如何在haskell中实现相同记录类型的不同实现？

Question

我理解为我们需要使用相同类型的不同实现 newtype

data Person = Person {
  name :: String
  , age :: Int
} deriving Show

class Describable a where describe :: a -> String

instance Describable Person where
  describe person = name person ++ " (" ++ show (age person) ++ ")" 

newtype AnotherPerson = AnotherPerson Person

但是,在相同字段名称的记录之间存在名称冲突的haskell问题

instance Describable AnotherPerson where
  describe person = name person ++ " - " ++ show (age person)

<interactive>:79:65: error:
    • Couldn't match expected type ‘Person’
                  with actual type ‘AnotherPerson’
    • In the first argument of ‘name’, namely ‘person’
      In the first argument of ‘(++)’, namely ‘name person’
      In the expression: name person ++ " - " ++ show (age person) 

我尝试使用pragma,DuplicateRecordFields但它没有帮助.我们应该怎么做？

Answer 1

你这里没有任何重复的记录字段; 唯一的记录字段name和age对Person类型.该AnotherPerson类型是不是一个记录,它没有记录的字段.AnotherPerson"包裹"一个Person价值; 它不会"继承"该Person类型的字段.

该AnotherPerson构造函数有型(非记录)字段Person.您可以模式匹配AnotherPerson以获取基础Person值:

instance Describable AnotherPerson where
  describe (AnotherPerson person) =
    name person ++ " - " ++ show (age person)