任意类型类的约束中的非类型变量参数

Question

对于 Haskell Programming From First Principles 第 15 章的练习，我正在尝试基于另一个 Arbitrary 实例编写一个 Arbitrary 实例：

module AccumulateRight where

import Data.Semigroup
import Test.QuickCheck

data Validation a b = Fail a | Pass b deriving (Eq, Show)

newtype AccumulateRight a b =
    AccumulateRight (Validation a b) deriving (Eq, Show)

type TestType = AccumulateRight String [Int]

instance Semigroup b => Semigroup (AccumulateRight a b) where
    _ <> (AccumulateRight (Fail x)) = Fail x
    (AccumulateRight (Fail x)) <> _ = Fail x
    (AccumulateRight (Success a)) <> (AccumulateRight (Success b)) =
        AccumulateRight . Success $ a <> b

instance (Arbitrary a, Arbitrary b) => Arbitrary (Validation a b) where
    arbitrary = oneof [Fail <$> arbitrary, Pass <$> arbitrary]

instance Arbitrary (Validation a b) => Arbitrary (AccumulateRight a b) where
    arbitrary = AccumulateRight <$> arbitrary

semigroupAssoc :: (Eq m, Semigroup m) => m -> m -> m -> Bool
semigroupAssoc a b c = (a <> (b <> c)) == ((a <> b) <> c)

type Assoc = TestType -> TestType -> TestType -> Bool

main :: IO ()
main = quickCheck (semigroupAssoc :: Assoc)

但出现以下错误：

    • Non type-variable argument
        in the constraint: Arbitrary (Validation a b)
      (Use FlexibleContexts to permit this)
    • In the context: Arbitrary (Validation a b)
      While checking an instance declaration
      In the instance declaration for ‘Arbitrary (AccumulateRight a b)’
   |
22 | instance Arbitrary (Validation a b) => Arbitrary (AccumulateRight a b) where
   |          ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Failed, no modules loaded.

那么我在这里做错了吗？为什么我不能在这里使用现有数据的 typeclass 作为约束？

Answer 1

这是一个愚蠢的限制，在人们了解实现类型类有多么困难之前就已经设置好了。事实证明它很容易支持，所以有一个语言扩展——在错误中提到——让你这么说。你可以通过添加来开启它

{-# LANGUAGE FlexibleContexts #-}

到您文件的顶部，并且随着扩展的进行，这被认为是完全良性的。但是，在这种情况下，您不应该打开它，而应该只写

instance (Arbitrary a, Arbitrary b) => Arbitrary (AccumulateRight a b)

-- 毕竟，(Arbitrary a, Arbitrary b) 正是 Arbitrary (Validation a b) 成立的条件。