Question

任何人都可以解释对象行为的模式匹配。同一类的两个对象是否相同？

object Solution extends App {
  case class EE() { }
  object EE1 extends EE
  object EE2 extends EE
  val k: EE = EE1
  println(k.getClass) // class Solution$EE1$
  println(k.isInstanceOf[EE2.type]) // returns FALSE
  println(k.getClass.isInstanceOf[EE2.type.getClass]) // returns FALSE
  k match {
    case EE2 => println("EE1 match EE2!!!") // MATCH THIS LINE/// WHY???
    case EE1 => println("EE1 match EE1. OK!")
  }
}

Answer 1

当你在匹配中写case EE2时，它会检查匹配的对象是否与EE2相等（请注意，如果标识符以小写字母开头，那么Lee的回答是正确的;此规则专门用于匹配object和常量，其名称通常以大写字母开头，有意义）。但由于EE1和EE2都延伸case class EE()，因此它们相等（您可以直接使用EE1 == EE2或k == EE2进行检查）。如果你使它们不等by removing case：

，它会按预期工作

object Solution extends App {
  class EE() { }
  object EE1 extends EE
  object EE2 extends EE
  val k: EE = EE1
  println(k.getClass) // class Solution$EE1$
  println(k.isInstanceOf[EE2.type]) // returns FALSE
  k match {
    case EE2 => println("EE1 match EE2!!!") // MATCH THIS LINE/// WHY???
    case EE1 => println("EE1 match EE1. OK!")
  }
}


Processing...
Reused last reload result
[info] Loading project definition from /tmp/renderer9oHZD8Bvx9/project/project
[info] Loading project definition from /tmp/renderer9oHZD8Bvx9/project
[info] Set current project to rendererWorker (in build file:/tmp/renderer9oHZD8Bvx9/)
[info] Reapplying settings...
[info] Set current project to rendererWorker (in build file:/tmp/renderer9oHZD8Bvx9/)

[info] Formatting 1 Scala source {file:/tmp/renderer9oHZD8Bvx9/}rendererWorker(compile) ...
[info] Reformatted 1 Scala source {file:/tmp/renderer9oHZD8Bvx9/}rendererWorker(compile).
[info] Compiling 1 Scala source to /tmp/renderer9oHZD8Bvx9/target/classes...
[success] Total time: 4 s, completed Aug 19, 2015 1:13:04 PM
Now running...
[info] Running Solution 
class Solution$EE1$
false
EE1 match EE1. OK!
[success] Total time: 0 s, completed Aug 19, 2015 1:13:04 PM

Scala模式匹配Object

1 个答案: