如何测试 Peano 数

Question

我正在学习课程 Functional Programming Principles in Scala on Coursera. There is an example of implementation of the Peano numbers 是这样的：

abstract class Nat {
  def isZero: Boolean
  def predecessor: Nat
  def successor: Nat = new Succ(this)
  def +(that: Nat): Nat
  def -(that: Nat): Nat
}

object Zero extends Nat {
  def isZero = true
  def predecessor: Nat = throw new Error("0.predecessor")
  def +(that: Nat): Nat = that
  def -(that: Nat): Nat = if (that.isZero) this else throw new Error("negative number")
}

class Succ(n: Nat) extends Nat {
  def isZero: Boolean = false
  def predecessor: Nat = n
  def +(that: Nat): Nat = new Succ(n + that)
  def -(that: Nat): Nat = if (that.isZero) this else n - that.predecessor
}

我写了几个单元测试。大多数通过，但以下那些 - 以天真方式编写 - 由于明显的原因（不同实例的比较）而失败：

trait Fixture {
  val one = new Succ(Zero)
  val two = new Succ(one)
}

test("successor of zero is one") {
  new Fixture {
    assert(Zero.successor == one)
  }
}

test("successor of one is two") {
  new Fixture {
    assert(one.successor == two)
  }
}

test("one plus zero is one") {
  new Fixture {
    assert((one + Zero) === one)
  }
}

test("one plus one is two") {
  new Fixture {
    assert((one + one) === two)
  }
}

我的问题是：应该如何实施单元测试以成功测试对 peano 数的 + 和 - 操作？

为了以防万一，您可以在这里找到 remaining unit tests。

Answer 1

查看您的测试，您似乎想要测试相等条件，您应该为此编写一个函数：

def eq(i: Nat, j: Nat): Boolean =
  if (i.isZero | j.isZero)
    i.isZero == j.isZero
  else
    eq(i.predecessor, j.predecessor)

通过调用 eq 替换您的 === 和 ==。您也可以考虑覆盖 equal 方法，而不是将其作为外部（测试）函数。

Answer 2

感谢 Cyrille Corpet 的提示，在我看来，使用 case class 和 "compares by structure, not by reference" 很优雅。现在所有单元测试都通过了，没有任何变化。

case class Succ(n: Nat) extends Nat {
  def isZero: Boolean = false
  def predecessor: Nat = n
  def +(that: Nat): Nat = new Succ(n + that)
  def -(that: Nat): Nat = if (that.isZero) this else n - that.predecessor
}

如何测试 Peano 数

How to test Peano numbers

unit-testing

scala

peano-numbers