Hi I just would like to know if there is any difference between giving abstract keyword or not like so.
// with
abstract class A {}
class B extends A {}
// without
class A {}
class B extends A {}
Should I give it?
With abstract you can omit implementations of methods and getters/setters
// with
abstract class A {
int foo();
String get bar;
set baz(String value);
}
var a = A(); // error about instantiating abstract class
class B extends A {
// error about missing implementations
}
var b = B(); // ok
// without
class A {
int foo(); // error about missing implementation
String get bar; // error about missing implementation
set baz(String value); // error about missing implementation
}
class B extends A {}
Related
Consider this code
final dynamic bar;
class Foo<T> {
const Foo(this.bla) : assert(T == A, T == B);
final T bla;
}
...
final Foo foo = Foo(bla); // Assert will trigerred because bla is dynamic even it is A type.
I don't know the bla type but i can assured its on a A type or B type.
I want to do something like this?
final Foo foo = Foo<bla.runtimeType>(bla);
You should write base class that encompasses class A and B. You should send class Foo that derives from this base class.
abstract class Base{
void printName();
}
class A extends Base{
A();
#override
printName(){
print("A class");
}
}
class B extends Base{
B();
#override
printName(){
print("b class");
}
}
class Foo {
Base bla;
Foo(this.bla);
}
void main(){
dynamic a=A();
Foo fooA = Foo(a);
fooA.bla.printName();
dynamic b=B();
Foo fooB = Foo(b as Base);
fooB.bla.printName();
}
You should be careful here Foo fooB = Foo(b as Base);. If this cast cannot be done, it will give an error.I suggest you use try catch block
look here for generic type check
because T==A (any className) in assert blog will not give correct result
for more information
I'm trying to specify a function parameter as a generic type T:
enum MyEnum {
Foo,
Bar
}
class DbColumn {
final Function<T>(T value) serializer;
const DbColumn({this.serializer});
}
class MyClass {
static final DbColumn rating = DbColumn(
serializer: <MyEnum>(v) {
var i = v.index;
}
);
}
However when trying to access index on v I get this type error message:
The getter 'index' isn't defined for the type 'Object'.
Try importing the library that defines 'index', correcting the name to the name of an existing getter, or defining a getter or field named 'index'.
When I hover over v in VSC it says that it's of type MyEnum.
If I instead remove the generic type and do a cast like this it works as expected:
class DbColumn {
final Function(dynamic value) serializer;
const DbColumn({this.serializer});
}
class MyClass {
static final DbColumn rating = DbColumn(
serializer: (v) {
var casted = v as MyEnum;
var i = casted.index;
}
);
}
Why is the generic type not working as expected?
EDIT:
What is even weirder is that this example works too if I put it inside MyClass:
x<T>(Function(T) c) {}
y() {
x<MyEnum>((v) {
print(v.index); // No error given and type of v is MyEnum
});
}
EDIT 2: The same problem happens when overriding methods:
abstract class MyInterface {
int someFunction<T>(T value);
}
class MyClass implements MyInterface {
#override
someFunction<MyEnum>(v) {
return v.index; // Gives same error and no intellisense happens in VSC
}
}
Instead of making the function generic, declare the class as generic and it will work as expected. Like this :
enum MyEnum {
Foo,
Bar
}
class DbColumn<T> {
final Function(T value) serializer;
const DbColumn({this.serializer});
}
class MyClass {
static final DbColumn<MyEnum> rating = DbColumn(
serializer: (v) {
var i = v.index;
print('Index : $i');
}
);
}
void main() {
MyClass.rating.serializer(MyEnum.Bar);
}
OUTPUT :
Index : 1
Here's the simplest example.
class MixA{
}
class MixB{
}
class Base{
}
class MyClass extends Base with MixA, MixB{
}
main(){
var m = new MyClass();
reflect(m).superclass.mixin; //This only gives MixB. Is there a way to get a list of all the mixins?
}
I can't find a way to just get a list of all the mixins that were applied to MyClass
Each mixin application creates a new subclass:
void main(){
final m = new MyClass();
final r = reflect(m);
print(r.type.superclass.mixin);
print(r.type.superclass.superclass.mixin);
}
How to check type of Super class with Child class instance? I have below example and don't want to use dart-mirrors.
class SomeClass{
}
class SomeOtherClass extends SomeClass{
}
void main() {
var s1 = new SomeOtherClass();
someMethod(SomeClass, s1);
}
void someMethod(Type t, dynamic instance){
print(instance.runtimeType == t);
//print(instance.runtimeType is t); Does not work!
}
Update
Just today the package reflectable was released which allows to do this like with mirrors, but a transformer generates code instead to avoid using mirrors in production.
import 'package:reflectable/reflectable.dart';
// Annotate with this class to enable reflection.
class Reflector extends Reflectable {
const Reflector()
: super(typeCapability); // Request the capability to invoke methods.
}
const reflector = const Reflector();
#reflector
class SomeClass{
}
#reflector
class SomeOtherClass extends SomeClass{
}
void someMethod(Type t, dynamic instance){
InstanceMirror instanceMirror = reflector.reflect(instance);
print(instanceMirror.type.isSubclassOf(reflector.reflectType(t)));
}
void main() {
var s1 = new SomeOtherClass();
someMethod(SomeClass, s1);
}
Original
It might be directly supported when https://github.com/gbracha/metaclasses is implemented.
Currently this workaround can be used:
class IsInstanceOf<E> {
bool check(t) => t is E;
}
void someMethod(Type t, dynamic instance){
print( new IsInstanceOf<t>().check(instance));
//print(instance.runtimeType is t); Does not work!
}
This runs fine and returns the correct result but the analyzer shows a warning because t can't be used as a type.
If you wrap SomeClass in a generic class it works without a warning
class SomeClass{
}
class SomeOtherClass extends SomeClass{
}
void main() {
var s1 = new SomeOtherClass();
someMethod(new IsInstanceOf<SomeClass>(), s1);
}
void someMethod(IsInstanceOf t, dynamic instance){
print(t.check(instance));
//print(instance.runtimeType is t); Does not work!
}
class IsInstanceOf<E> {
bool check(instance) => instance is E;
}
Try it at DartPad
I noticed PetitParserDart has a lot of #override in the code, but I don't know how do they be checked?
I tried IDEA dart-plugin for #override, but it has no effect at all. How can we use #override with Dart?
From #override doc :
An annotation used to mark an instance member (method, field, getter or setter) as overriding an inherited class member. Tools can use this annotation to provide a warning if there is no overridden member.
So, it depends on the tool you use.
In the current Dart Editor(r24275), there's no warning for the following code but it should (it looks like a bug).
import 'package:meta/meta.dart';
class A {
m1() {}
}
class B extends A {
#override m1() {} // no warning because A has a m1()
#override m2() {} // tools should display a warning because A has no m2()
}
The #override annotation is an example of metadata. You can use Mirrors to check for these in code. Here is a simple example that checks if the m1() method in the child class has the #override annotation:
import 'package:meta/meta.dart';
import 'dart:mirrors';
class A {
m1() {}
}
class B extends A {
#override m1() {}
}
void main() {
ClassMirror classMirror = reflectClass(B);
MethodMirror methodMirror = classMirror.methods[const Symbol('m1')];
InstanceMirror instanceMirror = methodMirror.metadata.first;
print(instanceMirror.reflectee); // Instance of '_Override#0x2fa0dc31'
}
it's 2021 . the override it's optional
Use the #override annotation judiciously and only for methods where the superclass is not under the programmer's control, the superclass is in a different library or package, and it is not considered stable. In any case, the use of #override is optional. from dart api https://api.dart.dev/stable/2.10.5/dart-core/override-constant.html
example
Class A {
void say (){
print ('Say something 1') ;
}
}
Class B extends A {
#override
void adds() { // when i don't type the same name of super class function show an
// warning not an error say 'The method doesn't override an inherited
// method.' because it's not same name but when type the same name must be
// overriding
print ('Say something 2 ')
}
Update : the main use of #override is when try to reach abstract method inside abstract class in sub class that inherited to the abstract super class . must use #override to access the abstract method .