haskell - 如何在没有缩进树的情况下处理嵌套的 case 语句？

Question

我偶尔会遇到一个小问题，即我有一组嵌套的 case 语句，处理起来很麻烦。我是否可以使用任何技术/模式来获得一个函数列表(这相当于示例的 case ... 语句)，评估所有函数，并选择第一个与模式(例如:右 x)？

我将它们放入列表时遇到的一个具体问题是它们不一定是同一类型(我认为是单态性限制)。例如以下内容:

  let possibleTags = [
          parse (parseFileReference) "file reference" xStr
        , parse (parseGitDiffReference) "git diff tag" xStr
        ]

产生以下错误:

• Couldn't match type ‘GitDiffReference’ with ‘FileReference’
  Expected type: Either ParseError FileReference
    Actual type: Either ParseError GitDiffReference
• In the expression:
    parse (parseGitDiffReference) "git diff tag" xStr
  In the expression:
    [parse (parseFileReference) "file reference" xStr,
     parse (parseGitDiffReference) "git diff tag" xStr]
  In an equation for ‘possibleTags’:
      possibleTags
        = [parse (parseFileReference) "file reference" xStr,
           parse (parseGitDiffReference) "git diff tag" xStr]

<小时/>

abc :: Int -> String
abc = undefined

abc2 :: Int -> Float
abc2 = undefined

abc3 :: Int -> Int
abc3 = undefined

example :: Int -> Maybe String
example x = case abc x of 
  ("yes") -> Just "abc"
  ("no")  -> case abc2 x of
               1.0 -> Just "abc2"
               2.0 -> case abc3 x of
                        100 -> Just "abc3"
                        200 -> Nothing

我理想地寻找类似下面的东西(无效代码):

-- Psuedo code
example :: Int -> Maybe String
example = 
 if (abc x && "yes") then Just "abc"
 if (abc2 x && 1.0) then Just "abc2"
 if (abc3 x && 100) then Just "abc3"

使用 if 条件的问题是我无法(据我所知)进行模式匹配，例如 if (Just x)。

Answer 1

翻译确切的示例(保持非详尽的模式匹配不变):

import Data.Foldable


example2 :: Int -> Maybe String
example2 x = asum tests
   where
   tests =
      [ case abc x of 
         "yes" -> Just "abc"
         "no"  -> Nothing
      , case abc2 x of
         1.0 -> Just "abc2"
         2.0 -> Nothing
      , case abc3 x of
         100 -> Just "abc3"
         200 -> Nothing
      ]