模式与守卫:否则不匹配？

Question

给定空字符串时,以下两个函数的行为有所不同:

guardMatch l@(x:xs) 
    | x == '-'        = "negative " ++ xs
    | otherwise       = l

patternMatch ('-':xs) = "negative " ++ xs
patternMatch l        = l

这是我的输出:

*Main> guardMatch ""
"*** Exception: matching.hs:(1,1)-(3,20): Non-exhaustive patterns in function guardMatch

*Main> patternMatch ""
""

问题:为什么'否则'关闭空接字符串？

Answer 1

的otherwise是该图案的范围内l@(x:xs),这只能匹配的非空字符串.可能有助于了解这(有效)内部转化为什么:

guardMatch   l = case l of
                   (x  :xs) -> if x == '-' then "negative " ++ xs else l
patternMatch l = case l of
                   ('-':xs) ->                  "negative " ++ xs
                   _        ->                                         l

(实际上,我认为它if被翻译成case+守卫而不是相反.)

Answer 2

始终在模式后评估防护.这是 - 如果模式成功,则尝试防护.在您的情况下,模式(x:xs)排除空字符串,因此甚至没有尝试防护,因为模式失败.