What is the role of `while`-loops in computation expressions in F#?
If you define a While
method of the builder-object, you can use while
-loops in your computation expressions. The signature of the While
method is:
member b.While (predicate:unit->bool, body:M<'a>) : M<'a>
For comparison, the signature of the For
method is:
member b.For (items:seq<'a>, body:unit->M<'a>) : M<'a>
You should notice that, in the While
-method, the body is a simple type, and not a function as in the For
method.
You can embed some other statements, like let
and function-calls inside your computation-expressions, but those can impossibly execute in a while
-l开发者_JAVA技巧oop more than once.
builder {
while foo() do
printfn "step"
yield bar()
}
Why is the while
-loop not executed more than once, but merely repeated? Why the significant difference from for-loops? Better yet, is there some intended strategy for using while-loops in computation-expressions?
If you look at how computation expressions are evaluated, you'll see that
while foo() do
printfn "step"
yield bar()
is translated to something like
builder.While(fun () -> foo(),
builder.Delay(fun () ->
printfn "step"
builder.Yield(bar()))))
This translation allows the body of the while loop to be evaluated multiple times. While your type signatures are accurate for some computation expressions (such as seq
or async
), note that the insertion of the call to Delay
may result in a different signature. For instance, you could define a list builder like this:
type ListBuilder() =
member x.Delay f = f
member x.While(f, l) = if f() then l() @ (x.While(f, l)) else []
member x.Yield(i) = [i]
member x.Combine(l1,l2) = l1 @ l2()
member x.Zero() = []
member x.Run f = f()
let list = ListBuilder()
Now you can evaluate an expression like:
list {
let x = ref 0
while !x < 10 do
yield !x
x := !x + 1
}
to get the equivalent of [0 .. 9]
.
Here, our While
method has the signature (unit -> bool) * (unit -> 'a list) -> 'a list
, rather than (unit -> bool) * 'a list -> 'a list
. In general, when the Delay
operation has type (unit -> M<'a>) -> D<M<'a>>
, the While
method's signature will be (unit -> bool) * D<M<'a>> -> M<'a>
.
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