快速检查简单函数:是否需要定义任意实例?为什么?怎样

我在用Functors和QuickCheck做运动。

我有一个超级简单的Functor,我想快速检查一下它的组成规律。 Functor只是一个 Identity a . 这是我迄今为止的代码:

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) 

不幸的是,这段代码无法编译,ghc抱怨编译错误:

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);
           .... }

所以我已经开始研究QuickCheck任意类型类,它需要定义 arbitrary :: Gen a 和 shrink :: a -> [a] .

我有一种(可能是错误的)感觉,我不需要为这样一个简单的函子定义任意实例。

如果我真的需要为身份定义任意实例,那么我不知道 arbitrary 和 shrink 应该是什么样子以及他们应该如何表现。

你能指导我吗?