Understanding byref, ref and &

Well, I came to understand that F# is able to manage references (some sort of C++ like references). This enables the possibilities to change value of parameters passed in functions and also enables the programmer to return more than a single value. However here's what I need to know:

Ref keyword: The keyword ref is used to create, from a value, a reference to that value of the inferred type. So
```
let myref = ref 10
```
This means that F# will create an object of type Ref<int> putting there (in the mutable field) my int 10.

OK. So I assume that ref is used to create instances of the Ref<'a> type. Is it correct?
Access value: In order to access a value stored in reference I can do this:
```
let myref = ref 10
let myval = myref.Value
let myval2 = !myref
```
While the := operator just lets me edit the value like this:
```
let myref = ref 10
myref.Value <- 30
myref := 40
```
So ! (Bang) dereferences my reference. And := edit it. I suppose this is correct too.
The & operator: What does this operator do? Is it to be applied to a reference type? No, I guess it must be applied to a mutable value and this returns what? The reference? The address? If using interactive:
```
let mutable mutvar = 10;;
&a;;
```
The last line throws an error so I do not understand what the & operator is for.
ByRef: What about byref? That's very important to me, but I realize I do not understand it. I understand it is used in function regarding parameter passing. One uses byref when he wants that the passed value can be edited (this is a bit against the functional languages' philosophy but f# is something more than that). Consider the following:
```
let myfunc (x: int byref) =
    x <- x + 10
```
This is strange. I know that if you have a reference let myref = ref 10 and then do this to edit the value: myref <- 10 it arises an error because it should be like this: myref := 10. However, the fact that in that function I can edit x using the <- operator means that x is not a reference, right?

If I assume that x is not a reference, then I assume also that, in functions, when using byref on a parameter, that parameter can have the mutable syntax applied to. So it is just a matter of syntax, if I assume this I am OK, and, in fact, everything works (no compiler errors). However, what is x?

Calling functions: How can I use a function utilizing byref parameters?

The & operator is involved but could you explain this better please? In this article: MSDN Parameters and Arguments the following example is provided:

type Incrementor(z) =
    member this.Increment(i : int byref) =
       i <- i + z

let incrementor = new Incrementor(1)
let mutable x = 10
// A: Not recommended: Does not actually increment the variable. (Me: why?)
incrementor.Increment(ref x)
// Prints 10.
printfn "%d" x  

let mutable y = 10
incrementor.Increment(&y) (* Me: & what does it return? *)
// Prints 11.
printfn "%d" y 

let refInt = ref 10
incrementor.Increment(refInt) (* Why does it not work in A, but here it does? *)
// Prints 11.
printfn "%d" !refInt

Ref keyword Yes, when you write let a = ref 10 you're essentially writing let a = new Ref<int>(10) where the Ref<T> type has a mutable field Value.

Access value The := and ! operators are just shortcuts for writing:

a.Value <- 10  // same as writing: a := 10
a.Value        // same as writing: !a

ByRef is a special type that can be (reasonably) used only in method parameters. It means that the argument should be essentially a pointer to some memory location (allocated on heap or stack). It corresponds to out and ref modifiers in C#. Note that you cannot create local variable of this type.

The & operator is a way to create a value (a pointer) that can be passed as an argument to a function/method expecting a byref type.

Calling functions the example with byref works because you're passing the method a reference to a local mutable variable. Via the reference, the method can change the value stored in that variable.

The following doesn't work:

let a = 10            // Note: You don't even need 'mutable' here
bar.Increment(ref a)

The reason is that you're creating a new instance of Ref<int> and you're copying the value of a into this instance. The Increment method then modifies the value stored on heap in the instance of Ref<int>, but you don't have a reference to this object anymore.

let a = ref 10
bar.Increment(a)

This works, because a is a value of type Ref<int> and you're passing a pointer to the heap-allocated instance to Increment and then get the value from heap-allocated reference cell using !a.

(You can use values created using ref as arguments for byref because the compiler handles this case specially - it will automatically take reference of the Value field because this is a useful scenario...).

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