基本的haskell：复制元素 [英] basic haskell : Copying elements

问题描述

我试图递归地调用一个函数来复制任何类型的指定次数的
，例如复制3'a'会给我['a'，'a'，'a']，同时复制3 2会给我[2,2,2]
这是我到目前为止，但我不知道我的字体是否正确，因为我认为我的代码应该运行良好。任何人都可以看到有什么问题吗？

  copy :: Int-> a-> [a] 
 copy x [] = [] 
 copy ya = a：（copy（y-1）a）
  

编辑：更新为：

  copy :: Int-> a-> [a] 
 copy 0 a = [] 
 copy ya = [a] ++（copy（y-1）a）
  

然而，这给了我aaa而不是[a，a，a]

解决方案

关于如何以更多的凯克利方式实现你的功能的建议。您正试图实现标准的复制函数，并且查看标准源代码总是有所帮助。这里是复制的代码（简化后）

$ pre $复制nx =取n（重复x）
重复x = xs其中xs = x：xs

源代码取自haskell上由lambdabot返回的简化src irc。

I'm trying to recursively call a function that copies any type a specified amount of times for example copy 3 'a' would give me ['a','a','a'] while copy 3 2 would give me [2,2,2] This is what I have so far but I'm not sure if my typeline is correct as I think my code should run fine. Can anyone see what's wrong?

copy :: Int->a->[a]
copy x [] = []
copy y a = a:(copy (y-1) a)

edit: updated to this:

copy :: Int->a->[a]
copy 0 a = [] 
copy y a = [a]++(copy (y-1) a)

However this gives me "aaa" instead of [a,a,a]

解决方案

This is just a suggestion about how to implement your function in a more haskelly way. You are trying to implement standard replicate function and looking at the standard source code always help. Here is the code for replicate (after simplification)

replicate n x = take n (repeat x)
repeat x = xs where xs = x : xs

The source code is taken from the simplified src returned by lambdabot on haskell irc.

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