Xod*_*rap 14 haskell quickcheck
当一个值未通过QuickCheck测试时,我想用它进行调试.有什么方法可以做我喜欢的事情:
let failValue = quickCheck' myTest
in someStuff failValue
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如果我的数据read
能够,那么我可能会破解某种方式从IO中获取它,但事实并非如此.
我在QuickCheck API中找不到任何方法来以一种很好的方式执行此操作,但这是我使用monadic QuickCheck API一起攻击的东西.它拦截并记录你的属性输入IORef
,并假设如果失败,最后一个是罪魁祸首并将其返回到Just
.如果测试通过,结果是Nothing
.这可能会有所改进,但对于简单的单参数属性,它应该可以完成这项工作.
import Control.Monad
import Data.IORef
import Test.QuickCheck
import Test.QuickCheck.Monadic
prop_failIfZero :: Int -> Bool
prop_failIfZero n = n /= 0
quickCheck' :: (Arbitrary a, Show a) => (a -> Bool) -> IO (Maybe a)
quickCheck' prop = do input <- newIORef Nothing
result <- quickCheckWithResult args (logInput input prop)
case result of
Failure {} -> readIORef input
_ -> return Nothing
where
logInput input prop x = monadicIO $ do run $ writeIORef input (Just x)
assert (prop x)
args = stdArgs { chatty = False }
main = do failed <- quickCheck' prop_failIfZero
case failed of
Just x -> putStrLn $ "The input that failed was: " ++ show x
Nothing -> putStrLn "The test passed"
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