是否可以使用延续使foldRight尾递归?

本文介绍了是否可以使用延续使foldRight尾递归?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

以下博客文章显示了如何在F#中设置为尾递归.

The following blog article shows how in F# foldBack can be made tail recursive using continuation passing style.

在Scala中，这意味着:

In Scala this would mean that:

def foldBack[T,U](l: List[T], acc: U)(f: (T, U) => U): U = { l match { case x :: xs => f(x, foldBack(xs, acc)(f)) case Nil => acc } }

通过执行以下操作，可以使

成为尾递归:

can be made tail recursive by doing this:

def foldCont[T,U](list: List[T], acc: U)(f: (T, U) => U): U = { @annotation.tailrec def loop(l: List[T], k: (U) => U): U = { l match { case x :: xs => loop(xs, (racc => k(f(x, racc)))) case Nil => k(acc) } } loop(list, u => u) }

不幸的是，对于长列表，我仍然会出现堆栈溢出.循环是尾部递归和优化的，但是我想堆栈堆积只是移到了延续调用中.

Unfortunately, I still get a stack overflow for long lists. loop is tail recursive and optimized but I guess the stack accumulation is just moved into the continuation calls.

为什么这不是F#的问题?还有什么方法可以解决Scala的问题?

Why is this not a problem with F#? And is there any way to work around this with Scala?

编辑:这是一些显示堆栈深度的代码:

Edit: here some code that shows depth of stack:

def showDepth(s: Any) { println(s.toString + ": " + (new Exception).getStackTrace.size) } def foldCont[T,U](list: List[T], acc: U)(f: (T, U) => U): U = { @annotation.tailrec def loop(l: List[T], k: (U) => U): U = { showDepth("loop") l match { case x :: xs => loop(xs, (racc => { showDepth("k"); k(f(x, racc)) })) case Nil => k(acc) } } loop(list, u => u) } foldCont(List.fill(10)(1), 0)(_ + _)

此打印:

loop: 50 loop: 50 loop: 50 loop: 50 loop: 50 loop: 50 loop: 50 loop: 50 loop: 50 loop: 50 loop: 50 k: 51 k: 52 k: 53 k: 54 k: 55 k: 56 k: 57 k: 58 k: 59 k: 60 res2: Int = 10

推荐答案

问题是延续函数(racc => k(f(x, racc)))本身.应该为整个业务进行尾调优化，但并非如此.

The problem is the continuation function (racc => k(f(x, racc))) itself. It should be tailcall optimized for this whole business to work, but isn't.

Scala无法对任意的尾部调用进行尾部调用优化，只能对其进行转换(例如，当函数调用自身而不是其他某些函数时)的循环进行优化.

Scala cannot make tailcall optimizations for arbitrary tail calls, only for those it can transform into loops (i.e. when the function calls itself, not some other function).