Another instances of object in Dart [duplicate] - dart

Question

Is there a Language supported way to make a full (deep) copy of an Object in Dart?
If multiple options exist, what are their differences?

Answer 1

Darts built-in collections use a named constructor called "from" to accomplish this. See this post: Clone a List, Map or Set in Dart
Map mapA = {
'foo': 'bar'
};
Map mapB = new Map.from(mapA);

Answer 2

No as far as open issues seems to suggest:
https://github.com/dart-lang/sdk/issues/3367
And specifically:
... Objects have identity, and you can only pass around references to them. There is no implicit copying.

Answer 3

Late to the party, but I recently faced this problem and had to do something along the lines of :-
class RandomObject {
RandomObject(this.x, this.y);
RandomObject.clone(RandomObject randomObject): this(randomObject.x, randomObject.y);
int x;
int y;
}
Then, you can just call copy with the original, like so:
final RandomObject original = RandomObject(1, 2);
final RandomObject copy = RandomObject.clone(original);

Answer 4

I guess for not-too-complex objects, you could use the convert library:
import 'dart:convert';
and then use the JSON encode/decode functionality
Map clonedObject = JSON.decode(JSON.encode(object));
If you're using a custom class as a value in the object to clone, the class either needs to implement a toJson() method or you have to provide a toEncodable function for the JSON.encode method and a reviver method for the decode call.

Answer 5

Unfortunately no language support. What I did is to create an abstract class called Copyable which I can implement in the classes I want to be able to copy:
abstract class Copyable<T> {
T copy();
T copyWith();
}
I can then use this as follows, e.g. for a Location object:
class Location implements Copyable<Location> {
Location({
required this.longitude,
required this.latitude,
required this.timestamp,
});
final double longitude;
final double latitude;
final DateTime timestamp;
#override
Location copy() => Location(
longitude: longitude,
latitude: latitude,
timestamp: timestamp,
);
#override
Location copyWith({
double? longitude,
double? latitude,
DateTime? timestamp,
}) =>
Location(
longitude: longitude ?? this.longitude,
latitude: latitude ?? this.latitude,
timestamp: timestamp ?? this.timestamp,
);
}

Answer 6

To copy an object without reference, the solution I found was similar to the one posted here, however if the object contains MAP or LIST you have to do it this way:
class Item {
int id;
String nome;
String email;
bool logado;
Map mapa;
List lista;
Item({this.id, this.nome, this.email, this.logado, this.mapa, this.lista});
Item copyWith({ int id, String nome, String email, bool logado, Map mapa, List lista }) {
return Item(
id: id ?? this.id,
nome: nome ?? this.nome,
email: email ?? this.email,
logado: logado ?? this.logado,
mapa: mapa ?? Map.from(this.mapa ?? {}),
lista: lista ?? List.from(this.lista ?? []),
);
}
}
Item item1 = Item(
id: 1,
nome: 'João Silva',
email: 'joaosilva#gmail.com',
logado: true,
mapa: {
'chave1': 'valor1',
'chave2': 'valor2',
},
lista: ['1', '2'],
);
// -----------------
// copy and change data
Item item2 = item1.copyWith(
id: 2,
nome: 'Pedro de Nobrega',
lista: ['4', '5', '6', '7', '8']
);
// -----------------
// copy and not change data
Item item3 = item1.copyWith();
// -----------------
// copy and change a specific key of Map or List
Item item4 = item1.copyWith();
item4.mapa['chave2'] = 'valor2New';
See an example on dartpad
https://dartpad.dev/f114ef18700a41a3aa04a4837c13c70e

Answer 7

With reference to #Phill Wiggins's answer, here is an example with .from constructor and named parameters:
class SomeObject{
String parameter1;
String parameter2;
// Normal Constructor
SomeObject({
this.parameter1,
this.parameter2,
});
// .from Constructor for copying
factory SomeObject.from(SomeObject objectA){
return SomeObject(
parameter1: objectA.parameter1,
parameter2: objectA.parameter2,
);
}
}
Then, do this where you want to copy:
SomeObject a = SomeObject(parameter1: "param1", parameter2: "param2");
SomeObject copyOfA = SomeObject.from(a);

Answer 8

Let's say you a have class
Class DailyInfo
{
String xxx;
}
Make a new clone of the class object dailyInfo by
DailyInfo newDailyInfo = new DailyInfo.fromJson(dailyInfo.toJson());
For this to work your class must have implemented
factory DailyInfo.fromJson(Map<String, dynamic> json) => _$DailyInfoFromJson(json);
Map<String, dynamic> toJson() => _$DailyInfoToJson(this);
which can be done by making class serializable using
#JsonSerializable(fieldRename: FieldRename.snake, includeIfNull: false)
Class DailyInfo{
String xxx;
}

Answer 9

It only works for object types that can be represented by JSON.
ClassName newObj = ClassName.fromMap(obj.toMap());
or
ClassName newObj = ClassName.fromJson(obj.toJson());

Answer 10

Trying using a Copyable interface provided by Dart.

Answer 11

there is an easier way for this issue
just use ... operator
for example, clone a Map
Map p = {'name' : 'parsa','age' : 27};
Map n = {...p};
also, you can do this for class properties.
in my case, I was needed to clone a listed property of a class.
So:
class P1 {
List<String> names = [some data];
}
/// codes
P1 p = P1();
List<String> clonedList = [...p.names]
// now clonedList is an unreferenced type

Answer 12

There is no built-in way of deep cloning an object - you have to provide the method for it yourself.
I often have a need to encode/decode my classes from JSON, so I usually provide MyClass fromMap(Map) and Map<String, dynamic> toJson() methods. These can be used to create a deep clone by first encoding the object to JSON and then decoding it back.
However, for performance reasons, I usually implement a separate clone method instead. It's a few minutes work, but I find that it is often time well spent.
In the example below, cloneSlow uses the JSON-technique, and cloneFast uses the explicitly implemented clone method. The printouts prove that the clone is really a deep clone, and not just a copy of the reference to a.
import 'dart:convert';
class A{
String a;
A(this.a);
factory A.fromMap(Map map){
return A(
map['a']
);
}
Map<String, dynamic> toJson(){
return {
'a': a
};
}
A cloneSlow(){
return A.fromMap(jsonDecode(jsonEncode(this)));
}
A cloneFast(){
return A(
a
);
}
#override
String toString() => 'A(a: $a)';
}
void main() {
A a = A('a');
A b = a.cloneFast();
b.a = 'b';
print('a: $a b: $b');
}

Answer 13

There's no API for cloning/deep-copying built into Dart.
We have to write clone() methods ourselves & (for better or worse) the Dart authors want it that way.
Deep copy Object /w List
If the Object we're cloning has a List of Objects as a field, we need to List.generate that field and those Objects need their own clone method.
Example of cloning method (copyWith()) on an Order class with a List field of objects (and those nested objects also have a copyWith()):
Order copyWith({
int? id,
Customer? customer,
List<OrderItem>? items,
}) {
return Order(
id: id ?? this.id,
customer: customer ?? this.customer,
//items: items ?? this.items, // this will NOT work, it references
items: items ?? List.generate(this.items.length, (i) => this.items[i].copyWith()),
);
}
Gunter mentions this here.
Note, we cannot use List.from(items) nor [...items]. These both only make shallow copies.

Answer 14

Dart does not share Memory within multiple threads (isolate), so...
extension Clone<T> on T {
/// in Flutter
Future<T> clone() => compute<T, T>((e) => e, this);
/// in Dart
Future<T> clone() async {
final receive = ReceivePort();
receive.sendPort.send(this);
return receive.first.then((e) => e as T).whenComplete(receive.close);
}
}

Answer 15

An example of Deep copy in dart.
void main() {
Person person1 = Person(
id: 1001,
firstName: 'John',
lastName: 'Doe',
email: 'john.doe#email.com',
alive: true);
Person person2 = Person(
id: person1.id,
firstName: person1.firstName,
lastName: person1.lastName,
email: person1.email,
alive: person1.alive);
print('Object: person1');
print('id : ${person1.id}');
print('fName : ${person1.firstName}');
print('lName : ${person1.lastName}');
print('email : ${person1.email}');
print('alive : ${person1.alive}');
print('=hashCode=: ${person1.hashCode}');
print('Object: person2');
print('id : ${person2.id}');
print('fName : ${person2.firstName}');
print('lName : ${person2.lastName}');
print('email : ${person2.email}');
print('alive : ${person2.alive}');
print('=hashCode=: ${person2.hashCode}');
}
class Person {
int id;
String firstName;
String lastName;
String email;
bool alive;
Person({this.id, this.firstName, this.lastName, this.email, this.alive});
}
And the output below.
id : 1001
fName : John
lName : Doe
email : john.doe#email.com
alive : true
=hashCode=: 515186678
Object: person2
id : 1001
fName : John
lName : Doe
email : john.doe#email.com
alive : true
=hashCode=: 686393765

Answer 16

// Hope this work
void main() {
List newList = [{"top": 179.399, "left": 384.5, "bottom": 362.6, "right": 1534.5}, {"top": 384.4, "left": 656.5, "bottom": 574.6, "right": 1264.5}];
List tempList = cloneMyList(newList);
tempList[0]["top"] = 100;
newList[1]["left"] = 300;
print(newList);
print(tempList);
}
List cloneMyList(List originalList) {
List clonedList = new List();
for(Map data in originalList) {
clonedList.add(Map.from(data));
}
return clonedList;
}

Answer 17

This works for me.
Use the fromJson and toJson from your Object's Class on JSON serializing
var copy = ObjectClass.fromJson(OrigObject.toJson());

Answer 18

make a helper class:
class DeepCopy {
static clone(obj) {
var tempObj = {};
for (var key in obj.keys) {
tempObj[key] = obj[key];
}
return tempObj;
}
}
and copy what you want:
List cloneList = [];
if (existList.length > 0) {
for (var element in existList) {
cloneList.add(DeepCopy.clone(element));
}
}

Answer 19

Let's say, you want to deep copy an object Person which has an attribute that is a list of other objects Skills. By convention, we use the copyWith method with optional parameters for deep copy, but you can name it anything you want.
You can do something like this
class Skills {
final String name;
Skills({required this.name});
Skills copyWith({
String? name,
}) {
return Skills(
name: name ?? this.name,
);
}
}
class Person {
final List<Skills> skills;
const Person({required this.skills});
Person copyWith({
List<Skills>? skills,
}) =>
Person(skills: skills ?? this.skills.map((e) => e.copyWith()).toList());
}
Keep in mind that using only this.skills will only copy the reference of the list. So original object and the copied object will point to the same list of skills.
Person copyWith({
List<Skills>? skills,
}) =>
Person(skills: skills ?? this.skills);
If your list is primitive type you can do it like this. Primitive types are automatically copied so you can use this shorter syntax.
class Person {
final List<int> names;
const Person({required this.names});
Person copyWith({
List<int>? names,
}) =>
Person(names: names ?? []...addAll(names));
}

Answer 20

The accepted answer doesn't provide an answer, and the highest-rated answer 'doesn't work' for more complex Map types.
It also doesn't make a deep copy, it makes a shallow copy which seems to be how most people land on this page. My solution also makes a shallow copy.
JSON-cloning, which a few people suggest, just seems like gross overhead for a shallow-clone.
I had this basically
List <Map<String, dynamic>> source = [{'sampledata', []}];
List <Map<String, dynamic>> destination = [];
This worked, but of course, it's not a clone, it's just a reference, but it proved in my real code that the data types of source and destination were compatible (identical in my case, and this case).
destination[0] = source[0];
This did not work
destination[0] = Map.from(source[0]);
This is the easy solution
destionation[0] = Map<String, dynamic>.from(source[0]);