How to set default argument value in F#? - f#

Question

Take this function as an example:
// 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
The randomSequence function takes two arguments: m, n. This works fine as a normal function. I would like to set the default for m, n, for example:
(m = 1, n = 100)
When there's no arguments given, the function take the default value. Is it possible in F#?

Answer 1

You can often achieve the same effect as overloading with a Discriminated Union.
Here's a suggestion based on the OP:
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) }
Notice the introduction of the Range Discriminated Union.
Here are some (FSI) examples of usage:
> 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]
Another option is to change the randomSequence ever so slightly to take a tuple instead of two values:
let randomSequence (m, n) =
seq {
let rng = new Random()
while true do
yield rng.Next(m, n) }
This would allow you to also define a default value, like this:
let DefaultRange = 1, 100
Here are some (FSI) examples of usage:
> 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]

Answer 2

Optional parameters are permitted only on members [...]
https://msdn.microsoft.com/en-us/library/dd233213.aspx
So, for your current example, I think it is impossible.
But you could wrap the functionality:
type Random =
static member sequence(?m, ?n) =
let n = defaultArg n 100
let rng = new System.Random()
match m with
| None -> Seq.initInfinite (fun _ -> rng.Next(1, n))
| Some(m) -> Seq.initInfinite (fun _ -> rng.Next(m, n))
and then use it like this:
let randomSequence = Random.sequence(2, 3)
let randomSequence' = Random.sequence(n = 200)
let randomSequence'' = Random.sequence()
Explanation: optional parameters can either be fully Optional (m) or defaultArgs (n). I like shadowing (reusing the same name) these parameters, but that's debatable.

Answer 3

This seems to be the most elegant solution to this problem:
//define this helper function
let (|Default|) defaultValue input =
defaultArg input defaultValue
//then specify default parameters like this
let compile (Default true optimize) =
optimize
//or as the OP asks
let randomSequence (Default 1 m) (Default 100 n) =
seq {
let rng = new System.Random()
while true do
yield rng.Next(m,n)
}
Credits:
http://fssnip.net/5z

Answer 4

let randomSequence m n=
seq {
let rng = new Random()
while true do
yield rng.Next(m,n)
}
let randomSequenceWithDefaults() =
randomSequence 1 100
So instead, call randomSequenceWithDefaults()