I have an abstract class Model with a static attribute and another generic class Controller<T extends Model>. I want to access the static attribute of Model in an instance of Controller. That should like this:
abstract class Model{
static hasStatus: boolean = false;
}
class MyModel extends Model{
static hasStatus = true;
}
class Controller<T extends Model>{
constructor(){
if(T.hasStatus)...
}
}
But TS says 'T' only refers to a type, but is being used as a value here.
Is there an easy way to achieve this? Or should i subclass Controller for each Heritage of Model and implement a method to retrieve the value?
There is no way to do that in typescript. Generic type parameters can only appear where types may appear in declarations, they are not accessible at runtime. The reason for that is simple - single javascript function is generated for each method of the generic class, and there is no way for that function to know which actual type was passed as generic type parameter.
If you need that information at runtime, you have to add a parameter to the constructor and pass a type yourself when calling it:
class Controller<T extends Model>{
constructor(cls: typeof Model){
if (cls.hasStatus) {
}
}
}
let c = new Controller<MyModel>(MyModel);
Here is how it looks when compiled to javascript to illustrate the point - there is nothing left of generic parameters there, and if you remove cls parameter there is no information about where hasStatus should come from.
var Controller = (function () {
function Controller(cls) {
if (cls.hasStatus) {
}
}
return Controller;
}());
var c = new Controller(MyModel);
Related
I have an abstract base class validator with a method which takes a generic type as parameter.
I will be passing generic type parameter to base class from the subclass inheriting the base class.
Base Class:
abstract class BaseValidator {
bool isValid<T>(T obj);
}
Child Class:
class IPv4Validator extends BaseValidator{
final IPV4_REGEX = "^((25[0-5]|(2[0-4]|1d|[1-9]|)d).?\b){4}\$";
#override
bool isValid<String>(String obj) {
bool hasMatch = RegExp(IPV4_REGEX).hasMatch(obj);
return hasMatch;
}
}
Here hasMatch takes in non nullable string. When I directly pass some string hasMatch doesn't throw an error.
But when I try to pass the generic value in the method parameter, it shows an error.
The argument type 'String' can't be assigned to the parameter type
'String'.
I couldn't able to understand why generic type is not accepting, even though its compile-time type.
The following code solves this particular problem. But it may be different from what you intended to implement. On the other hand, the code will be cleaner if you create a new concrete class for different data types.
abstract class BaseValidator<T> {
bool isValid(T obj);
}
class IPv4Validator extends BaseValidator<String>{
final IPV4_REGEX = "^((25[0-5]|(2[0-4]|1d|[1-9]|)d).?\b){4}\$";
#override
bool isValid(String obj) {
bool hasMatch = RegExp(IPV4_REGEX).hasMatch(obj);
return hasMatch;
}
}
Explanation.
In the line class IPv4Validator extends BaseValidator<String> we are not declaring a new class BaseValidator, it is already declared as BaseValidator<T>. Here we are inheriting the specialization of the existing generic class BaseValidator. While in the line bool isValid<String>(String obj), we declare a new function, so the compiler understands it as if we were declaring a new generic function with a parameter type named String. So, here bool isValid<String>(String obj) is equivalent to bool isValid<T>(T obj), just instead of name T we used name String, which is not an object String.
another fix that you can do is to use the covariant keyword, to implement that, try this:
abstract class BaseValidator<T> {
bool isValid(T obj);
}
class IPv4Validator extends BaseValidator {
final IPV4_REGEX = "^((25[0-5]|(2[0-4]|1d|[1-9]|)d).?\b){4}\$";
#override
bool isValid(covariant String obj) {
bool hasMatch = RegExp(IPV4_REGEX).hasMatch(obj);
return hasMatch;
}
}
In PHP there is a way of accessing a static property value that is defined/overridden on an inheritor.
e.g.
class Foo {
public static $name='Foo';
public function who(){
echo static::$name;//the static operator
}
}
class Bar extends Foo {
public static $name='Bar';
}
$bar = new Bar();
$bar->who();
//Prints "Bar";
Is there ANY way of doing the exact same thing in Dart language?
Addressing comments:
About making it instance prop/method: There's a reason for the existence of static properties and methods and it's not having to create a new instance of the object to access a value or functionality that is not mutable.
Yes, but that's not how you are using it. Your use case is to invoke the method on an object, and therefore you really want an instance method. Now, some languages automatically allow invoking class methods as instance methods, and I see two choices for a language that offers that ability:
Statically transform fooInstance.classMethod() to ClassFoo.classMethod() based on the declared type (not the runtime type) of the object. This is what Java and C++ do.
Implicitly generate virtual instance methods that call the class method. This would allow fooInstance.classMethod() to invoke the appropriate method based on the runtime type of the object. For example, given:
class Foo {
static void f() => print('Foo.f');
}
You instead could write:
class Foo {
static void classMethod() => print('Foo.f');
final instanceMethod = classMethod;
}
and then you either could call Foo.classMethod() or Foo().instanceMethod() and do the same thing.
In either case, it's syntactic sugar and therefore isn't anything that you couldn't do yourself by being more verbose.
About the "meaning of static" and "only work because they allow invoking class methods as instance methods" : That affirmation is actually wrong. In the case of PHP, as per the example above, the Language is providing a way to access the TYPE of the class calling the method in the inheritance chain. A(methodA) >B > C. When C calls methodA, PHP allows you to know that the class type you're in is indeed C, but there's no object instance attached to it. the word "static" there is a replacement for the caller class type itself
All of that is still known at compilation time. That C derives from B derives from A is statically known, so when you try to invoke C.methodA, the compiler knows that it needs to look for methodA in B and then in A. There's no dynamic dispatch that occurs at runtime; that is still compile-time syntactic sugar. That is, if you wanted, you could explicitly write:
class A {
static void methodA() {}
}
class B extends A {
static void methodA() => A.methodA();
}
class C extends B {
static void methodA() => B.methodA();
}
Anyway, in your example, you could write:
class Foo {
static String name = 'Foo';
String get nameFromInstance => name;
void who() {
print(nameFromInstance);
}
}
class Bar extends Foo {
static String name = 'Bar';
#override
String get nameFromInstance => name;
}
void main() {
var bar = Bar();
bar.who(); // Prints: Bar
}
class ExampleClass {
//default constructor
ExampleClass() {
//do stuff
}
//named constructor
ExampleClass.namedConstructor() {
//do stuff
}
}
void main() {
//is there a way to create a variable with datatype to store an object that is constructed only with a specific constructor?
//I have tried something like this, but it returns an error
ExampleClass.namedConstructor variable_1;
}
Is there any way to do this or an alternative? because I need to be able to differentiate between an object that is constructed with the default constructor or with a named constructor.
You can add some identification to classes builded with different constructors and compare entities by unique parameters.
If instances of your classes creating once (Singleton design pattern), you can create entities as constants and compare it by reference:
const administrator = User.administrator();
class User {
final int id;
User(this.id);
factory User.administrator() {
return User(0);
}
factory User.administrator(int id) {
return User(id);
}
}
I have multiple class like this:-
Class A {
static int xyz = 10;
int c;
int d;
static A getData() {
// Do something
return A()..c = xyz*5;
}
Class B {
static int abc = 10;
int c;
static B getData() {
// Do something
return B()..c = xyz*5;
}
So, here you can see that the the getData() is doing the same thing, but have different return types.
Is there any way to avoid duplicate implementation like this, can it be done by defining a single function which can reference the class and have multiple return type?
This has two parts: creating the object, and assigning to a field of the object.
Creating the object, you are mostly out of luck. The only way to create an object of a specific type in a generic method is by using reflection via dart:mirrors. However, you have indicated that this is for a Flutter project, and Flutter doesn't support reflection, so that isn't an option. The only way you are going to be able to dynamically create an object is to pass in a factory method that the generic method can call to construct the object.
Assigning to a field of the object is easier, but it requires that you either lose static type checking by using dynamic or by tying your classes together with inheritance. The latter is the preferable choice, but if you are working with a library than it isn't always an option.
Combining these two things, the code will look like this:
class Foo {
static int xyz = 10;
int c;
}
class A extends Foo {
int d;
static A getData() {
return modifyObject(() => A());
}
}
class B extends Foo {
static B getData() {
return modifyObject(() => B());
}
}
T modifyObject<T extends Foo>(T create()) {
return create()..c = Foo.xyz * 5;
}
Before doing this, though, I'd take a look at whether your project actually needs it. If your use case is as simple as your example, I would argue that this level of generalization is overkill and you are hurting your code's readability more than you are helping its modularity.
Class1 has a property List<ClassA> xyz = [];. ClassA has a number of properties and methods. ClassB extends ClassA adding additional properties and methods. Is there a way for me to create Class2 extending Class1 but change the TYPE of xyz to List<ClassB>? If that is confusing hopefully code below will give an example of what I'm trying to accomplish. Basically the same as overriding a method but overriding a property.
class Game {
int points;
String opponent;
int opponentPoints;
}
class FootballGame extends Game {
int touchdowns;
int opponentstouchdowns;
}
class BaseballGame extends Game {
int homeruns;
int opponentsHomeruns;
}
class Team {
String name;
List<Game> games;
double winningPercentage() {
int wins = 0;
for(var game in games){
wins += (game.points > game.opponentPoints) ? 1 : 0;
}
return wins / games.length;
}
}
class FootballTeam extends Team {
// How do I change the TYPE of the games property to <FootballGame>
}
You could use the covariant keyword in this case:
class FootballTeam extends Team {
#override
covariant List<FootballGame> games;
}
However, be aware that doing so is potentially unsafe; the reason why you need the covariant keyword is to suppress the type error that arises because the override can violate the contract of the base class: the base class advertises that games can be assigned a List<Game>, but such an assignment would be invalid for the derived class. By using the covariant keyword, you disable the type-check and take responsibility for ensuring that you do not violate the contract in practice.
Note that if the games member were final (or were only a getter), then the override (which uses a more specific type) would be safe and wouldn't need to use covariant.
Edit
I had forgotten that I had written this answer when writing a more detailed answer for a similar question.