如何验证 Haskell "public safe" 构造函数的参数？

Question

在我的 Python package 中，我有一个函数可以用来创建我的 class 的经过验证的实例，例如

@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。但是以同样的方式（使用断言）这样做是行不通的，因为 Haskell 断言通常被禁用（unless certain compiler flags are set). For example, in something like

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

不能依赖它来工作，因为在大多数情况下断言将被删除。

强制在 Haskell 中验证我的所有实例（使用 "public safe" 构造函数）的惯用且可靠的方法是什么？

Answer 1

正常的事情是明确地表达失败。例如，可以写

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 Error，您甚至可以考虑从 Maybe 升级到 Except Error，以便与调用者沟通在构造过程中出了什么问题。然后，您可以使用如下结构代替 guard：

    unless (all (inRange (1,26)) rngs') (throwError OutOfRange)

configEnigma 的调用者必须说明如何处理失败。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

而使用 Except 时，您会得到有关错误信息：

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

如何验证 Haskell "public safe" 构造函数的参数？

How do I validate arguments to a Haskell "public safe" constructor?

validation

haskell

assertions

python-2.7