为什么下面的函数不是尾递归的？

Question

下面的函数在我看来是尾递归的
但是如果我在它上面放一个 @tailrec 编译器仍然会抱怨：

def loop(newInterests: Set[Interest], oldInterests: Set[Interest]): Set[Interest] = {
  newInterests.headOption.fold(oldInterests){ ni =>
    val withSameKeyWord = oldInterests.find(oi => oi.keyword == ni.keyword)

    withSameKeyWord.fold(loop(newInterests.tail, oldInterests + ni)){ k => 
      loop(newInterests.tail,
      oldInterests - k + k.copy(count = k.count + 1))
    }
  }
}

Answer 1

如 thefourtheye 所述，您的函数 returns 是 fold 的结果。尾递归版本（带有模式匹配）看起来像：

@tailrec
def loop(newInterests: Set[Interest], oldInterests: Set[Interest]): Set[Interest] = {
  newInterests.headOption match {
    case None => oldInterests
    case Some(ni) =>
      oldInterests.find(oi => oi.keyword == ni.keyword) match {
        case None => loop(newInterests.tail, oldInterests + ni)
        case Some(k) => loop(newInterests.tail, oldInterests - k + k.copy(count = k.count + 1))
      }
  }
}

为什么下面的函数不是尾递归的？

Why the following function is not tail recursive?

scala

tail-recursion