However, I'll take a quick stab.

On Tue, Nov 08, 2005 at 04:46:56PM +0000, Joel Reymont wrote:

...

props :: Props -> PU Props props m = props' $ sort $ M.toList m where props' [] = lift [] props' ((_, (Attr _ _ pp := _)):xs) = wrap (\(a, b) -> a : b, \(a : b) -> (a, b)) (pair pp (props' xs))

./Script/Prop.hs:80:10: Inferred type is less polymorphic than expected Quantified type variable `a' is mentioned in the environment: props' :: [(a1, Prop)] -> PU [a] (bound at ./Script/Prop.hs: 79:10)

Recall that pp is a (PU a) for some a. However, the inferred type for props' is fully polymorphic in a. It might help first of all to write down the type you expect for props', and then add type annotations to different pieces until you find the one that's going awry.

Andrew

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