clone
만 얕은 복사본을 만듭니다, 당신은 깊은 사본이 필요합니다 :
scala> import collection.mutable.ListBuffer
import collection.mutable.ListBuffer
scala> var a = ListBuffer(ListBuffer(1, 2), ListBuffer(3,4))
a: scala.collection.mutable.ListBuffer[scala.collection.mutable.ListBuffer[Int]] = ListBuffer(ListBuffer(1, 2), ListBuffer(3, 4))
scala> var b = a.clone
b: scala.collection.mutable.ListBuffer[scala.collection.mutable.ListBuffer[Int]] = ListBuffer(ListBuffer(1, 2), ListBuffer(3, 4))
scala> b(0)(0) = 100
scala> a
res1: scala.collection.mutable.ListBuffer[scala.collection.mutable.ListBuffer[Int]] = ListBuffer(ListBuffer(100, 2), ListBuffer(3, 4))
scala> b
res2: scala.collection.mutable.ListBuffer[scala.collection.mutable.ListBuffer[Int]] = ListBuffer(ListBuffer(100, 2), ListBuffer(3, 4))
scala> var c = a.clone.map(_.clone)
c: scala.collection.mutable.ListBuffer[scala.collection.mutable.ListBuffer[Int]] = ListBuffer(ListBuffer(100, 2), ListBuffer(3, 4))
scala> c(0)(0) = 1000
scala> c
res3: scala.collection.mutable.ListBuffer[scala.collection.mutable.ListBuffer[Int]] = ListBuffer(ListBuffer(1000, 2), ListBuffer(3, 4))
scala> a
res4: scala.collection.mutable.ListBuffer[scala.collection.mutable.ListBuffer[Int]] = ListBuffer(ListBuffer(100, 2), ListBuffer(3, 4))
깊은 사본을 달성하기 위해, var에 C = a.clone.map는 이상적인 방법에 (. _ 클론)''입니까? –
간단한 예를 들어, 나는 네가 더 복잡하거나 더 깊은 뭔가를 원한다면 [예] (https://github.com/kostaskougios/cloning)와 같은 멋진 솔루션을 원할 수도 있습니다. –