haskell - 可变参数列表构造函数，如何默认为正确的类型并获得类型安全

Question

这是我所拥有的:

{-# 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

以下是它的工作原理:

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

这是失败的原因

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

与 printf 对比:

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

所以这就是我想要做的:

类型安全。我在定义 ListResultMult r a => ListResultMult (a -> r) a 时想和 ListResultMult [a] a ，它只允许您建立同构列表，并在您不这样做时注意到类型错误。 为什么没有呢？

违约。我不知道 listM 1 :: [Int] 发生了什么奇怪的事情. 这是怎么回事？

Answer 1

类型函数似乎只是解决这个问题的原因。这是一个示例文件:

{-# 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)

这是您在 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"