![]() |
От: |
nikov
|
http://www.linkedin.com/in/nikov |
Дата: | 09.06.09 11:46 | ||
Оценка: |
{-# LANGUAGE RankNTypes #-}
f :: [forall a. t a -> t a] -> t b -> t b
f = foldr (.) id
Couldn't match expected type `forall a. f a -> f a'
against inferred type `b -> c'
In the first argument of `foldr', namely `(.)'
{-# LANGUAGE RankNTypes #-}
f :: [forall a. t a -> t a] -> t b -> t b
f = foldr (\g -> (.) g) id