哈斯克爾快速檢查最小的反例

Question

考慮下面的測試爲distributivity law between reverse and ++，

import Test.QuickCheck 

test :: [Int] -> [Int] -> Bool 
test xs ys = reverse (xs ++ ys) == reverse xs ++ reverse ys 

test2 :: (Eq a) => [a] -> [a] -> Bool 
test2 xs ys = reverse (xs ++ ys) == reverse xs ++ reverse ys 

注意有關Int列出了

*Main> quickCheck test 
*** Failed! Falsifiable (after 5 tests and 3 shrinks):  
[1] 
[0] 

然而，測試equatable項目的列表，

*Main> quickCheck test2 
+++ OK, passed 100 tests. 

什麼讓第二次測試通過？

更新在與main = quickCheck test2編譯曖昧類型變量的後續錯誤的提示問題（如答案已經描繪），

No instance for (Eq a0) arising from a use of `test2' 
The type variable `a0' is ambiguous 
Possible fix: add a type signature that fixes these type variable(s) 

Answer 1

當你真正評估test2，GHCI必須選擇一個類型a使用。如果沒有更多的信息，GHCi的擴展默認規則默認爲()，對此法律適用。

Answer 2

> verboseCheck test2 

Passed: 
[] 
[] 
Passed: 
[] 
[] 
Passed: 
[(),()] 
[()] 
Passed: 
[(),(),()] 
[()] 
Passed: 
[()] 
[(),(),(),()] 
... 

多態參數默認爲()，當然所有這些值都是相等的。