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Mutable locals and parameters
f#

Are local variables and parameters mutable by default? I typed in the following function definition to Visual Studio 2010 Beta 2:

let foo bar =

let bar = bar - 1

let fubar = 2

let fubar = 3

let bar = bar - 1

bar

I compiled this and got no errors. I thought locals and parameters were immutable. What's going on here?

.

It is in fact quite easy to expose the fact that there are several variables "bar" here, shadowing one another, if you close over them with a lambda:

> let x = 1;;

val x : int = 1

> let f1 = fun() -> x;;

val f1 : unit -> int

> let x = 2;;

val x : int = 2

> let f2 = fun() -> x;;

val f2 : unit -> int

> f1();;

val it : int = 1

> f2();;

val it : int = 2

>

By on 11/17/2009 1:29 PM ()

Unlike some languages, F# supports and encourages shadowing. So when you do "let a = 1; let a = a + 2", you're simply "re-binding" the name a to a new value. The definition of the second binding can use the first binding because the second binding isn't defined yet.

In fact, the types don't even need to match:

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8

 

> let a = 1;;
val a : int = 1

> let a = string a + " oh";;
val a : string = "1 oh"

The second a is completely independent from the original a. The only link is that the second a happened to use the first a's value.

By on 11/17/2009 11:23 AM ()

However, it won't allow me to do this at

the module level. For example:

let f = 1

let f = 3

The VS editor will flag a "duplicate defintion of value f" warning.

By on 11/17/2009 11:43 AM ()

Right because it's exporting those values from the module. In a class constructor, you're shadowing them and they are not exported directly. But you can do this in a module:

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let f = 
   let f = 1
   let f = f + 3
   f
By on 11/17/2009 11:49 AM ()

Also, direct mutability will happen with <-

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let mutable a = 1
a <- a + 1

In that case, we're directly updating the value of binding a; just like = does in C style languages.

By on 11/17/2009 11:25 AM ()

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