Haskell中的"无法创建无限类型"错误

Question

我想知道为什么Haskell会接受这个

perms xs = [ x:y | i <- [0..(length xs - 1)], x <- [xs!!i], y <- perms (takeOut i xs)]

但不会接受:

perms xs = [ x:(perms y) | i <- [0..(length xs - 1)], x <- [xs!!i], y <- (takeOut i xs)]

它抱怨说

[1/1]编译Main(abc.hs,解释)

Occurs check: cannot construct the infinite type: t = [t]
  Expected type: t -> [t]
  Inferred type: [t] -> [[a]]
In the second argument of `(:)', namely `(perms y)'
In the expression: x : (perms y)

我能理解它的内容,我不能理解为什么第一个是好的而第二个不是!

编辑:啊,我当然也有

perms [] = [[]]

在顶部.

谢谢

Answer 1

在第一个表达式中你x:y有意思,如果x :: a那么y :: [a].在x : perms y如果x :: a那么它必须是perms y :: [a],但perms y :: [[a]](排列的列表).Typechecker试图统一[a]和[[a]]和失败.