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

我想做那样的事情

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))

将其解析为

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

告诉编译器drop 1 fl必须是应用于(f1 a)的函数。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，你想要像

这样的东西

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

虽然我合并了他的前两个案例。您还可以为此函数提供更通用的类型

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

也可以使用foldr作为

来实现

applyAll fs a = foldr ($) a fs

然后你根本不用担心基本情况。这是有效的，因为foldr t b获取了一个列表并用:替换了其中的所有t，并将[]替换为b，因此在此示例中：

  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]

Answer 2

看起来你想要：

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

Answer 3

看起来像组成列表中的函数，翻转参数，与通常的组合(.)

相比

所以我写道： applyAll = foldr1 (flip (.))

我一直认为使用($)作为一个部分有点奇怪，而且难以理解

为什么我有类型错误？哈斯克尔

3 个答案: