Flutter 如何在没有旧参考的情况下在dart中复制 List

发布于05月22日

代码将解释所有:

//modal for list
class MyModal
{
 int myField1;
 List<MyModal> adjacentNodes;
 MyModal(this.myField1)
 {
  adjacentNodes= new List<MyModal>();
 }
}

//pre code
List<MyModal> originalList = new List<MyModal>();
originalList.add(new MyModal(1,"firstBuddy"));

//copying list
List<MyModal> secondList = new List<MyModal>();
secondList.addAll(originalList);

//Modifing copy list
secondList.adjacentNodes.add(new MyModal(2,"anotherBuddy"));

//Also modifies original list
print(originalList[0].childs.length); //prints 1, it should prints 0

如何在不影响原始列表的情况下在第二个列表中执行更改？

推荐答案

class MyModal {
  int myField1;
  String myField2;
  List<MyModal> adjacentNodes;
  MyModal(this.myField1,this.myField2);

  MyModal.clone(MyModal source) : 
      this.myField1 = source.myField1, 
      this.myField2 = source.myField2,
      this.adjacentNodes = source.adjacentNodes.map((item) => new MyModal.clone(item)).toList();
}

var secondList = originalList.map((item) => new MyModal.clone(item)).toList();

如果MyModal的成员是非基元类型，如String、int、double、num、bool，则clone()方法还需要克隆引用所指向的实例.

我认为对于您的用例，使用不可变值是更好的方法，例如使用https://pub.dartlang.org/packages/built_value