我正在尝试建立一种可以逆转的关系模型。例如,North 的反面可能是 South。左的反面可能是右。我想用一个案例类来表示我的关系。我在这里找到了一个使用案例对象的类似解决方案,但这并不是我想要的,here .
这是我的非功能代码:
case class Relationship(name: String, opposite:Relationship)
def relationshipFactory(nameA:String, nameB:String): Relationship = {
lazy val x:Relationship = Relationship(nameA, Relationship(nameB, x))
x
}
val ns = relationshipFactory("North", "South")
ns // North
ns.opposite // South
ns.opposite.opposite // North
ns.opposite.opposite.opposite // South
是否可以更改此代码以便:
- 不会崩溃
- 我可以根据需要成对地创建这些东西。
最佳答案
如果你真的想构建具有循环依赖的不可变对象(immutable对象)的图,你必须将 opposite
声明为 def
,并且(最好)再向混合:
abstract class Relationship(val name: String) {
def opposite: Relationship
}
object Relationship {
/** Factory method */
def apply(nameA: String, nameB: String): Relationship = {
lazy val x: Relationship = new Relationship(nameA) {
lazy val opposite = new Relationship(nameB) {
def opposite = x
}
}
x
}
/** Extractor */
def unapply(r: Relationship): Option[(String, Relationship)] =
Some((r.name, r.opposite))
}
val ns = Relationship("North", "South")
println(ns.name)
println(ns.opposite.name)
println(ns.opposite.opposite.name)
println(ns.opposite.opposite.opposite.name)
如果您在这个循环依赖环上运行几百万轮,您可以很快说服自己不会发生任何不好的事情:
// just to demonstrate that it doesn't blow up in any way if you
// call it hundred million times:
// Should be "North"
println((1 to 100000000).foldLeft(ns)((r, _) => r.opposite).name)
它确实打印了“North”。它不适用于案例类,但您始终可以添加自己的提取器,因此它有效:
val Relationship(x, op) = ns
val Relationship(y, original) = op
println(s"Extracted x = $x y = $y")
它为 x
和 y
打印“North”和“South”。
然而,更明显的做法是只保存关系的两个组成部分,并添加 opposite
作为构造相反对的方法。
case class Rel(a: String, b: String) {
def opposite: Rel = Rel(b, a)
}
实际上,这已经在标准库中实现了:
scala> val rel = ("North", "South")
rel: (String, String) = (North,South)
scala> rel.swap
res0: (String, String) = (South,North)
关于scala - Scala 中案例类的不可变配对实例?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/48973051/