我认为除了模块Test.QuickCheck.Arbitrary
和Test.QuickCheck.Gen
.
只有一个参数
下面是一些简单的代码,它只提供了一个参数的函数所需的内容:
import Test.QuickCheck.Arbitrary
import Test.QuickCheck.Gen
import System.Random
randomEvaluate :: (Arbitrary a, Show a, Show b) => (a -> b) -> IO (String, String)
randomEvaluate f = do
stdGen <- newStdGen
let x = unGen arbitrary stdGen 1000
let y = f x
return (show x, show y)
在这里你可以看到它的实际效果:
*Main> randomEvaluate (\(a,b) -> a + b)
("(-292,-655)","-947")
*Main> randomEvaluate (\(a,b) -> a + b)
("(586,-905)","-319")
*Main> randomEvaluate (\(a,b) -> a + b)
("(547,-72)","475")
如您所见,如果您对它进行 uncurry,则可以将它与具有多个参数的函数一起使用。如果这还不够的话,事情会变得有点困难,但应该可以通过一些类型类技巧来实现。
多个参数,显式标记的返回类型
这是一种需要“仅”将函数的返回值包装在新类型中的方法。(使用非 Haskell98 功能可能可以避免这种情况):
class RandEval a where
randomEvaluate :: StdGen -> a -> ([String], String)
newtype Ret a = Ret a
instance Show a => RandEval (Ret a) where
randomEvaluate _ (Ret x) = ([], show x)
instance (Show a, Arbitrary a, RandEval b) => RandEval (a -> b) where
randomEvaluate stdGen f = (show x : args, ret)
where (stdGen1, stdGen2) = split stdGen
x = unGen arbitrary stdGen1 1000
(args, ret) = randomEvaluate stdGen2 (f x)
doRandomEvaluate :: RandEval a => a -> IO ([String], String)
doRandomEvaluate f = do
stdGen <- newStdGen
return $ randomEvaluate stdGen f
在这里查看它的实际效果:
*Main> doRandomEvaluate (\a b -> Ret (a && b))
(["False","True"],"False")
*Main> doRandomEvaluate (\a b -> Ret (a + b))
(["944","758"],"1702")
*Main> doRandomEvaluate (\a b c -> Ret (a + b + c))
(["-274","413","865"],"1004")
*Main> doRandomEvaluate (\a b c d -> Ret (a + b + c + d))
(["-61","-503","-704","-877"],"-2145")
带有语言扩展的多个参数
如果还不需要显式标记返回值,这可行,但使用语言扩展:
{-# LANGUAGE FlexibleInstances, UndecidableInstances, OverlappingInstances #-}
import Test.QuickCheck.Arbitrary
import Test.QuickCheck.Gen
import System.Random
import Control.Arrow
class RandEval a where
randomEvaluate :: StdGen -> a -> ([String], String)
instance (Show a, Arbitrary a, RandEval b) => RandEval (a -> b) where
randomEvaluate stdGen f = first (show x:) $ randomEvaluate stdGen2 (f x)
where (stdGen1, stdGen2) = split stdGen
x = unGen arbitrary stdGen1 1000
instance Show a => RandEval a where
randomEvaluate _ x = ([], show x)
doRandomEvaluate :: RandEval a => a -> IO ([String], String)
doRandomEvaluate f = do
stdGen <- newStdGen
return $ randomEvaluate stdGen f
这是帖子中的原始用例:
*Main> doRandomEvaluate ( (+) :: Int -> Int -> Int )
(["-5998437593420471249","339001240294599646"],"-5659436353125871603")
但现在您对 GHC 如何解决重叠实例一时心血来潮。例如,即使使用这个不错的(但也不是 Haskell98)实例来显示布尔函数:
type BoolFun a = Bool -> a
instance Show a => Show (BoolFun a) where
show f = "True -> " ++ show (f True) ++ ", False -> " ++ show (f False)
aBoolFun :: Bool -> BoolFun Bool
aBoolFun x y = x && y
您看不到此实例在使用中doRandomEvaluate
:
*Main> doRandomEvaluate aBoolFun
(["False","False"],"False")
使用原始解决方案,您可以:
*Main> doRandomEvaluate (Ret . aBoolFun)
(["False"],"True -> False, False -> False")
*Main> doRandomEvaluate (Ret . aBoolFun)
(["True"],"True -> True, False -> False")
一个警告
但请注意,这是一个滑坡。对上面的代码做一个小改动,它在 GHC 7.6.1 中停止工作(但在 GHC 7.4.1 中仍然有效):
instance (Show a, Arbitrary a, RandEval b) => RandEval (a -> b) where
randomEvaluate stdGen f = (show x:args, ret)
where (stdGen1, stdGen2) = split stdGen
x = unGen arbitrary stdGen1 1000
(args, ret) = randomEvaluate stdGen2 (f x)
SPJ 解释了为什么这不是一个真正的错误——对我来说这是一个明显的迹象,表明这种方法将类型类黑客技术推得太远了。