haskell - 为什么 `and` 为空可折叠返回 True 而 `or` 在 Haskell 中返回 False？

Question

这个问题在这里已经有了答案:





Why "and []" is True and "or []" is False

(7 个回答)


2年前关闭。




是否有理由可以解释是否可以预期这样的结果？它们比未定义更好吗？

>>> any (const True) []
False

>>> any (const False) []
False

>>> or []
False

>>> and []
True

我真的不明白报告想说什么:

-- and returns the conjunction of a Boolean list. For the result to be
-- True, the list must be finite; False, however, results from a False
-- value at a finite index of a finite or infinite list. or is the
-- disjunctive dual of and. 
and, or     :: [Bool]    -> Bool
and         =  foldr    (&&)    True
or          =  foldr    (||)    False

Answer 1

为了扩展 Dan 的答案，真值的空列表的合取为真 True允许您将连接的预期属性扩展到这种情况。

例如，我们希望，

and (xs ++ ys) = (and xs) && (and ys) 

和所有 xs我们有， xs = xs ++ []所以，

  and xs 
= and (xs ++ [])
= (and xs) && (and []) 

考虑到 and xs可能是 True或 False它遵循，

True  = True  && (and [])
False = False && (and [])

因此我们必须有 and []是 True .