haskell - Type构造函数接受相同类型的几种数据类型，如何获得同名访问器？

Question

简短:我有一个解决我的问题的方法，但它似乎有点矫枉过正，所以我想知道我是否遗漏了什么。

长 :
我有 2 个数据类型，一个动物类型和一个所有者类型。两者都具有相同的“属性”年龄和名称。为了简单起见，我希望能够在 Animal 和 Owner 上无差别地调用年龄和名称

type Age = Int
type Name = String

data AnimalType = Dog | Cat | Snake
 deriving (Read, Show,Eq)

--with  datatype and pattern matching
data Animal = Animal AnimalType Name Age
 deriving(Show, Eq)
name (Animal _ name _) = name 
age (Animal _ _ age) = age 
animalType (Animal animalType _ _) = animalType 

data Owner = Owner Name Age [Animal]
 deriving(Show,Eq)
name    (Owner name _      _) = name 
age     (Owner _    age    _) = age
animals (Owner _    _      animals) = animals

garfield = Animal Cat "Garfield" 8
rantanplan = Animal Dog "Rantanplan " 4
kaa = Animal Snake "Kaa" 15

dupond = Owner "Dupont" 28 [garfield, rantanplan]
bob = Owner "Bob" 35 [kaa]

这不编译，

  Multiple declarations of `age'

同样的事情也不适用于 Record 语法。

这样做，一个人被迫为所有者和动物的年龄命名不同的年龄。

然后我挖了一下，发现我可以使用 typeclass 来实现这一点。

type Age = Int
type Name = String

class Nameable a where
 name:: a -> Name

class Ageable a where
 age:: a -> Age


data AnimalType = Dog | Cat | Snake
 deriving (Read, Show,Eq)

--with  datatype and pattern matching
data Animal = Animal AnimalType Name Age
 deriving(Show, Eq)

instance Nameable Animal where
 name (Animal _ name _) = name 

instance Ageable Animal where
 age (Animal _ _ age) = age

animalType (Animal animalType _ _) = animalType 


data Owner = Owner Name Age [Animal]
 deriving(Show,Eq)

instance Nameable Owner where 
 name (Owner name _ _) = name

instance Ageable Owner where
 age (Owner _ age _) = age

animals (Owner _    _      animals) = animals



garfield = Animal Cat "Garfield" 8
rantanplan = Animal Dog "Rantanplan " 4
kaa = Animal Snake "Kaa" 15

dupond = Owner "Dupont" 28 [garfield, rantanplan]
bob = Owner "Bob" 35 [kaa]

这种方法几乎与 Java 中接口(interface)的使用相同。第一个不起作用的方法更接近于旧的 C 结构方法。

有没有更快的方法来达到同样的效果？

Answer 1

Haskell 中的记录(及其访问器)是……次优的。也就是说，这个关于重复记录字段的特殊问题有一个解决方法(从 GHC 8.0 开始)，形式为 DuplicateRecordFields 。扩大。请注意，必须以明确的方式使用记录访问器(这里没有发生花哨的多态性)。

{-# LANGUAGE DuplicateRecordFields #-}

type Age = Int
type Name = String

data AnimalType = Dog | Cat | Snake
  deriving (Read, Show, Eq)

data Animal = Animal
  { animalType :: AnimalType
  , name :: Name
  , age :: Age
  } deriving(Show, Eq)

data Owner = Owner
  { name :: Name
  , age :: Age
  , animals :: [Animal]
  } deriving(Show, Eq)

garfield = Animal Cat "Garfield" 8
rantanplan = Animal Dog "Rantanplan " 4
kaa = Animal Snake "Kaa" 15

dupond = Owner "Dupont" 28 [garfield, rantanplan]
bob = Owner "Bob" 35 [kaa]