Deserialization of abstract types in Dart - dart

Question

Please consider the following code segment, declaration of one base class and two extenders.
abstract class Animal {
final bool flag;
Animal(this.flag);
}
class Cat extends Animal {
final String name;
Cat(this.name, bool flag) : super(flag);
Cat.fromJson(Map<String, dynamic> json)
: name = json['name'],
super(json['flag']);
}
class Fish extends Animal {
final int memory;
Fish(this.memory, bool flag) : super(flag);
Fish.fromJson(Map<String, dynamic> json)
: memory = json['memory'],
super(json['flag']);
}
What is the best practice of serializing and deserializing the following list?
List<Animal> animals = [Cat('Nancy', true), Fish(10, false)];
So, how to serialize the type information for each class in order to construct the appropriate instance upon deserialization?
The serialization code is ommited for the sake of simplicity, however it would return a map of the member fields. The code above is only an example - there is no meaning in the structure or the data.
EDIT: I would prefer not using any library or code generator, but using the internal capabilities of the language.
Any suggestion is appretiated, thank you.

Answer 1

I had to figure that out, so I provide my naive solution for the problem. However, if anyone gives a better approach, I will mark that as an answer.
My naive solution is to decorate every JSON result with a 'type' field that contains the name of the subclass. Then I could create a factory for the base type that constructs the appropriate instance depending on that 'type' field. The full code:
abstract class Animal {
final bool flag;
Animal(this.flag);
factory Animal.fromJson(Map<String, dynamic> json) {
switch (json['type']) {
case 'cat':
return Cat.fromJson(json);
case 'fish':
return Fish.fromJson(json);
default:
throw 'Invalid animal type';
}
}
}
class Cat extends Animal {
final String name;
Cat(this.name, bool flag) : super(flag);
Cat.fromJson(Map<String, dynamic> json)
: name = json['name'],
super(json['flag'].toString() == 'true');
Map<String, dynamic> toJson() {
return {'type': 'cat', 'name': name, 'flag': flag.toString()};
}
}
class Fish extends Animal {
final int memory;
Fish(this.memory, bool flag) : super(flag);
Fish.fromJson(Map<String, dynamic> json)
: memory = json['memory'],
super(json['flag'].toString() == 'true');
Map<String, dynamic> toJson() {
return {'type': 'fish', 'memory': memory, 'flag': flag.toString()};
}
Obviously, during deserialization the list items should be mapped using the factory method as follows:
var data = jsonDecode("[{'type':'cat','name':'Nancy','flag':'true'},{'type':'fish','memory':10,'flag':'false'}]");
List<Animal> list = data.map((e) => Animal.fromJson(e)).cast<Animal>().toList();