Question

如何将Haskell对象作为有效Haskell代码的字符串输出到输出中，省略现有的show声明？

例如，我有以下类型声明和相应的show声明（check the code out in an online IDE）：

{-# LANGUAGE TypeSynonymInstances, FlexibleInstances #-}

import Data.Maybe

type VersionCompound = Maybe Int

data VersionNumber = VersionNumber [VersionCompound] deriving (Show)

data MaturityLevel = Dev
                   | Test
                   | User
                   | ReleaseCandidate
                   | Prod
                   deriving (Show)

data Version = MaturityVersion MaturityLevel VersionNumber  
     | Version VersionNumber

class ToString a where
    toString :: a -> String

instance ToString VersionCompound where
    toString (Just n) = (show n)
    toString Nothing = "x"

instance ToString [VersionCompound] where
    toString [] = "" 
    toString (x:[]) = (toString x)
    toString (x:xs) = (toString x) ++ "." ++ (toString xs) 

instance ToString VersionNumber where
    toString (VersionNumber []) = "" 
    toString (VersionNumber (x:[])) = (toString x) 
    toString (VersionNumber (x:xs)) = (toString x) ++ "." ++ (toString xs)

instance ToString Version where
    toString (MaturityVersion maturityLevel versionNumber) = (show maturityLevel) ++ "/" ++ (toString versionNumber)
    toString (Version versionNumber) = (toString versionNumber)

instance Show Version where
    show version = toString version

main = putStrLn $ show (Version $ VersionNumber [ Just 1, Just 2, Nothing])

该计划的输出是：

1.2.x

但是有没有办法以有效的Haskell代码形式输出对象？例如，对于上面的代码，showIntact $ Version $ VersionNumber [ Just 1, Just 2, Nothing]会输出如下内容：

Version ( VersionNumber [ Just 1, Just 2, Nothing] )

Answer 1

您可以定义自己的功能：

showIntact :: Version -> String
showIntact (Version v) = "Version " ++ show v
showIntact (MaturityVersion ml vn) = "MaturityVersion " ++ show ml ++ " " ++ show vn

然后，您可以使用putStrLn $ showIntact $ ....。

输出haskell对象省略＆＃39;显示＆＃39;声明

1 个答案: