如何从递归函数中得到终止原因?

通常,您需要返回一个包含停止原因和结果的值。使用 (A, Boolean) 您提议的返回签名允许这样做。

您的代码将变成:

import scala.annotation.tailrec

object SquareRootAlg {
  def next(a: Double)(x: Double): Double = (x + a/x)/2
  def repeat[A](f: A=>A, a: A): Stream[A] = a #:: repeat(f, f(a))

  @tailrec // Checks function is truly tail recursive.
  def loopConditional[A](stop: (A, A) => Boolean)(s: => Stream[A] ): (A, Boolean) = {
    val a = s.head
    val t = s.tail
    if(t.isEmpty) (a, false)
    else if(stop(a, t.head)) (t.head, true)
    else loopConditional(stop)(t)
  }
}

只需返回布尔值而不修改任何其他内容:

object SquareRootAlg {
  def next(a: Double)(x: Double): Double = (x + a/x)/2
  def repeat[A](f: A => A, a: A): Stream[A] = a #:: repeat(f, f(a))

  def loopConditionalInfo[A]
    (stop: (A, A)=> Boolean)
    (s: => Stream[A])
  : (A, Boolean) = s match {
    case a #:: t if t.isEmpty => (a, false)
    case a #:: t => 
      if (stop(a, t.head)) (t.head, true) 
      else loopConditionalInfo(stop)(t)
  }
}

import SquareRootAlg._
val cond = (a: Double, b: Double) => (a-b).abs < 0.01
val alg = loopConditionalInfo(cond) _
val s = repeat(next(4.0), 4.0)

println(alg(s.take(3))) // = 2.05, "maxIters exceeded"
println(alg(s.take(5)))

(2.05,false)
(2.0000000929222947,true)