在我的Python package我有一个函数,我用它来创建我的类的验证实例,类似
@staticmethod
def config_enigma(rotor_names, window_letters, plugs, rings):
comps = (rotor_names + '-' + plugs).split('-')[::-1]
winds = [num_A0(c) for c in 'A' + window_letters + 'A'][::-1]
rngs = [int(x) for x in ('01.' + rings + '.01').split('.')][::-1]
assert all(name in rotors for name in comps[1:-1])
assert comps[-1] in reflectors
assert len(rngs) == len(winds) == len(comps)
assert all(1 <= rng <= 26 for rng in rngs)
assert all(chr_A0(wind) in LETTERS for wind in winds)
#...
我想在Haskell中强制执行相同的行为。但是以相同的方式这样做 - 使用断言 - 不起作用,因为Haskell断言一般被禁用(unless certain compiler flags are set)。例如,在something like
中configEnigma rots winds plug rngs =
assert ((and $ (==(length components')) <$> [length winds', length rngs']) &&
(and $ [(>=1),(<=26)] <*> rngs') &&
(and $ (`elem` letters) <$> winds') &&
(and $ (`M.member` comps) <$> tail components'))
-- ...
where
rngs' = reverse $ (read <$> (splitOn "." $ "01." ++ rngs ++ ".01") :: [Int])
winds' = "A" ++ reverse winds ++ "A"
components' = reverse $ splitOn "-" $ rots ++ "-" ++ plug
不能依赖于工作,因为断言将在大多数情况下被删除。
什么是强制所有实例在Haskell中验证的惯用且可靠的方法(使用“公共安全”构造函数)?
答案 0 :(得分:5)
正常的事情是明确地表达失败。例如,可以写一个
configEnigma :: ... -> Maybe ...
configEnigma ... = do
guard (all (((==) `on` length) components') [winds', rngs'])
guard (all (inRange (1,26)) rngs')
guard (all (`elem` letters) winds')
guard (all (`M.member` comps) (tail components'))
return ...
where
...
对于某些自定义类型Maybe
,您甚至可以考虑从Except Error
升级到Error
,以便向调用者传达构建期间出错的内容。然后,您可以使用类似以下的结构而不是guard
。
unless (all (inRange (1,26)) rngs') (throwError OutOfRange)
configEnigma
的来电者必须表达如何处理失败。对于Maybe
,这看起来像
case configEnigma ... of
Just v -> -- use the configured enigma machine v
Nothing -> -- failure case; perhaps print a message to the user and quit
与Except
同时,您可以获得有关错误的信息:
case runExcept (configEnigma ...) of
Right v -> -- use the configured enigma machine v
Left err -> -- failure case; tell the user exactly what went wrong based on err and then quit