Haskell:无法构造无限类型:a~也许是a

Question

我对二叉树和获取最左边元素的函数有以下定义，或者如果树是Empty树，则返回Nothing，但它说x(类型为a(是无限类型，因为它返回Maybe a？

感谢任何帮助都将不胜感激：(

data Tree a = EmptyTree | Node a (Tree a) (Tree a) deriving (Show, Eq)
leftistElement :: (Ord a) => Tree a -> Maybe a
leftistElement EmptyTree = Nothing
leftistElement (Node x left _) =
if left == EmptyTree
then x
else leftistElement left

types.hs:55:10: error:
• Occurs check: cannot construct the infinite type: a ~ Maybe a
• In the expression: x
In the expression:
if left == EmptyTree then x else leftistElement left
In an equation for ‘leftistElement’:
leftistElement (Node x left _)
= if left == EmptyTree then x else leftistElement left
• Relevant bindings include
left :: Tree a (bound at types.hs:53:24)
x :: a (bound at types.hs:53:22)
leftistElement :: Tree a -> Maybe a (bound at types.hs:52:1)
|
55 |     then x
|          ^
Failed, no modules loaded.

Answer 1

x的类型为a，但您尝试将其作为Maybe a类型的值返回。由于a是类型变量，因此类型检查器会尝试查看是否存在使a和Maybe a统一的Maybe实例，从而导致无限类型错误。

使用pure(或Just(返回正确类型的值。

leftistElement (Node x left _) =
if left == EmptyTree
then pure x
else leftistElement left