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我正在用Functors和QuickCheck做运动。QuickChecking simple Functors:是否需要定义任意实例?为什么?怎么样?
我有一个超级简单的函数,其组成法我希望quickCheck。 该Functor只是一个Identity a
。 这是我的代码至今:
import Data.Functor
import Test.QuickCheck
newtype Identity a = Identity a
instance (Eq a) => Eq (Identity a) where
(==) (Identity x) (Identity y) = x == y
(/=) (Identity x) (Identity y) = x /= y
instance Functor Identity where
fmap f (Identity x) = Identity (f x)
propertyFunctorCompose ::(Eq (f c), Functor f) => (a -> b) -> (b -> c) -> f a -> Bool
propertyFunctorCompose f g fr = (fmap (g . f) fr) == (fmap g . fmap f) fr
main = do
quickCheck $ \x -> propertyFunctorCompose (+1) (*2) (x :: Identity Int)
遗憾的是这段代码不能编译,GHC与此编译错误抱怨:
functor_exercises.hs:43:5:
No instance for (Arbitrary (Identity Int))
arising from a use of `quickCheck'
Possible fix:
add an instance declaration for (Arbitrary (Identity Int))
In the expression: quickCheck
In a stmt of a 'do' block:
quickCheck $ \ x -> propertyFunctorId (x :: Identity Int)
In the expression:
do { quickCheck $ \ x -> propertyFunctorId (x :: [Int]);
quickCheck
$ \ x -> propertyFunctorCompose (+ 1) (* 2) (x :: [Int]);
quickCheck (propertyFunctorCompose' :: IntFC);
quickCheck $ \ x -> propertyFunctorId (x :: Identity Int);
.... }
所以我已经开始看快速检查任意类型类,它需要定义arbitrary :: Gen a
和shrink :: a -> [a]
。
我有(也许是错误的)感觉,我不应该为这样一个简单的函子定义任意实例。
如果我确实需要定义Identity的实例Arbitrary,那么我不知道arbitrary
和shrink
应该是什么样子以及它们应该如何表现。
你能指导我吗?