当对象是动态的时，如何在Scala中对列表进行排序？

Question

I have the following hierarchy in Scala

abstract class MyAbstractClass
case class Child1(alpha: MyAbstractClass, points: Int) extends MyAbstractClass
case class Child2(beta: Char, points: Int) extends MyAbstractClass

Now I have a list List(MyAbstractClass type objects ... either Chil1 or Child2 )

I want to sort the above list based on points. How to write this in Scala?

Answer 1

You should add a method called points to your base class:

abstract class MyAbstractClass { def points:Int }

The subclasses already implement this method, since they are case classes, so you don't need to change them.

case class Child1(alpha: MyAbstractClass, points: Int) extends MyAbstractClass
case class Child2(beta: Char, points: Int) extends MyAbstractClass

Then you can sort the list using this method.

println(List(Child2('a',0),Child1(Child2('b',2),3)).sortBy(_.points))

Answer 2

You can use structural typing helps you here

list.sortBy{
  case x: {def points: Int} => x.points
}

sortBy takes a function that when applied to each element will return the value to sort by. The case statement says "if x is a type that has a "point" method, then return x.points

Answer 3

Used a slightly different approach 1) Without adding points var to parent class 2) Without changing the hierarchy of classes

list containing objects of type MyAbstractClass

list.sortWith((x,y) => point(x) < point(y))

def point(mylist: MyAbstractClass): Int = mylist match {
case Child1(a, p) => p
case Child2(b, p) => p
}