Question

我想创建一个具有类型约束的函数定义的实例，但我不想将类型约束添加到类中。

class Foo a where
  f :: a b -> b

instance Foo Maybe where
  f = fMaybe

fMaybe :: (Num a) => Maybe a -> a
fMaybe (Just i) = i+i
fMaybe _ = 0

如何指定这是包含Maybe的{{1}}的实例？

这有效：

Nums

但我不想为每种{-# LANGUAGE MultiParamTypeClasses #-} class Foo a b where f :: a b -> b instance Foo Maybe Int where f = fMaybe fMaybe :: (Num a) => Maybe a -> a fMaybe (Just i) = i+i fMaybe _ = 0

声明实例

我试过了：

Num

但是我收到了错误：

class Foo a where
  f :: a -> b

instance (Num b) => Foo (Maybe b) where
  f = fMaybe

fMaybe :: (Num a) => Maybe a -> a
fMaybe (Just i) = i+i
fMaybe _ = 0

Answer 1

如果不改变类或类型，则无法实现。

MultiParamTypeClasses方法实际上并不要求您为所有Num类型编写单独的实例 - 以下工作：
```
class Foo a b where
  f :: a b -> b

instance (Num b) => Foo Maybe b where
  f = fMaybe
```
尽管如此，我并不认为这是一个特别好的方法。

您可以使用ConstraintKinds允许每个实例有选择地约束所包含的类型。

{-# LANGAUGE TypeFamilies, ConstraintKinds #-}

import GHC.Exts (Constraint)

class Foo a where
  type FooCstrt a b :: Constraint
  type FooCstrt a b = () -- default to unconstrained
  f :: FooCstrt a b => a b -> b

instance Foo Maybe where
  type FooCstrt Maybe b = Num b
  f = fMaybe

您可以切换到仅允许包含Num类型的类型。

{-# LANGUAGE GADTs #-}

data NMaybe where
  NJust :: Num b => b -> NMaybe b
  NNothing :: Num b => NMaybe b

然后，

class Foo a where
  f :: a b -> b

instance Foo NMaybe where
  f (NJust i) = i+1
  f (NNothing) = 0

在实例声明中键入约束但不在类中键入

1 个答案: