Scala:为什么这个函数不是尾递归？

Question

我有Merge Sort的这种实现:

import scala.annotation.tailrec

object MergeSort {
  def sortBy[T]: ((T, T) => Int) => Seq[T] => Seq[T] = comparator => seqToSort => {
    @tailrec
    def merge(xs : Seq[T], ys : Seq[T], accum : Seq[T] = Seq()) : Seq[T] = (xs, ys) match {
      case (Seq(), _) => ys ++ accum
      case (_, Seq()) => xs ++ accum
      case (x::rx, y::ry) =>
        if(comparator(x, y) < 0)
          merge(xs, ry, y +: accum)
        else
          merge(rx, ys, x +: accum)
    }

    @tailrec
    // Problem with this function
    def step : Seq[Seq[T]] => Seq[T] = {
      case Seq(xs) => xs
      case xss =>
        val afterStep = xss.grouped(2).map({
          case Seq(xs) => xs
          case Seq(xs, ys) => merge(xs, ys)
        }).toSeq
        // Error here
        step(afterStep)
    }

    step(seqToSort.map(Seq(_)))
  }
}

它不编译.它表示步进函数中的递归调用不在尾部位置.但它处于尾部位置.没有蹦床有没有办法解决它？

Answer 1

原因是,这step是一个返回签名函数的函数:Seq[Seq[T]] => Seq[T].因此递归调用不会直接调用相同的方法,而是首先获取此函数,然后为给定的参数调用它,这不是尾递归.

要解决此错误,您必须以step这种方式声明:

@tailrec
def step(seq: Seq[Seq[T]]): Seq[T] = seq match {
  ...
}