通过使用traits和with-keyword减少代码

Question

我有一些具有相同超类型的类.因此,所有这些类都必须覆盖相同的方法.现在我可以调用一个方法并将其提交给普通超类型的对象.但是对每个提交的类型做出反应并不总是有用的,因此会引发异常.首先,我尝试解决此问题,如下所示:

def operation(s: SuperType) = s match {
  case t: SubType1 => ...
  case t: SubType2 => ...
  case _ => ...
}

由于很多子类型,这将导致很多代码(在每个方法和每个类中),我试图解决这个问题traits.每个特征应该只测试一种类型,然后将对象转发到堆栈上的更高方法.下面的代码描述了我的想象.但这不起作用,因为编译器无法解散类型.另一个问题是我必须在每个行为类中声明类的每个属性.

object TraitWithTest {
  def main(args: Array[String]) {
    val e1 = Even(2, 4)
    val e2 = Even(1, 3)
    val o1 = Odd(1.25, 3.75)
    val o2 = Odd(7.25, 9.25)
    val a1 = All(5.5)
    val a2 = All(3.5)

    println("e1 + e2: " + (e1 + e2))
    println("o1 + o2: " + (o1 + o2))
    try { println("e1 + o2: " + (e1 + o2)) } catch { case e => println(e) }
    println("o1 + e2: " + (o1 + e2))
    println("a1 + e1: " + (a1 + e2))
  }
}

abstract class Num {
  def +(n: Num): Num
}

trait OddBehaviour extends Num {
  val e1, e2: Int // here I don't want to declare all attributes
  val a1: Double
  abstract override def +(n: Num) = n match {
    case o: Odd => throw new UnsupportedOperationException("Even#+(Odd)")
    case _ => super.+(n)
  }
}

trait EvenBehaviour extends Num {
  val o1, o2: Double
  val a1: Double
  abstract override def +(n: Num) = n match {
    case e: Even => Odd(o1 + e.e1, o2 + e.e2)
    case _ => super.+(n)
  }
}

trait AllBehaviour extends Num {
  val o1, o2: Double
  val e1, e2: Int
  abstract override def +(n: Num) = n match {
    case a: All => Odd(o1 + a.a1, o2 + a.a1)
    case _ => super.+(n)
  }
}

object Even {
  def apply(e1: Int, e2: Int) = new Even(e1, e2) with OddBehaviour with AllBehaviour
}

abstract case class Even(e1: Int, e2: Int) extends Num {
  override def +(n: Num) = n match {
    case c: Even => Even(e1 + c.e1, e2 + c.e2)
    case _ => throw new IllegalArgumentException
  }
}

object Odd {
  def apply(o1: Double, o2: Double) = new Odd(o1, o2) with EvenBehaviour with AllBehaviour
}

abstract case class Odd(o1: Double, o2: Double) extends Num {
  override def +(n: Num) = n match {
    case o: Odd => Odd(o1 + o.o1, o2 + o.o2)
    case _ => throw new IllegalArgumentException
  }
}

object All {
  def apply(a1: Double) = new All(a1) with EvenBehaviour with OddBehaviour
}

abstract case class All(a1: Double) extends Num {
  override def +(n: Num) = n match {
    case a: All => All(a1 + a.a1)
    case _ => throw new IllegalArgumentException
  }
}

有人可以告诉我是否可以通过使用特征来减少代码行数？或者是我目前使用最好的全匹配解决方案？

编辑:

在你的帮助下,我找到了一个半工作的解决方案.我的主要问题是我试图通过使用Scala功能来减少代码行.所以我忽略了最简单的方法:外包代码!我只需要创建一个检查对象组合的新对象.对象本身只处理自己的类型.

这是代码:

final object TraitWithTest {
  def main(args: Array[String]) {
    import traitwith.operations._
    val e1 = Even(2, 4)
    val e2 = Even(1, 3)
    val o1 = Odd(1.25, 3.75)
    val o2 = Odd(7.25, 9.25)
    val a1 = All(5.5)
    val a2 = All(3.5)

    val n1 = NumHolder(o1)
    val n2 = NumHolder(a1)

    println("e1 + e2: " + add(e1, e2))
    println("o1 + o2: " + add(o1, o2))
    try { println("e1 + o2: " + add(e1, o2)) } catch { case e => println(e) }
    println("o1 + e2: " + add(o1, e2))
    try { println("a1 + e2: " + add(a1, e2)) } catch { case e => println(e) }
    println("n1 + n2: " + add(n1, n2))
  }
}

final object operations {
  def add(a: Num, b: Num) = a -> b match {
    case (a1: Odd, b1: Odd) => a1 + b1
    case (a1: Odd, b1: Even) => Odd(a1.x + b1.x, a1.y + b1.y)
    case (a1: Odd, b1: All) => Odd(a1.x + b1.x, a1.y + b1.x)
    case (a1: Even, b1: Even) => a1 + b1
    case (a1: All, b1: All) => a1 + b1
    case _ => error("can't add " + b + " to " + a)
  }
}

abstract class Num {
  type A <: Num
  def +(a: A): A
}

final case class Odd(x: Double, y: Double) extends Num {
  override type A = Odd
  override def +(a: Odd) = Odd(x + a.x, y + a.y)
}

final case class Even(x: Int, y: Int) extends Num {
  override type A = Even
  override def +(a: Even) = Even(x + a.x, y + a.y)
}

final case class All(x: Double) extends Num {
  override type A = All
  override def +(a: All) = All(x + a.x)
}

final case class NumHolder(x: Num) extends Num {
  override type A = NumHolder
  override def +(a: NumHolder) = NumHolder(x + a.x)
}

我稍微扩展了代码并插入了对象NumHolder.现在,只有一个小缺陷:在NumHolder中,如果没有在add-method中出现编译错误,我就无法提交超类型.我尝试使用Generics而不是type-keyword,但这是不方便的,因为我总是将类型设置为Num(也在对象操作中).

我怎样才能解决这个小编译错误？

Answer 1

您的问题是您正在尝试使用面向对象的功能,例如类和继承,其设计不是面向对象的.

整点 OOP的是,你不反思什么是类.相反,使用多态来实现结果.我特别喜欢这篇论文来说明OO应该如何运作,但在这方面并不缺乏资源.

编辑

例如,提供的代码大致转换为以下内容,减去不起作用的内容(提供的代码因为它们而无法精确编译).

abstract class Num {
  def +(n: Num): Num
  def plus(n1: Int, n2: Int): Num
  def plus(n1: Double, n2: Double): Num
  def plus(n: Double): Num
}

case class Even(e1: Int, e2: Int) extends Num {
  override def +(n: Num) = n.plus(e1, e2)
  override def plus(n1: Int, n2: Int) = Even(e1 + n1, e2 + n2)
  override def plus(n1: Double, n2: Double) = Odd(n1 + e1, n2 + e2)
  // the code provided references o1 and o2, which are not defined anywhere for Even
  // so I'm providing an alternate version
  override def plus(n: Double) = Odd(n + e1, n + e2)
}

case class Odd(o1: Double, o2: Double) extends Num {
  override def +(n: Num) = n.plus(o1, o2)
  override def plus(n1: Int, n2: Int) = throw new UnsupportedOperationException("Even#+(Odd)")
  override def plus(n1: Double, n2: Double) = Odd(o1 + n1, o2 + n2)
  override def plus(n: Double) = throw new UnsupportedOperationException("Even#+(Odd)")
}

case class All(a1: Double) extends Num {
  override def +(n: Num) = n.plus(a1)
  // the code provided references o1 and o2, which are not defined anywhere for All
  // so I'm providing an alternate version
  override def plus(n1: Int, n2: Int) = Odd(a1 + n1, a1 + n2)
  override def plus(n1: Double, n2: Double) = Odd(n1 + a1, n2 + a1)
  override def plus(n: Double) = All(a1 + n)
}

它看起来可以通过访问者模式进一步改进,这是有道理的,因为它针对类型匹配通常所做的相同问题.