Haskell 变量列表构造函数，如何默认为正确的类型并获得类型安全性

以下是我得到的：

{-# LANGUAGE MultiParamTypeClasses
           , FlexibleInstances #-}

class ListResultMult r a where
  lstM :: a -> [a] -> r

listM :: ListResultMult r a => a -> r
listM a = lstM a []


instance ListResultMult r a => ListResultMult (a -> r) a where
  lstM a as x = lstM x $ a:as

instance ListResultMult [a] a where
  lstM a as = reverse $ a:as

下面是它的工作原理：

> listM 'a' 'b' 'c' :: String
"abc"
> putStrLn $ listM 'a' 'b' 'c'
abc
> listM (1::Int) (2::Int) :: [Int]
[1,2]

下面是它失败的原因

> sum $ listM 1 2
No instance for (ListResultMult (a2 -> [a0]) a1) ...
> listM 1 :: [Int]
No instance for (ListResultMult [Int] a0) ...

与printf相比：

instance Show a => ListResultMult (IO ()) a where
  lstM a as = print . reverse $ a:as

> listM "foo" "bar" -- boo
No instance for (ListResult t0 [Char]) ...
> printf "%s %s" "foo" "bar"
foo bar
> listM "foo" "bar" :: IO () -- yay
["foo","bar"]

类型不安全：

> :t listM 2 "foo"
Some weird type is actually inferred

下面是我想做的：

• 类型安全。我想，当我定义

  listrestmult r a=>listrestmult（a->r）a

  和

  listrestmult[a]a

  时，它只允许您建立同构列表，如果您不这样做，就会注意到一个类型错误为什么没有呢？
• 违约。我不知道

  listm1:：[Int]

  到底发生了什么奇怪的事情怎么了？

---

类型函数似乎正是解决此问题的关键。下面是一个示例文件：

{-# LANGUAGE TypeFamilies #-}

class ListResultMult r where
    type Elem r
    lstM :: Elem r -> [Elem r] -> r

listM a = lstM a []

instance (ListResultMult r, Elem r ~ a) => ListResultMult (a -> r) where
    type Elem (a -> r) = a
    lstM a as x = lstM x (a:as)

instance ListResultMult [a] where
    type Elem [a] = a
    lstM a as = reverse (a:as)

以下是您在ghci中的示例：

*Main> listM 'a' 'b' 'c' :: String
"abc"
*Main> putStrLn $ listM 'a' 'b' 'c'
abc
*Main> listM 1 2 :: [Int]
[1,2]
*Main> sum $ listM 1 2
3
*Main> listM 1 :: [Int]
[1]
*Main> :t listM 'a' True

<interactive>:1:7:
    Couldn't match type `Bool' with `Char'
    In the first argument of `listM', namely 'a'
    In the expression: listM 'a' True
*Main> :t listM 2 "foo"

<interactive>:1:7:
    No instance for (Num [Char])
      arising from the literal `2'
    Possible fix: add an instance declaration for (Num [Char])
    In the first argument of `listM', namely `2'
    In the expression: listM 2 "foo"

*Main>listma'b'c'：：字符串
“abc”
*Main>putStrLn$listM'a''b''c'
abc
*Main>listm12:：[Int]
[1,2]
*Main>sum$listm1 2
3.
*Main>listm1:：[Int]
[1]
*Main>：t listM'a'True
:1:7:
无法将类型“Bool”与“Char”匹配
在'listM'的第一个参数中，即'a'
在表达式中：listM'a'True
*Main>：t listm2“foo”
:1:7:
没有（Num[Char]的实例）
源于字面上的“2”
可能的修复方法：为（Num[Char]）添加实例声明
在'listM'的第一个参数中，即'2'
在表达式中：listm2“foo”

键入安全和理智默认值！这完美地回答了这个问题，同时也巧合地回答了这个问题