Question

测试（“ok”）是由Noel Welsh和Dave Gurnell pag.254（“D.4更安全折叠使用Eval”）中的书“scala with cats”复制的 “），代码运行正常，这是trampolined foldRight

import cats.Eval
test("ok") {
val list = (1 to 100000).toList


def foldRightEval[A, B](as: List[A], acc: Eval[B])(fn: (A, Eval[B]) => Eval[B]): Eval[B] =
  as match {
    case head :: tail =>
      Eval.defer(fn(head, foldRightEval(tail, acc)(fn)))
    case Nil =>
      acc
  }

def foldRight[A, B](as: List[A], acc: B)(fn: (A, B) => B): B =
  foldRightEval(as, Eval.now(acc)) { (a, b) =>
    b.map(fn(a, _))
  }.value

val res = foldRight(list, 0L)(_ + _)

assert(res == 5000050000l)
}

测试（“ko”）为小列表返回相同的测试值（“ok”），但对于长列表，值不同。为什么呢？

test("ko") {
val list = (1 to 100000).toList

def foldRightSafer[A, B](as: List[A], acc: B)(fn: (A, B) => B): Eval[B] = as match {
  case head :: tail =>
    Eval.defer(foldRightSafer(tail, acc)(fn)).map(fn(head, _))
  case Nil => Eval.now(acc)
}

val res = foldRightSafer(list, 0)((a, b) => a + b).value

assert(res == 5000050000l)
}

Answer 1

这是@ OlegPyzhcov的评论，转换为社区wiki答案

您忘记L中的0L作为第二个参数传递给foldRightSafer。因此，调用的推断泛型类型是

foldRightSafer[Int, Int]((list : List[Int]), (0: Int))((_: Int) + (_: Int))

因此您的添加会溢出并为您提供小于2000000000（9个零，Int.MaxValue = 2147483647）的内容。

斯卡拉猫蹦床

1 个答案: