compress xs@(_:_:_) = (ifte <$> ((==) <$> head <*> head.tail) <$> ((compress.).(:) <$> head <*> tail.tail) <*> ((:) <$> head <*> compress.tail) ) xs
导致类型错误,但我不明白为什么。它应该相当于
compress xs@(_:_:_) = (ifte (((==) <$> head <*> head.tail) xs) (((compress.).(:) <$> head <*> tail.tail) xs) (((:) <$> head <*> compress.tail) xs))
,这不是。
注意ifte = (\ x y z -> if x then y else z)
:<$>
和<*>
来自Control.Applicative
。
编辑:错误是:
Couldn't match expected type `[a]' with actual type `[a] -> [a]'
In the expression:
(ifte <$> ((==) <$> head <*> head . tail)
<$>
((compress .) . (:) <$> head <*> tail . tail)
<*>
((:) <$> head <*> compress . tail))
$ xs
In an equation for `compress':
compress xs@(_ : _ : _)
= (ifte <$> ((==) <$> head <*> head . tail)
<$>
((compress .) . (:) <$> head <*> tail . tail)
<*>
((:) <$> head <*> compress . tail))
$ xs
我在尝试编写九十九个 Haskell 问题的第 8 题的无点解决方案时遇到了这个问题。我试图通过修改我写的有意义的解决方案来做到这一点,这是
compress::Eq a => [a]->[a]
compress [] = []
compress (x:[]) = (x:[])
compress (x:y:xs) = ifte ((==) x y) (compress (x:xs)) (x:(compress (y:xs)))