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F#检查列表是否为空

functional-programming f#

1

Tartar · 技术社区 · 7 年前

作为一个F#新手,我正在尝试实现一个简单的函数,它将索引和列表作为参数,然后返回给定索引的列表值。

let rec getElementAtIndex (index : int) (list : 'a list) = 
  match index, list with
    | _ ,[] -> failwith "index is greater than number of elements in list.."
    | _, _ when index < 0 -> failwith "index is less than 0." 
    | 0, (first::_) -> first
    | _, (_::rest') -> getElementAtIndex (index - 1) rest'

| _ ,[] -> failwith "index is greater than number of elements in list."

任何帮助都将不胜感激

1 回复 | 直到 7 年前

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4

CaringDev 7 年前

检查的模式 全球的 实际的 工作。通过这种方式,递归函数变得更简单,并且不需要 when 警卫或 length :

let getElementAtIndex index list =
  if index < 0 then failwith "index is less than 0"
  if List.isEmpty list then failwith "list is empty"
  let rec get = function
    | _ , [] -> failwith "index is greater than number of elements in list"
    | 0, first::_ -> first
    | _, _::rest -> get(index - 1, rest)
  get(index, list)

这个 function

  let rec get i l =
    match i, l with
    | _ , [] -> failwith "index is greater than number of elements in list"
    | 0, first::_ -> first
    | _, _::rest -> get (index - 1) rest

更新

if List.isEmpty list then failwith "list is empty" 你可以用 match list with [] -> failwith "..." | _ -> () 或 if list = [] then failwith "..." 后者只适用于支持平等的元素列表。