函数返回带有修改字段的Haskell记录 [英] Function to return a Haskell record with a modified field

问题描述

  data MyRecord a = MyRecord {list :: [a]，other_fields :: Char，...} 
  

我正在编写一个函数，它将一个新的 a 列表并返回一个新的 MyRecord ：

  pushOntoList :: a  - > MyRecord  - > MyRecord 
  

问题：

有没有办法写 pushOntoList 是这样一种方式，它不依赖于记录其余部分的内容，而只是简单地将它返回而未修改？

另一个问这个问题的方法是，你可以在没有看到的其余部分的情况下编写 pushOntoList MyRecord 定义？

解决方案

是的，非常容易使用记录访问器/标签语法：

  b = a {list ='x'：list a} 
  

  pushOntoList ca = a {list = c：list a} 
 
 code> 

eg

  data MyRecord a = MyRecord {list :: [a]，other_fields :: Char} 
派生显示
 
 main = do 
 let a = MyRecord []'x'
b = a {list ='x'：list a} 
 return（a，b）
  




Given:

data MyRecord a = MyRecord{list :: [a], other_fields :: Char, …}

I am trying to write a function which puts a new a on list and returns a new MyRecord:

pushOntoList :: a -> MyRecord -> MyRecord

Question:

Is there a way to write pushOntoList is such a way that it does not depend on what is in the rest of the record, but simply gives it back unmodified?

Another way to ask this is can you write pushOntoList without seeing the rest of the MyRecord definition?

解决方案

Yes, very easily using the record accessor/label syntax:

b = a { list = 'x' : list a }

as in the function:

pushOntoList c a = a { list = c : list a }

e.g.

data MyRecord a = MyRecord {list :: [a], other_fields :: Char}
    deriving Show

main = do
    let a = MyRecord [] 'x'
        b = a { list = 'x' : list a }
    return (a,b)

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