Using the following continuation monad:
type ContinuationMonad() =
member this.Bind (m, f) = fun c -> m (fun a -> f a c)
member this.Return x = fun k -> k x
let cont = ContinuationMonad()
I fail to see why the following gives me a stack overflow:
let map f xs =
let rec map xs =
cont {
match xs with
| [] -> return []
| x :: xs ->
let! xs = map xs
return f x :: xs
}
map xs id;;
let q = [1..100000] |> map ((+) 1)
While the following doesn't:
let map f xs =
let rec map xs =
cont {
match xs with
| [] -> return []
| x :: xs ->
let! v = fun g -> g(f x)
let! xs = map xs
return v :: xs
}
map xs id;;
let q = [1..100000] |> map ((+) 1)
To fix your example, add this method to your definition of the monad:
member this.Delay(mk) = fun c -> mk () c
Apparently the part that overflows is the destruction of the large input list in the recursive call of map
. Delaying it solves the problem.
Note that your second version puts the recursive call to map
behind another let!
which desugars to Bind
and an extra lambda, in effect delaying the recursive call to map
.
I had to pursue a few false trails before reaching this conclusion. What helped was observing that StackOverflow
is thrown by OCaml
as well (albeit at a higher N
) unless the recursive call is delayed. While F#
TCO has some quirks, OCaml
is more proven, so this convinced me that the problem is indeed with the code and not the compiler:
let cReturn x = fun k -> k x
let cBind m f = fun c -> m (fun a -> f a c)
let map f xs =
(* inner map loop overflows trying to pattern-match long lists *)
let rec map xs =
match xs with
| [] -> cReturn []
| x :: xs ->
cBind (map xs) (fun xs -> cReturn (f x :: xs)) in
map xs (fun x -> x)
let map_fixed f xs =
(* works without overflowing by delaying the recursive call *)
let rec map xs =
match xs with
| [] -> cReturn []
| x :: xs ->
cBind (fun c -> map xs c) (fun xs -> cReturn (f x :: xs)) in
map xs (fun x -> x)
let map_fused f xs =
(* manually fused version avoids the problem by tail-calling `map` *)
let rec map xs k =
match xs with
| [] -> k []
| x :: xs ->
map xs (fun xs -> k (f x :: xs)) in
map xs (fun x -> x)
The F# compiler is sometimes not very clever - in the first case it computes map xs
then f x
and then joins them, so map xs
is not in a tail position. In the second case, it can reorder the map xs
to be in tail position easily.
来源:https://stackoverflow.com/questions/11188779/stackoverflow-in-continuation-monad