当使用Bool和有限2时,为什么非详尽模式警告？

Question

我希望定义一个新类的实例,因为它Bool没有创建部分函数Finite 2,但它不起作用.

我的代码:

-- SO test case, re: my HasFin instance for Bool.
--
-- David Banas <capn.freako@gmail.com>
-- February 9, 2018

{-# OPTIONS_GHC -Wall #-}

{-# LANGUAGE DataKinds #-}
{-# LANGUAGE FlexibleContexts #-}
{-# LANGUAGE LambdaCase #-}
{-# LANGUAGE TypeFamilies #-}

module Bogus.BoolHasFin where

import GHC.TypeLits
import Data.Finite
import Data.Finite.Internal (Finite(..))

class KnownNat (Card a) => HasFin a where
  type Card a :: Nat
  toFin :: a -> Finite (Card a)
  unFin :: Finite (Card a) -> a

instance HasFin Bool where
  type Card Bool = 2

  toFin False = finite 0
  toFin True  = finite 1

  unFin = \case
    Finite 0 -> False
    Finite 1 -> True

GHC编译结果如下:

Davids-Air-2:test dbanas$ stack ghc -- -c so_BoolHasFin.hs 

so_BoolHasFin.hs:30:11: warning: [-Wincomplete-patterns]
    Pattern match(es) are non-exhaustive
    In a case alternative:
        Patterns not matched: (Finite p) where p is not one of {1, 0}

任何人都可以帮助我理解为什么我会收到此警告？看起来,已经界定的参数unFin,通过Finite 2,应该已经足够了.

添加于2018-02-10:

根据Conal私下提出的建议,此代码:

unFin (Finite 0) = False
unFin _          = True

消除了警告.

Answer 1

我认为这是至少部分是因为没有什么的定义Finite限制了载有Integer被假定的范围内(0到n-1)newtype Finite (n :: Nat) = Finite Integer.