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Encapsulation boundaries. How do they work? (Expert FSharp p75)
f#functional

Context on your OP here: I'm a C# developer trying to learn functional programming via F#.

On Page 75 of Expert F#, there's a brief example of an encapsulation boundary:

let generateStamp =
let count = ref 0
(fun () -> count := !count + 1; !count)

The explanation for this function says that the line "let count = ref 0" is executed once.

Here's the question:

How is this statement executed only once if the method "generateStamp" is called repeatedly? Wouldn't "let count = ref 0" be executed more than once as well?

Obviously, the answer is no, because the text says so, but <i>why<i> is it no? When does this statement get executed, and why would it be executed only once? I suppose I'm trying to figure out the "why" behind the "how", so I can recognize and properly interpret this pattern in the future.

.

generateStamp isn't a method. It's a value. let-bound functions have the form

let funcName args = ...

whereas let-bound values have the form

let someValue = ...

There are no args here. generateStamp is a value. It's a function value, of type unit->int. So you can call it like it's a function/method. But the thing you're calling is the lambda that's bound to the name. Each time you call it, the body of the 'fun' is executed, but the body of 'generateStamp' is only executed once when the program/module is initialized. You can see this by adding 'printf's to the code (e.g. before 'let count' and before 'count :=').

Does it make sense?

By on 4/6/2009 6:33 PM ()

Absolutely. Never quite realized the "functional" difference between when the arguments are present and when they're not.

I appreciate the help.

By on 4/6/2009 7:09 PM ()

To create a function with no parameters (instead of a value), declare an argument of type "unit" (similar to null) via a set of empty parens:

let funcName () = ...

By on 4/7/2009 7:50 AM ()

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