Hiding a System.Random instance by returning a function

Question

The following code comes from Stylish F# 6: Crafting Elegant Functional Code for .NET 6 listing 9-13:

let randomByte =
    let r = System.Random()
    fun () ->
        r.Next(0, 255) |> byte
// E.g. A3-52-31-D2-90-E6-6F-45-1C-3F-F2-9B-7F-58-34-44-
for _ in 0..15 do
    printf "%X-" (randomByte())
printfn ""

The author states, "Although we call randomByte() multiple times, only one System.Random() instance is created."

I understand randomByte returns a function that does not create a System.Random() instance, but it seems to me multiple System.Random() instances would be created each time through the for-do-loop anyway.

I would appreciate an explanation of how multiple instances of System.Random() are not created in this case.

Answer 1

The key point is that randomByte is not a function. It's a value with some complex initialization logic. Like, for example, I could write:

let x = 5

Or I could write:

let x =
  let fourtyTwo = 42
  let thirtySeven = 37
  fourtyTwo - thirtySeven

And these would be equivalent. Both declare a value named x and equal to 5. I hope you can see how the expression fourtyTwo - thirtySeven is evaluated only once, not every time somebody gets the value of x.

And so it works with randomByte too: it's a value with non-trivial initialization logic. During that value's initialization, first it creates an instance of System.Random, and then it creates an anonymous function that closes over that instance, and this anonymous function becomes the value of randomByte.