Haskell在列表中显示实例 [英] Haskell show instance on list

问题描述

  data结构=结构字符串[结构] 
  

我想要这样的输出：

  strct 
 strct1 
 strct2 
 strct3 
  

我一直在试试这个

  instance显示结构其中
 show（Structure a （xs））= show a ++\\\
++++ show xs 
  

但是它的输出是

$ $ $ $ $ $ $ $ $ $ $ $ $ $ $'
[]]

所以，我不需要括号，也不需要逗号和引号。有任何想法吗？谢谢

解决方案

我确定有更好的库例程，但不会工作吗？ b
$ b

  unlines $ a：[++ show x | x < -  xs] 
  

但是，这仅涵盖一个级别。您可能想要定义与show不同的功能来维护缩进，或者您必须使用 lines 来分割子节目以查找注入缩进的位置。

这种缩进插入函数的草稿是：

pre> 前缀ps = unlines [p ++ l | l< - 行s]


再次，我确信库中有更好的东西。即使这些简短的片段也经历了几个细化步骤（ foldl1（++） - > concat - > unlines ，然后以：的形式加入第一行。

Im having issues with adding a show instance to my data structure, wich is:

data Structure = Structure String [Structure]  

and I would like to have this output:

strct  
    strct1  
    strct2  
    strct3  

I've been trying this

instance Show Structure where  
    show (Structure a (xs)) = show a ++ "\n" ++ "  " ++ show xs  

But its output is

"strct"  
    ["strct1"  
    [], "strct2"  
    []]  

So, I would need no brackets, no commas and no quotation marks. Any ideas? Thanks

解决方案

I'm sure there are better library routines for this, but wouldn't this work?

unlines $ a : ["  " ++ show x | x <- xs]

However, that covers only one level. You probably want to define a different function than show to maintain the indentation, or you'd have to keep splitting sub-shows with lines to find where to inject indentation.

A rough draft of such an indentation insertion function is:

prefix p s = unlines [p ++ l | l <- lines s]

Again, I'm sure there's something better in a library. Even these short snippets have gone through a few steps of refinement (foldl1 (++) -> concat -> unlines, then join the first line as head with :).

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