Flutter&Dart: How to remove references between parameter-CodePudding

I have a global parameter template for any parameter that needs to copy these data to show in any widget. Some widgets have a function that changes the data in parameters. The issue is when the function changes the data of a parameter by using .add() or .addAll(), the global parameter data also changes. The data type that will change is List and Map.

Example:

dynamic a;
dynamic b;
a = [];
b = a;
print("List");
print("a = $a , b = $b"); //a = [] , b = []
b.addAll([1]);
print("a = $a , b = $b"); //a = [1] , b = [1]
b = [
  ...b,
  ...[2]
];
print("a = $a , b = $b"); //a = [1] , b = [1, 2]

a = {};
b = a;
print("Map");
print("a = $a , b = $b"); //a = {} , b = {}
b.addAll({'1': 1});
print("a = $a , b = $b"); //a = {1: 1} , b = {1: 1}
b = {
  ...b,
  ...{'2': 2}
};
print("a = $a , b = $b"); //a = {1: 1} , b = {1: 1, 2: 2}

How to solve this with using .add() or .addAll()?

CodePudding user response：

Instead of assigning the same value you should assign a copy. To do that you can call toList() on it. In this example you also need to cast it to List first because you declared them as dynamic. So like

b = (a as List).toList();

For the map you can do this instead:

b = {...(a as Map)};

Edit: it seems the casting to List isn't necessary, but the casting to Map is

CodePudding user response：

Firstly, try to avoid using dynamic whenever possible (of course, sometimes it is not avoidable). In your example List<int> seems to be good.

Secondly, lists (and maps) are passed by reference. For example:

final a = <int>[];
final b = a;

If you add an element to b, a changes as well because they refer to the same list. To prevent this, do final b = List.of(a), which creates a new list that has the same elements as a, then you can do whatever you want to b without touching a.

CodePudding user response：

You can create a new instance while assigning items like

  dynamic a;
  dynamic b;
  a = [];
  b = a.toList();
  print("List");
  print("a = $a , b = $b"); //a = [] , b = [1]

CodePudding user response：

Update: Thank Yeasin Sheikh for easy step with list -> b = a.toList();. Thank Ivo for easy step with map -> b = {...(a as Map)};.

Aw, I notice in my example and see that. Just use ..., or .addAll()

dynamic a;
    dynamic b;
    a = [0];
    b = [...a];
    print("List 1");
    print("a = $a , b = $b");
    b.addAll([1]);
    print("a = $a , b = $b");

    a = [0];
    b = []   a;
    print("List 2");
    print("a = $a , b = $b");
    b.addAll([1]);
    print("a = $a , b = $b");

    a = [0];
    b = [];
    b.addAll(a);
    print("List 3");
    print("a = $a , b = $b");
    b.addAll([1]);
    print("a = $a , b = $b");

    a = {'0': 0};
    b = {
      ...{},
      ...a,
    };
    print("Map");
    print("a = $a , b = $b");
    b.addAll({'1': 1});
    print("a = $a , b = $b");

result:

List 1
a = [0] , b = [0]
a = [0] , b = [0, 1]
List 2
a = [0] , b = [0]
a = [0] , b = [0, 1]
List 3
a = [0] , b = [0]
a = [0] , b = [0, 1]
Map
a = {0: 0} , b = {0: 0}
a = {0: 0} , b = {0: 0, 1: 1}