递归列表逻辑错误 [英] Error in recursive list logic

问题描述

我正在尝试在给定输入(长度和函数)的 Scala 中构建一个列表，输出将是一个从 0 到长度为 1 的列表.

I am trying to build a list in scala that given input (length,and a function) the output would be a list from 0 up to that length-1.

例如:

listMaker(3,f) = List(0,1,2)

到目前为止，我已经创建了一个辅助类，它接受 2 个 int 并返回该范围内的 list.

so far I have created a helper class that takes 2 int and returns a list in that range.

listMaker 函数如下:

def listMaker[A](length:Int, f:Int =>A):List[A] = length match{
  case 0 => List()
  case _ => listMaker(length,f)
}

我的 f 函数只需要一个变量 x 并返回:

my f function just takes a variable x and returns that:

 def f(x:Int)=x 

下面的评论是有道理的，但它仍然让我出错.我认为编辑后的代码更容易到达我想要的地方

但是，现在我得到了一个无限循环.我遗漏了哪一部分逻辑?

However, now I get an infinite loop. What part of the logic am I missing?

推荐答案

递归函数通常必须逐渐咬掉"输入数据的片段，直到没有任何剩余 - 否则它永远不会终止.

A recursive function typically has to gradually "bite off" pieces of the input data until there is nothing left - otherwise it can never terminate.

在您的特定情况下，这意味着 length 必须在每次递归调用时减少，直到达到零.

What this means in your particular case is that length must decrease on each recursive call until it reaches zero.

def listMaker[A](length:Int, f:Int =>A):List[A] = length match{
  case 0 => List()
  case _ => listMaker(length,f)
}

但是您并没有减少 length - 您将它原封不动地传递给下一个递归调用，因此，您的函数无法终止.

But you are not reducing length - you are passing it unchanged to the next recursive call, so, your function cannot terminate.

(还有其他问题——你需要在递归时建立你的结果列表，但你当前的代码只是返回一个空列表.我认为这是一个学习练习，所以我没有提供工作代码...).

(There are other problems too - you need to build up your result list as you recurse, but your current code simply returns an empty list. I assume this is a learning exercise, so I'm not supplying working code...).

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