我是一名Haskell noob,在阅读Haskell Road中的含义时,我开始关注这个难题。
verdict :: Bool -> Bool -> (Bool, String)
verdict p q | not result = (False, "LIAR")
| result && p = (True, "telling the truth")
| result = (True, "innocent unless proven guilty")
| otherwise = (False, "we shouldn't get here")
where result = (not p) || q
-- map (\x -> verdict (fst x == 1) (snd x == 1)) [(1,1),(1,0),(0,1),(0,0)]
是否有工具可以警告我其他类似的逻辑错误?
答案 0 :(得分:8)
我想我会以不同的方式编写这个函数:
-- not result
verdict True False = (False, "LIAR")
-- result && p
verdict True True = (True , "telling the truth")
-- result
verdict False _ = (True , "innocent unless proven guilty")
verdict _ True = (True , "innocent unless proven guilty")
-- otherwise
verdict _ _ = (False, "we shouldn't get here")
那么不仅人类可以忽略哪些条款(最后两个),而且机器也是如此; ghc
在默认警告级别上说明了这一点:
test.hs:2:5: Warning:
Pattern match(es) are overlapped
In an equation for ‘verdict’:
verdict _ True = ...
verdict _ _ = ...
检查警卫重叠一般当然是不可判定的;此外,我不知道会尝试给出一个近似答案的工具。
答案 1 :(得分:4)
这可能更清楚地表达了你的意图:
implies :: Bool -> Bool -> Bool
p `implies` q = not p || q -- The backticks allow infix usage.
-- The assumption is that p `implies` q is a known fact.
verdict :: Bool -> Bool -> (Bool, String)
verdict p q = (result, remark)
where
result = p `implies` q
remark
| not result = "LIAR"
| p = "telling the truth"
| otherwise = "innocent until proven guilty"
Guards是Bool
值上模式匹配的语法糖。有关安排模式匹配的一般提示,请参阅Daniel Wagner的回答。