Question

假设

type family F (x :: Nat) (y :: Nat) :: (Nat, Nat) where
  F x y = '(x, y)

然后

:t Proxy :: Proxy '[ (F 1 2) ]
Proxy :: Proxy '[ (F 1 2) ] :: Proxy '['(1, 2)]

但现在假设

type family ZipWith (f :: a -> b -> c) (as :: [a]) (bs :: [b]) :: [c] where
  ZipWith f (a ': as) (b ': bs) = (f a b) ': (ZipWith f as bs)
  ZipWith f as        bs        = '[]

:t Proxy :: Proxy (ZipWith F ('[1,2]) ('[3,4]))
Proxy :: Proxy (ZipWith F ('[1,2]) ('[3,4])) :: Proxy '[F 1 3, F 2 4]

为什么不是类型

Proxy :: Proxy (ZipWith F ('[1,2]) ('[3,4])) :: Proxy '['(1, 3), '(2, 4)]

如果我说

:t (Proxy :: Proxy (ZipWith F ('[1,2]) ('[3,4]))) :: Proxy '[ '(1, 3), '(2, 4)]

然后我收到类型错误

    Couldn't match type ‘F’ with ‘'(,)’
Expected type: Proxy (ZipWith F '[1, 2] '[3, 4])
  Actual type: Proxy '['(1, 3), '(2, 4)]
In the expression:
    (Proxy :: Proxy (ZipWith F ('[1, 2]) ('[3, 4]))) ::
      Proxy '['(1, 3), '(2, 4)]

但是F是'（，）所以发生了什么？这与注入性有关吗？

编辑回应评论，这是一个真正的答案：

{-# LANGUAGE DataKinds             #-}
{-# LANGUAGE TypeOperators         #-}
{-# LANGUAGE KindSignatures        #-}
{-# LANGUAGE GADTs                 #-}
{-# LANGUAGE Rank2Types            #-}
{-# LANGUAGE ScopedTypeVariables   #-}
{-# LANGUAGE MultiParamTypeClasses #-}
{-# LANGUAGE FlexibleInstances     #-}
{-# LANGUAGE TypeFamilies          #-}
{-# LANGUAGE UndecidableInstances  #-}
{-# LANGUAGE PolyKinds             #-}

为什么不扩展所有类型？

0 个答案: