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Hidden ref variables in functions
f#

Hi,

why will the internal state of the ref variables in the following examples stay after each call :

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let f =
  let x = ref 0
  let y = ref 0
  fun a ->
    if a = 0 then
      x := 1
      y := 1
    System.Console.WriteLine("{0} \t {1} ", !x, !y)
f 0 //prints 1 1
f 100 // prints 1 1

and, with two tupled arguments :

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let f =
  let x = ref 0
  let y = ref 0
  fun (a, b) ->
    if a = 0 then
      x := 1
      y := 1
    if b then ()
    System.Console.WriteLine("{0} \t {1} ", !x, !y)
f (0, true) // prints 1 1
f (100, false) // prints 1 1

But not when arguments are separated as follows :

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let f a =
  let x = ref 0
  let y = ref 0
  fun b ->
    if a = 0 then
      x := 1
      y := 1
    System.Console.WriteLine("{0} \t {1} ", !x, !y)
f 0 100 // prints 1 1
f 100 100 // prints 0 0

Thank you

.

In the first two cases, 'f' is a value, it gets evaluated 'right now' as the code is executed top to bottom.

In the third case, 'f' is a function, its body is evaluated when it is called. It is called twice (f 0, then f 100), and so body is evaluated each time there.

Does it make sense?

By on 3/13/2009 6:06 PM ()

yes, thank you

By on 3/14/2009 1:35 AM ()

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