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实例实现的模式匹配

typeclass pattern-matching haskell

bzim · 技术社区 · 8 年前

我正在尝试实现数据类型的show方法。

data OptionList a b = EmptyOpt | OptionList { optListHead :: a, optListTail :: b } deriving (Read)

instance (Show a, Show b) => Show (OptionList a b) where
    show (OptionList a EmptyOpt) = "{" ++ (show a) ++"}"
    show (OptionList EmptyOpt b) = "{" ++ (show b) ++"}"
    show (OptionList a b) = "{"++ (show a) ++ ", " ++ (show b) ++"}"
    show EmptyOpt = ""

我希望选项列表不显示逗号,如果: a 或 b 具有由EmptyOpt构造的值。但编译器显示以下错误:

OptionList.hs:11:28:
    Couldn't match expected type âbâ
                with actual type âOptionList t0 t1â
      âbâ is a rigid type variable bound by
          the instance declaration at OptionList.hs:10:10
    Relevant bindings include
      show :: OptionList a b -> String (bound at OptionList.hs:11:9)
    In the pattern: EmptyOpt
    In the pattern: OptionList a EmptyOpt
    In an equation for âshowâ:
        show (OptionList a EmptyOpt) = "{" ++ (show a) ++ "}"

OptionList.hs:12:26:
    Couldn't match expected type âaâ
                with actual type âOptionList t2 t3â
      âaâ is a rigid type variable bound by
          the instance declaration at OptionList.hs:10:10
    Relevant bindings include
      show :: OptionList a b -> String (bound at OptionList.hs:11:9)
    In the pattern: EmptyOpt
    In the pattern: OptionList EmptyOpt b
    In an equation for âshowâ:
        show (OptionList EmptyOpt b) = "{" ++ (show b) ++ "}"

更新 :OptionList应该类似于无类型列表。

(+:) :: a -> b -> (OptionList a b)
infixr 5 +:
t1 +: t2 = OptionList t1 t2

因此,列表如下: 0 +: "test" +: True 将被定义为: OptionList Int (OptionList String (OptionList Bool EmptyOpt)) 而且会 show n as {0, {"test", {True}}}

3 回复 | 直到 8 年前

ase 8 年前

您的更新的更新。 如果您愿意打开一些扩展,您可以使其工作:

{-# LANGUAGE FlexibleInstances #-}

data EmptyOpt = EmptyOpt

data OptionList a b =
  OptionList a b
  deriving (Read)

instance (Show a, Show b) => Show (OptionList a b) where
  show (OptionList a b) = "{ " ++ show a ++ ", " ++ show b ++ " }"

instance {-# OVERLAPPING  #-} (Show a) => Show (OptionList a EmptyOpt) where
  show (OptionList a EmptyOpt) = "{ " ++ show a ++ " }"

(+:) :: a -> b -> (OptionList a b)
infixr 5 +:
t1 +: t2 = OptionList t1 t2

test = 0 +: "test" +: True +: EmptyOpt

但就我个人而言,我会尝试做一些类似的事情

data Option = B Bool | I Int | S String
data OptionsList = Empty | OptionsList Option OptionsList

你的问题是实例头( (Show a, Show b) => Show (OptionList a b) )表示您正在执行 Show 对于 OptionList a b 哪里 a 和 b 有任何类型的 显示 实例,但在您的实现中,您需要 一 和 b 实际上是这类人 OptionList .

也许您可以将类型更改为更像普通列表:

data OptionList a
  = EmptyOpt
  | OptionList { optListHead :: a
              ,  optListTail :: OptionList a}
  deriving (Read)

然后您可以有一个实例:

instance (Show a) => Show (OptionList a) where
  show (OptionList a EmptyOpt) = "{" ++ show a ++"}"
  show (OptionList a b) = "{"++ show a ++ ", " ++ show b ++"}"
  show EmptyOpt = ""

willeM_ Van Onsem 8 年前

问题是在实例声明中:

instance (Show a, Show b) => Show (OptionList a b) where
    show (OptionList a EmptyOpt) = "{" ++ (show a) ++"}"
    show (OptionList EmptyOpt b) = "{" ++ (show b) ++"}"
    show (OptionList a b) = "{"++ (show a) ++ ", " ++ (show b) ++"}"
    show EmptyOpt = ""

你 使用数据构造函数 因此,Haskell正确地推导出,您实际上定义了 show 结束 OptionList (OptionList c d) (OptionList e f) A:毕竟 EmptyOpt 是 OptionList a b 所以 不能将这些用作参数 因为在实例的开头,您说要为 选项列表a b 带有泛型 a 和 b .

所以我不清楚你的目标是什么;将其修改为:

instance (Show a, Show b, Show c, Show d) => Show (OptionList (OptionList a b) (OptionList c d)) where

也不会有帮助,因为结构是递归的,因此您将定义具有无限递归深度的实例。

在我看来,唯一合理的是,您的数据定义是错误的,应该是这样的:

data OptionList a = EmptyOpt | OptionList { optListHead :: a, optListTail :: OptionList a }

在这种情况下,您可以将其定义为:

instance Show a => Show (OptionList a) where
    show (OptionList a b) = "{"++ show a ++ ", " ++ show b ++"}"
    show EmptyOpt = ""

或者沿着这些线的东西。

bzim 8 年前

我的想法是创建一个类型独立的列表。这个列表的元素应该包含声明的任何类型。但我会放弃,做一些类似C联合的事情:

data Option = OptStr String | OptInt Int | OptBool Bool