Haskell函数组成 - 错误推断类型

Question

在下面的Haskell代码中,函数typeError没有进行类型检查.

wrap x = [x]

listf :: [[a]] -> [[a]]
listf = id

typeCheck :: [a] -> [[a]]
typeCheck x = listf (wrap x)

typeError :: [a] -> [[a]]
typeError = wrap . listf

如果GHC取消注释,GHC会产生此错误:

Couldn't match type `a' with `[a0]'
  `a' is a rigid type variable bound by
      the type signature for typeError :: [a] -> [[a]] at tim.hs:10:1
Expected type: [a] -> [a]
  Actual type: [[a0]] -> [[a0]]
In the second argument of `(.)', namely `listf'
In the expression: wrap . listf

我不明白为什么.a应该能够统一[a0]- 它们是独立的类型变量.这正是它所推断的类型typeCheck- 但不是在使用.运算符时.

Hugs会产生一个非常相似且类似的虚假错误消息:

ERROR "repro.hs":10 - Inferred type is not general enough
*** Expression    : typeError
*** Expected type : [a] -> [[a]]
*** Inferred type : [[a]] -> [[[a]]]

此外,这工作正常:

listf' :: [a] -> [a]
listf' = id

typeCheck' :: [a] -> [[a]]
typeCheck' = wrap . listf'

只有[[a]]或[[[a]]]或更大时才会出现此问题.这是什么交易？

Answer 1

你似乎已经扭转了这里的功能构成.

-- This
typeCheck :: [a] -> [[a]]
typeCheck x = listf (wrap x)

-- is the same as
typeCheck' :: [a] -> [[a]]
typeCheck' = listf . wrap

-- while this
typeError :: [a] -> [[a]]
typeError = wrap . listf

-- is the same as
typeError' :: [a] -> [[a]] 
typeError' x = wrap (listf x)

现在应该明白为什么这不起作用.listf要求它的论点是[[a]],但typeError声称它适用于任何声明[a].