为什么此F#序列函数不能尾部递归? [英] Why is this F# sequence function not tail recursive?

问题描述

披露:这是在FsCheck(我维护的F#随机测试框架)中提出的.我有一个解决方案，但我不喜欢它.而且，我不明白这个问题-它只是被规避了.

Disclosure: this came up in FsCheck, an F# random testing framework I maintain. I have a solution, but I do not like it. Moreover, I do not understand the problem - it was merely circumvented.

(monadic，如果我们要使用大词)序列的一个相当标准的实现是:

A fairly standard implementation of (monadic, if we're going to use big words) sequence is:

let sequence l = 
    let k m m' = gen { let! x = m
                       let! xs = m'
                       return (x::xs) }
    List.foldBack k l (gen { return [] })

其中gen可以由所选的计算构建器代替.不幸的是，该实现会占用堆栈空间，因此如果列表足够长，最终堆栈就会溢出.问题是:为什么呢?我知道foldBack原则上不是尾递归，但是F#团队的聪明兔子已经在foldBack实现中规避了这一点.计算生成器实现中存在问题吗?

Where gen can be replaced by a computation builder of choice. Unfortunately, that implementation consumes stack space, and so eventually stack overflows if the list is long enough.The question is: why? I know in principle foldBack is not tail recursive, but the clever bunnies of the F# team have circumvented that in the foldBack implementation. Is there a problem in the computation builder implementation?

如果我将实现更改为以下内容，则一切正常:

If I change the implementation to the below, everything is fine:

let sequence l =
    let rec go gs acc size r0 = 
        match gs with
        | [] -> List.rev acc
        | (Gen g)::gs' ->
            let r1,r2 = split r0
            let y = g size r1
            go gs' (y::acc) size r2
    Gen(fun n r -> go l [] n r)

为完整起见，可以在在FsCheck源文件中

推荐答案

基于Tomas的答案，让我们定义两个模块:

Building on Tomas's answer, let's define two modules:

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()

为简化起见，让我们将原始序列函数重写为

To simplify things a bit, let's rewrite your original sequence function as

let rec sequence = function
| [] -> gen { return [] }
| m::ms -> gen {
    let! x = m
    let! xs = sequence ms
    return x::xs }

现在，无论sequence是用Kurt.gen还是Tomas.gen定义的，sequence [for i in 1 .. 100000 -> unit i]都将运行完成.问题不是sequence在使用您的定义时导致堆栈溢出，而是从sequence的调用返回的函数在调用 时导致堆栈溢出.

Now, sequence [for i in 1 .. 100000 -> unit i] will run to completion regardless of whether sequence is defined in terms of Kurt.gen or Tomas.gen. The issue is not that sequence causes a stack overflow when using your definitions, it's that the function returned from the call to sequence causes a stack overflow when it is called.

要了解为什么会这样，让我们​​根据基本的单子运算扩展sequence的定义:

To see why this is so, let's expand the definition of sequence in terms of the underlying monadic operations:

let rec sequence = function
| [] -> unit []
| m::ms ->
    bind (fun x -> bind (fun xs -> unit (x::xs)) (sequence ms)) m

将Kurt.unit和Kurt.bind值内联并疯狂简化，我们得到了

Inlining the Kurt.unit and Kurt.bind values and simplifying like crazy, we get

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，希望很清楚为什么调用let (Kurt.Gen f) = sequence [for i in 1 .. 1000000 -> unit i] in f 0会使堆栈溢出:f需要一个非尾递归调用来对结果函数进行排序和求值，因此每个递归调用都会有一个堆栈帧.

Now it's hopefully clear why calling let (Kurt.Gen f) = sequence [for i in 1 .. 1000000 -> unit i] in f 0 overflows the stack: f requires a non-tail-recursive call to sequence and evaluation of the resulting function, so there will be one stack frame for each recursive call.

将Tomas.unit和Tomas.bind内联到sequence的定义中，我们得到以下简化版本:

Inlining Tomas.unit and Tomas.bind into the definition of sequence instead, we get the following simplified version:

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))))

对这种变体进行推理很棘手.您可以凭经验验证它不会对某些任意大的输入造成损失(如Tomas在他的回答中所示)，并且您可以逐步进行评估以使自己相信这一事实.但是，堆栈消耗取决于传入列表中的Gen实例，并且 可能会为本身不是尾部递归的输入而炸毁堆栈:

Reasoning about this variant is tricky. You can empirically verify that it won't blow the stack for some arbitrarily large inputs (as Tomas shows in his answer), and you can step through the evaluation to convince yourself of this fact. However, the stack consumption depends on the Gen instances in the list that's passed in, and it is possible to blow the stack for inputs that aren't themselves tail recursive:

// 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)

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