我如何验证Haskell的参数“public safe”构造函数？

本文介绍了我如何验证Haskell的参数“public safe”构造函数？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 在我的 Python包中，我有一个函数用于创建经过验证的实例类，类似于 @staticmethod def config_enigma（rotor_names，window_letters，plugs，rings）：'b $ b comps =（rotor_names +' - '+ plugs）.split（' - '）[:: - 1] winds = [num_A0（c）for'A'+ window_letters +' A'] [:: - 1] rngs = [int（x）for x in（'01。'+ rings +'.01'）。split（'。'）] [:: - 1] 声明所有（名字在旋转中的名称在comps [1：-1]）$ ​​b $ b在反射器中声明comps [-1] 声明len（rngs）== len风）== len（comps）断言所有（rng中1 断言所有风（风中的chr_A0（风））风 ＃... 我想在Haskell中执行相同的行为。但是以相同的方式这样做（带断言）不起作用，因为一般情况下Haskell断言是被禁用的（除非设置了某些编译器标志）。例如，在像 $ b $ pref =lang-hs prettyprint-override> configEnigma rots winds plug rngs = ==（长度分量'））< $> [长度风'，长度rng']）& （和$ [（> = 1），（（和$（`M.member` comps ）< $>尾部组件'）） - ... 其中 rngs'=反向$（读< $>（splitOn。$01。 ++ rngs ++.01）:: [Int]） winds'=A++反向风++A组件'=反向$ splitOn - $不能依赖于工作，因为断言将会在大多数情况下被删除。 什么是惯用的a nd可靠的方式来强制所有的实例在Haskell中验证（使用公共安全构造函数）？解决方案正常事情是明确表达失败。例如，可以写成： configEnigma :: ... - >也许... configEnigma ... = do guard（全部（（（==）`on长度）组件'）[winds'，rngs']） guard（all （inRange（1,26））rngs'） guard（all（`elem` letters）winds'） guard（all（`M.member` comps）（tail components'）） return ... 其中 ... 您甚至可能考虑对某些自定义类型错误从 Maybe 升级到 Except Error $ c>，以告诉来电者在施工过程中出了什么问题。然后，您可以使用像 $ b guard 除非（all（inRange（1,26））rngs'）（throwError OutOfRange） configEnigma 的调用者必须表达如何处理失败。对于可能，这看起来像 $ b 案例configEnigma ... of Just v - > - 使用配置的enigma机器v Nothing - > - 失败案例;可能会打印一条消息给用户，并退出 ，同时使用除你得到了有关错误的信息： case runExcept（configEnigma ... ） Right v - > - 使用配置的enigma机器v Left err - > - 失败案例;根据错误告知用户究竟出了什么问题，然后退出 In my Python package I have a function I use to create validated instances of my class, with something like @staticmethoddef 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) #...and I would like to enforce the same behavior in Haskell. But doing so in the same way — with assertions — does not work, because Haskell assertions are disabled in general (unless certain compiler flags are set). For example, in something likeconfigEnigma 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 ++ "-" ++ plugcan't be relied on to work because the assertions will be removed in most contexts.What is an idiomatic and reliable way to force all my instances to be validated in Haskell (using a "public safe" constructor)? 解决方案 The normal thing is to express failure explicitly. For example, one might writeconfigEnigma :: ... -> 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 ...You might even consider upgrading from Maybe to Except Error for some custom-defined type Error, to communicate to the caller what it was that went wrong during construction. Then instead of guard, you could use a construction like: unless (all (inRange (1,26)) rngs') (throwError OutOfRange)The caller to configEnigma will have to express how to handle failures. For Maybe, this looks likecase configEnigma ... of Just v -> -- use the configured enigma machine v Nothing -> -- failure case; perhaps print a message to the user and quitwhile with Except you get information about what went wrong: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 这篇关于我如何验证Haskell的参数“public safe”构造函数？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！