f# - 如何在 F# 中设置默认参数值？

Question

以这个函数为例:

// Sequence of random numbers
open System

let randomSequence m n = 
    seq { 
        let rng = Random ()
        while true do
            yield rng.Next (m, n)
    }


randomSequence 8 39

randomSequence函数有两个参数: m, n .这作为正常功能工作正常。我想为 m, n 设置默认值， 例如:

(m = 1, n = 100)

当没有给出参数时，函数采用默认值。在 F# 中可能吗？

Answer 1

您可以经常achieve the same effect as overloading with a Discriminated Union .

这是基于OP的建议:

type Range = Default | Between of int * int

let randomSequence range = 
    let m, n =
        match range with
        | Default -> 1, 100
        | Between (min, max) -> min, max

    seq {
        let rng = new Random()
        while true do
            yield rng.Next(m, n) }

注意 Range的介绍歧视联盟。

以下是一些 (FSI) 使用示例:

> randomSequence (Between(8, 39)) |> Seq.take 10 |> Seq.toList;;
val it : int list = [11; 20; 36; 30; 35; 16; 38; 17; 9; 29]

> randomSequence Default |> Seq.take 10 |> Seq.toList;;
val it : int list = [98; 31; 29; 73; 3; 75; 17; 99; 36; 25]

另一种选择是稍微更改 randomSequence 以采用元组而不是两个值:

let randomSequence (m, n) = 
    seq {
        let rng = new Random()
        while true do
            yield rng.Next(m, n) }

这将允许您还定义一个默认值，如下所示:

let DefaultRange = 1, 100

以下是一些 (FSI) 使用示例:

> randomSequence (8, 39) |> Seq.take 10 |> Seq.toList;;
val it : int list = [30; 37; 12; 32; 12; 33; 9; 23; 31; 32]

> randomSequence DefaultRange |> Seq.take 10 |> Seq.toList;;
val it : int list = [72; 2; 55; 88; 21; 96; 57; 46; 56; 7]