如何为封装函数的类型编写任意实例？

Question

我有这种类型的包装功能

newtype Transaction d e r = Transaction (d -> Either e (d,r)) 

...我想为它的函子&应用型实例做快速检查测试，但编译器抱怨说，它不具有任意实例。

我试图这样做，但我坚持生成随机函数。

谢谢！

的快速检查属性这样

type IdProperty f a = f a -> Bool 
functorIdProp :: (Functor f, Eq (f a)) => IdProperty f a 
functorIdProp x = (fmap id x) == id x 

type CompositionProperty f a b c = f a -> Fun a b -> Fun b c -> Bool 
functorCompProp :: (Functor f, Eq (f c)) => CompositionProperty f a b c 
functorCompProp x (apply -> f) (apply -> g) = (fmap (g . f) x) == (fmap g . fmap f $ x) 

instance (Arbitrary ((->) d (Either e (d, a)))) => Arbitrary (DbTr d e a) where 
    arbitrary = do 
     f <- ...??? 
     return $ Transaction f 

...和测试定义== == UPDATE看起来是这样的：

spec = do 
    describe "Functor properties for (Transaction Int String)" $ do 
    it "IdProperty (Transaction Int String) Int" $ do 
     property (functorIdProp :: IdProperty (Transaction Int String) Int) 

    it "CompositionProperty (Transaction Int String) Int String Float" $ do 
     property (functorCompProp :: CompositionProperty (Transaction Int String) Int String Float) 

Answer 1

您应该使用Test.QuickCheck.Function包装器测试功能。如果你只需要为你的Transaction类型（但如果你真的需要它，你可以找到my answer to this question）测试类型分类法则，那么Arbitrary或CoArbitrary例子对于Transaction似乎没有意义。

要测试法，你可以在这样的方式写属性：

{-# LANGUAGE DeriveFunctor #-} 

import Test.QuickCheck 
import Test.QuickCheck.Function 

newtype Transaction d e r = Transaction (d -> Either e (d,r)) 
    deriving (Functor) 

-- fmap id ≡ id 
prop_transactionFunctorId :: Int 
          -> Fun Int (Either Bool (Int,String)) 
          -> Bool 
prop_transactionFunctorId d (Fun _ f) = let t    = Transaction f 
              Transaction f' = fmap id t 
             in f' d == f d 

那么，这可能看起来不漂亮，漂亮，只要你想。但是这是测试任意函数的好方法。例如，我们可以最后一行in f' d == f d更换到in f' d == f 1，看看它是如何失败的，如果它失败：

ghci> quickCheck prop_transactionFunctorId 
*** Failed! Falsifiable (after 2 tests and 4 shrinks):  
0 
{0->Left False, _->Right (0,"")}