F#有循环退出语句吗？

Question

我知道递归函数是F#中一种强大的技术.我的问题是:是否有退出语句,它可以跳出递归函数,就像命令式语言一样.例如,将节点插入二叉树.

type Tree<'a> when 'a :> IComparable<'a> =
           | Nil
           | Leaf of 'a
           | Node of Tree<'a> * 'a * Tree<'a>

let tt2 = Node(
              Node(Leaf "D", "B",Node(Leaf "G", "E", Leaf "H" )),
              "A",
              Node(Nil, "C", Node(Nil, "F", Leaf "I")))

let rec contains (x : #IComparable<'a>) = function
    | Nil -> false
    | Leaf y -> if x.CompareTo(y) = 0 then true else false
    | Node(l, y, r) -> 
         match l, y, r with
             | l, y, Nil -> if x.CompareTo(y) = 0 then true else contains x  l
             | Nil,y, r -> if x.CompareTo(y) = 0 then true else contains x r 
             | _ -> if x.CompareTo(y) = 0 then true
                    else contains x r |>ignore
                         contains x l

let xx = contains "C"  tt2  //It is wrong answer.

Answer 1

是否有退出语句,它可以跳出递归函数,就像命令式语言一样.

不是.原因是您可以通过递归函数和模式匹配来编码命令性中断/返回.如果您想要中断,只需返回一个值,否则调用另一个递归调用.

这个问题更适合要求高阶函数.当您需要提前退出高阶函数时,编写自定义递归函数是可行的方法.如果您对F#中的命令式构造感兴趣,请查看@Tomas 的优秀系列.

确定条件后,您的函数将在某个分支处退出.唯一的问题是你不应该contain x r在倒数第二行丢弃.

if/else为清晰起见,您可以删除多余的

let rec contains (x : #IComparable<'a>) = function
    | Nil -> false
    | Leaf y -> x.CompareTo(y) = 0
    | Node(l, y, r) -> 
         match l, y, r with
         | l, y, Nil -> x.CompareTo(y) = 0 || contains x l
         | Nil,y, r -> x.CompareTo(y) = 0 || contains x r 
         | _ -> x.CompareTo(y) = 0 || contains x l || contains x r