如何在Scala中匹配嵌套类？

发布时间：2020-12-16 21:30:48 所属栏目：安全来源：网络整理

导读：我试过下面的代码(相同的方法写在 Programming in Scala书之后) class Person() { class Room(r: Int,c: Int) { val row = r val col = c override def hashCode: Int = 41 * ( 41 + row.hashCode ) + col.hashCode override def equals(other: Any) = other

我试过下面的代码(相同的方法写在 Programming in Scala书之后)

class Person() {
    class Room(r: Int,c: Int) {
        val row = r
        val col = c

        override def hashCode: Int =
            41 * (
                41 + row.hashCode
            ) + col.hashCode

        override def equals(other: Any) =
            other match {
                case that: Room =>
                   (that canEqual this) &&
                   this.row == that.row &&
                   this.col == that.col
               case _ => false
            }

        def canEqual(other: Any) =
            other.isInstanceOf[Room]
    }

    val room = new Room(2,1)
}

val p1 = new Person()
val p2 = new Person()

println(p1.room == p2.room)
>>> false

经过一番分析,我发现Scala重新定义了每个人的实例的课室,这就是两个房间不平等的原因.

解决这个问题的一个可能之处就是将课程放在课外人员中,但这并不总是最简单的. (例如,如果类必须访问Person的某些参数.)

有什么替代方法可以写平等的方法？

解决方法

问题是您的两个房间是路径相关类型的实例：它们的类型是p1.Room和p2.Room：

scala> :type p1.room
p1.Room

做这项工作的一种方式是参考房间使用类型选择,即作为人#房.

class Person() {
    class Room(r: Int,c: Int) {
        val row = r
        val col = c

        override def hashCode: Int = // omitted for brevity

        override def equals(other: Any) =
            other match {
                case that: Person#Room =>
                   (that canEqual this) &&
                   this.row == that.row &&
                   this.col == that.col
               case _ => false
            }

        def canEqual(other: Any) =
            other.isInstanceOf[Person#Room]
    }

    val room: Room = new Room(2,1)
}

val p1 = new Person()
val p2 = new Person()

scala> p1.room == p2.room
res1: Boolean = true

（编辑：李大同）

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