作为 Monoid 实例的函数类型

Question

我有一个看起来像这样的函数

transition :: State -> ([State], [State])

鉴于我的问题的特定领域，我知道如何将两个连续的 transition 函数调用链接在一起，如下所示：

transition `chain` trainsition ... `chain` transition

但是，我想将其表示为 Monoid 并执行与 <> 和 mappend 的链接。不幸的是，我似乎无法使以下或类似的变体起作用：

instance Monoid (State -> ([State], [State])) where
    mempty  = ...
    mappend = ...

返回错误如下：

• Illegal instance declaration for
    ‘Monoid (State -> ([State], [State]))’
    (All instance types must be of the form (T a1 ... an)
     where a1 ... an are *distinct type variables*,
     and each type variable appears at most once in the instance head.
     Use FlexibleInstances if you want to disable this.)
• In the instance declaration for
    ‘Monoid (State -> ([State], [State]))’

一般来说，函数如何表示为Monoid的实例？

Answer 1

Functions are already instances of monoids in a different way。您希望 Haskell 如何决定使用该实例还是您的实例？解决问题的通常方法是声明一个 newtype 包装器，例如

newtype Transition a = Transition { runTransition :: a -> ([a], [a]) }

然后，你可以让你的幺半群实例就好了：

instance Monoid (Transition a) where
  mempty  = ...
  mappend = ...

完成此操作后，您甚至可能会发现 foldMap 很有用。而不是写类似

的东西

runTransition (Transition  transition `chain`
               Transition  transition `chain`
               ...
               Transition  transition)

您可以使用foldMap

runTransition (foldMap Transition [transition, transition, ... transition])

作为 Monoid 实例的函数类型

Function Types as Monoid Instances

haskell

functional-programming

monoids