为什么我有类型错误？哈斯克尔

Question

applyAll :: [[a] -> [a]]  -> [a] -> [a]
applyAll [] [] = []
applyAll [] a = a   
applyAll (f1:fl) a = applyAll( (drop 1 fl)(f1 a))

我收到了这个错误

Expression     : drop 1 fl (f1 a)         
Term           : drop         
Type           : Int -> [e] -> [e]         
Does not match : a -> b -> c -> d          

我想做那样的事

applyAll [tail, tail, tail, tail] [1,2,3,4,5] = [5],         
applyAll [(map (* 2)), (map (+ 1))] [1,2,3,4,5]) = [3,5,7,9,11]

Answer 1

问题是你有

applyAll( (drop 1 fl)(f1 a))

这被解析为

  applyAll ((drop 1 fl) (f1 a))
= applyAll (drop 1 fl (f1 a))

这告诉编译器drop 1 fl必须是一个应用的函数(f1 a).但是,我们知道drop 1 fl必须返回一个列表,所以这显然是个问题.正如@Lee指出的那样,你需要类似的东西

applyAll :: [[a] -> [a]] -> [a] -> [a]
applyAll [] a = a
applyAll (f:fs) a = applyAll fs (f a)

虽然我已经合并了他的前两个案例.您也可以为此函数提供更通用的类型

applyAll :: [a -> a] -> a -> a

这可以替代地使用foldras 来实现

applyAll fs a = foldr ($) a fs

然后你根本不用担心基本情况.这工作,因为foldr t b接受一个列表,并取代所有:与它t,并替换[]用b在这个例子中,因此:

  foldr ($) [1, 2, 3, 4] (replicate 3 tail)
= foldr ($) [1, 2, 3, 4] (tail : tail : tail : [])
--            replace these    ^      ^      ^  ^
--            with $ and replace the empty list |
--            with [1, 2, 3, 4]
= (tail $ (tail $ (tail $ [1, 2, 3, 4])))
= (tail $ (tail $ (tail $ [1, 2, 3, 4])))
= (tail $ (tail $ [2, 3, 4]))
= (tail $ [3, 4])
= [4]