我的 Haskell 类型同义词有什么问题？

Question

我有两个控制循环的函数，continue and break:

type Control a = (a -> a) -> a -> a

continue :: Control a
continue = id

break :: Control a
break = const id

然后，我想简化Control类型的同义词。因此，我写道：

type Endo a = a -> a

type Control a = Endo (Endo a)

continue :: Control a
continue = id

break :: Control a
break = const id

然而，当我试图进一步简化它时，我得到了一个错误：

GHCi, version 7.10.2: http://www.haskell.org/ghc/  :? for help
Prelude> type Endo a = a -> a
Prelude> type Duplicate w a = w (w a)
Prelude> type Control a = Duplicate Endo a

<interactive>:4:1:
    Type synonym ‘Endo’ should have 1 argument, but has been given none
    In the type declaration for ‘Control’

我不明白为什么会出现此错误。也许你能启发我。

Answer 1

必须始终完全应用类型同义词。您不能部分应用它们。

如果您打算这样做，您可能需要重新键入它。

Answer 2

正如 Fraser 所说，这种东西通常无法工作，因为类型部分应用了类型同义词 make everything undecidable。

但是，如果您加入 -XLiberalTypeSynonyms 扩展，GHC 将内联同义词直到它可以得出推论：

Prelude> type Endo a = a -> a
Prelude> type Duplicate w a = w (w a)
Prelude> type Control a = Duplicate Endo a

<‌interactive>:4:1:
    Type synonym ‘Endo’ should have 1 argument, but has been given none
    In the type declaration for ‘Control’
Prelude> :set -XLiberalTypeSynonyms
Prelude> type Control a = Duplicate Endo a

我的 Haskell 类型同义词有什么问题？

What's wrong with my Haskell type synonym?

haskell

ocaml

type-inference

hindley-milner

type-synonyms