haskell - 发生检查: cannot construct the infinite type

Question

我要处理的任务如下:

pipe [f1,...,fn] x should return f1(f2(...(fn x)))

这是我的代码:

pipe :: [(a -> a)] -> (a -> a)
pipe fs   = foldl' f base fs
  where
    f a x = a (x)
    base  = (\x -> x)

我认为这在遍历函数列表时是有意义的，但是，编译器告诉我:

tutorial.hs:45:20: error:
    • Occurs check: cannot construct the infinite type: a ~ a -> a
      Expected type: (a -> a) -> (a -> a) -> a -> a
      Actual type: ((a -> a) -> a -> a) -> (a -> a) -> a -> a
    • In the first argument of ‘foldl'’, namely ‘f’
      In the expression: foldl' f base fs
      In an equation for ‘pipe’:
         pipe fs
            = foldl' f base fs
            where
                f a x = a (x)
                base = (\ x -> x)
    • Relevant bindings include
        fs :: [a -> a] (bound at tutorial.hs:45:6)
        pipe :: [a -> a] -> a -> a (bound at tutorial.hs:45:1)
   |
45 | pipe fs   = foldl' f base fs
   |                    ^
Failed, no modules loaded.

我是Haskell的新手，感谢您的提前帮助:)

Answer 1

你可以做

pipe :: [a -> a] -> (a -> a)
pipe = foldr (flip (.)) id

foldr和 foldl之间有区别。

您的 base函数等效于 id。

在您的版本中， f的定义不正确。它应该具有两个函数并返回一个新函数。

你可以写成

pipe :: [(a -> a)] -> (a -> a)
pipe fs   = foldl f id fs
  where
    f g h = g . h
    base  = (\x -> x)

要么

pipe :: [(a -> a)] -> (a -> a)
pipe fs   = foldl f id fs
  where
    f g h a = h (g a)
    base  = (\x -> x)

如果所有函数都是可交换的，则使用 foldr或 foldl没什么区别，否则您需要正确选择 foldr或 foldl。