Question

我正在课程中学习Functional Programming Principles in Scala课程。有一个Peano numbers的实现示例，如下所示：

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。大多数传递，但以下 - 以天真方式编写 - 都是由于明显的原因（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：如何实施单元测试以成功测试+和 - 对peano数字的操作？

以防万一，您可以在这里找到remaining unit tests。

Answer 1

感谢来自Cyrille Corpet的提示，我认为使用case class是优雅的＆＃34;按结构进行比较，而不是通过引用进行比较＆＃34; < / em>的。所有单元测试现在都没有任何变化。

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 }

Answer 2

看看你的测试看起来你想测试相等条件，你应该写一个函数：

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

替换您的===＆amp; ==致电eq。您也可以考虑重写等于方法，而不是将其作为外部（测试）函数。

如何测试Peano数字

2 个答案: