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. Solet myref = ref 10
This means that F# will create an object of type
Ref<int>
putting there (in the mutable field) myint 10
.OK. So I assume that
ref
is used to create instances of theRef<'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
!
(B开发者_如何学编程ang) 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 editx
using the<-
operator means thatx
is not a reference, right?If I assume that
x
is not a reference, then I assume also that, in functions, when usingbyref
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 isx
?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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