在Tomas的回答基础上,让我们定义两个模块:
module Kurt =
type Gen<'a> = Gen of (int -> 'a)
let unit x = Gen (fun _ -> x)
let bind k (Gen m) =
Gen (fun n ->
let (Gen m') = k (m n)
m' n)
type GenBuilder() =
member x.Return(v) = unit v
member x.Bind(v,f) = bind f v
let gen = GenBuilder()
module Tomas =
type Gen<'a> = Gen of (int -> ('a -> unit) -> unit)
let unit x = Gen (fun _ f -> f x)
let bind k (Gen m) =
Gen (fun n f ->
m n (fun r ->
let (Gen m') = k r
m' n f))
type GenBuilder() =
member x.Return v = unit v
member x.Bind(v,f) = bind f v
let gen = GenBuilder()
为了简化问题,让我们把您原来的序列函数改写为:
let rec sequence = function
| [] -> gen { return [] }
| m::ms -> gen {
let! x = m
let! xs = sequence ms
return x::xs }
现在,无论是基于Kurt.gen还是Tomas.gen定义的sequence [for i in 1 .. 100000 -> unit i],都将运行到完成。问题不在于使用您的定义时sequence会导致堆栈溢出,而在于从对sequence调用返回的函数在其被调用时会导致堆栈溢出。
为了看清楚这一点,让我们基于底层单子操作扩展sequence的定义:
let rec sequence = function
| [] -> unit []
| m::ms ->
bind (fun x -> bind (fun xs -> unit (x::xs)) (sequence ms)) m
内联
Kurt.unit
和
Kurt.bind
的值,并进行极度简化,我们得到
let rec sequence = function
| [] -> Kurt.Gen(fun _ -> [])
| (Kurt.Gen m)::ms ->
Kurt.Gen(fun n ->
let (Kurt.Gen ms') = sequence ms
(m n)::(ms' n))
现在很清楚为什么调用
let (Kurt.Gen f) = sequence [for i in 1 .. 1000000 -> unit i] in f 0
会导致堆栈溢出了:
f
需要对sequence进行非尾递归调用并评估结果函数,因此每个递归调用都会有一个堆栈帧。
将
Tomas.unit
和
Tomas.bind
内联到
sequence
的定义中,我们得到以下简化版本:
let rec sequence = function
| [] -> Tomas.Gen (fun _ f -> f [])
| (Tomas.Gen m)::ms ->
Tomas.Gen(fun n f ->
m n (fun r ->
let (Tomas.Gen ms') = sequence ms
ms' n (fun rs -> f (r::rs))))
理解这个变量是很棘手的。你可以通过经验验证它不会在一些任意大的输入情况下爆栈(正如Tomas在他的回答中所展示的那样),并且你可以逐步评估以确信这一事实。然而,堆栈消耗取决于传入列表中的Gen
实例,并且对于不是自身尾递归的输入,可能会导致堆栈溢出:
// ok
let (Tomas.Gen f) = sequence [for i in 1 .. 1000000 -> unit i]
f 0 (fun list -> printfn "%i" list.Length)
// not ok...
let (Tomas.Gen f) = sequence [for i in 1 .. 1000000 -> Gen(fun _ f -> f i; printfn "%i" i)]
f 0 (fun list -> printfn "%i" list.Length)