I saw simple class which was look like:
class SomeClass extends Object{
int a;
int b;
...
...
}
Why this class was extended an Object class? As in documentation was written "Because Object is the root of the Dart class hierarchy, every other Dart class is a subclass of Object." in https://api.dartlang.org/stable/2.4.0/dart-core/Object-class.html.
What will happened if we will not extends Object? Or maybe it will be useful in some specific problems?
All dart classes implicitly extend Object, even if not specified.
This can be verified using the following code:
class Foo {}
void main() {
var foo = Foo();
print(foo is Object); // true
}
Even null implements Object, which allows doing:
null.toString()
null.hashCode
null == something
Related
https://dart.dev/guides/language/language-tour#extending-a-class
Argument types must be the same type as (or a supertype of) the
overridden method’s argument types. In the preceding example, the
contrast setter of SmartTelevision changes the argument type from int
to a supertype, num.
I was looking at the above explanation and wondering why the arguments of subtype member methods need to be defined more "widely"(generally) than the original class's one.
https://en.wikipedia.org/wiki/Covariance_and_contravariance_(computer_science)#Function_types
class AnimalShelter {
Animal getAnimalForAdoption() {
// ...
}
void putAnimal(Animal animal) {
//...
}
}
class CatShelter extends AnimalShelter {
//↓ Definitions that are desirable in the commentary
void putAnimal(Object animal) {
// ...
}
//↓Definitions that are not desirable in the commentary
void putAnimal(Cat animal) {
// ...
}
//I can't understand why this definition is risky.
//What specific problems can occur?
}
I think this wikipedia sample code is very easy to understand, so what kind of specific problem (fault) can occur if the argument of the member method of the subtype is defined as a more "narrower"(specific) type?
Even if it is explained in natural language, it will be abstract after all, so it would be very helpful if you could give me a complete working code and an explanation using it.
Let's consider an example where you have a class hierarchy:
Animal
/ \
Mammal Reptile
/ \
Dog Cat
with superclasses (wider types) above subclasses (narrower types).
Now suppose you have classes:
class Base {
void takeObject(Mammal mammal) {
// ...
}
Mammal returnObject() {
// ...
}
}
class Derived extends Base {
// ...
}
The public members of a class specify an interface: a contract to the callers. In this case, the Base class advertises a takeObject method that accepts any Mammal argument. Every instance of a Base class thus is expected to conform to this interface.
Following the Liskov substitution principle, because Derived extends Base, a Derived instance is a Base, and therefore it too must conform to that same Base class interface: its takeObject method also must accept any Mammal argument.
If Derived overrode takeObject to accept only Dog arguments:
class Derived extends Base {
#override
void takeObject(Dog mammal) { // ERROR
// ...
}
}
that would violate the contract from the Base class's interface. Derived's override of takeObject could be invoked with a Cat argument, which should be allowed according to the interface declared by Base. Since this is unsafe, Dart's static type system normally prevents you from doing that. (An exception is if you add the covariant keyword to disable type-safety and indicate that you personally guarantee that Derived.takeObject will never be called with any Mammals that aren't Dogs. If that claim is wrong, you will end up with a runtime error.)
Note that it'd be okay if Derived overrode takeObject to accept an Animal argument instead:
class Derived extends Base {
#override
void takeObject(Animal mammal) { // OK
// ...
}
}
because that would still conform to the contract of Base.takeObject: it's safe to call Derived.takeObject with any Mammal since all Mammals are also Animals.
Note that the behavior for return values is the opposite: it's okay for an overridden method to return a narrower type, but returning a wider type would violate the contract of the Base interface. For example:
class Derived extends Base {
#override
Dog returnObject() { // OK, a `Dog` is a `Mammal`, as required by `Base`
// ...
}
}
but:
class Derived extends Base {
#override
Animal returnObject() { // ERROR: Could return a `Reptile`, which is not a `Mammal`
// ...
}
}
void main(){
Animal a1 = Animal();
Cat c1 = Cat();
Dog d1 = Dog();
AnimalCage ac1 = AnimalCage();
CatCage cc1 = CatCage();
AnimalCage ac2 = CatCage();
ac2.setAnimal(d1);
//cc1.setAnimal(d1);
}
class AnimalCage{
Animal? _animal;
void setAnimal(Animal animal){
print('animals setter');
_animal = animal;
}
}
class CatCage extends AnimalCage{
Cat? _cat;
#override
void setAnimal(covariant Cat animal){
print('cats setter');
_cat = animal;
/*
if(animal is Cat){
_cat = animal;
}else{
print('$animal is not Cat!');
}
*/
}
}
class Animal {}
class Cat extends Animal{}
class Dog extends Animal{}
Unhandled Exception: type 'Dog' is not a subtype of type 'Cat' of 'animal'
In the above code, even if the setAnimal method receives a Dog instance, a compile error does not occur and a runtime error occurs, so making the parameter the same type as the superclass's one and checking the type inside the method is necessary.
Suppose I have an abstract superclass A. That class has a property abstract val predicate: (ModelClass) -> Boolean.
Let B be a subclass.
I want to be able to do both of the following:
Use the predicate from an instance aInstance.predicate
Also use that predicate elsewhere, without having to create an instance to get that predicate B.predicate
How can I do this.
I don't think this is possible.
There is no such thing as an abstract static method in Kotlin or Java.
Perhaps this will give more insight.
Does your class need to be abstract? Maybe the code below can be useful:
open class MyClass {
companion object myCompanion {
val myStatic = "MyClass"
}
open val myStatic = myCompanion.myStatic
}
class MySubClass : MyClass() {
companion object myCompanionSubClass {
val myStatic = "MySubClass"
}
override var myStatic = myCompanionSubClass.myStatic
}
fun main() {
println(MyClass.myStatic)
val t = MyClass()
println(t.myStatic)
println(MySubClass.myStatic)
val subClass = MySubClass()
println(subClass.myStatic)
}
In this code you can define a static property and use it from the class or any instance. It is also possible to override the property in a subclass and use it in the same way.
In the following code I would like to set a reference to the class instance so that static functions can return a reference to it:
open class TestRunner {
init {
instance = this
}
companion object {
private lateinit var instance: TestRunner
fun addTestSetups(vararg testSetups: () -> TestSetup): TestRunner {
for (setup in testSetups) {
testsSetups.add(setup)
}
return instance
}
}
}
But setting instance = this is not allowed. How can I return an instance of the class from a function while keeping the class as a singleton?
If I get you right, you want something like this:
abstract class TestRunner {
companion object : TestRunner()
}
This seems to work. Instead of keeping a variable that holds a reference to the class, simply referencing the name of the class is sufficient. However, to return an instance of the class from functions, the return type must be Companion:
open class TestRunner {
companion object {
fun addTestSetups(vararg testSetups: () -> TestSetup): Companion {
for (setup in testSetups) {
testsSetups.add(setup)
}
return TestRunner
}
}
}
This is not a true singleton because you can still create a new instance if you did this:
val testRunner = TestRunner()
However, if you never create an instance but only refer to the functions statically, it does behave like a singleton and the state of any private variables inside the companion object will still be maintained.
Update:
I came across this code on the Android developer site that shows an example of a class that is setup as a singleton:
class StockLiveData(symbol: String) : LiveData<BigDecimal>() {
private val stockManager: StockManager = StockManager(symbol)
private val listener = { price: BigDecimal ->
value = price
}
override fun onActive() {
stockManager.requestPriceUpdates(listener)
}
override fun onInactive() {
stockManager.removeUpdates(listener)
}
companion object {
private lateinit var sInstance: StockLiveData
#MainThread
fun get(symbol: String): StockLiveData {
sInstance = if (::sInstance.isInitialized) sInstance else StockLiveData(symbol)
return sInstance
}
}
}
But it should be pointed out that this example requires functions that need to return an instance to first check if the instance variable is set and if not, create a new instance. I'm not sure what the point of that is since to call the function you already have an instance. So why bother create a new instance? Doesn't seem to make any sense.
object in Kotlin is the singleton, not the class its defined within. A companion object has the extra convenience of allowing you to call it by the name of that outer class. But it otherwise shares no hierarchy with it.
To make your class subclassable, you can't define the functions in the companion object. But you can make the class abstract so it can't be instantiated unless subclassed. Then make your companion object extend the abstract class so it will have all those functions available.
abstract class TestRunner{
open fun addTestSetups(vararg testSetups: () -> TestSetup): TestRunner{
//...
return this
}
companion object: TestRunner()
}
Usage:
TestRunner.addTestSetups(someTestSetup)
Note that your singleton is not an instance of TestRunner. It is a singleton instance of a subclass of TestRunner. But since you define no extra functions and override nothing, it behaves exactly like a TestRunner.
If you want a subclass:
abstract class ExtendedTestRunner: TestRunner() {
fun someOtherFunction() {}
companion object: ExtendedTestRunner()
}
The companions are not being subclassed, but their abstract parents can be.
Type 1:
class TestExample {
object Bell {
fun add(){
}
}
Class B{
TestExample.Bell.add()
}
Type 2:
class TestExample {
companion object Bell {
fun add(){
}
}
Class B{
TestExample.add()
}
In this type 1 and type 2, which is static example and which is singleton example? Both behaves similar behavior right?
From official Kotlin docs:
Object declarations
If you need a singleton - a class that only has got one instance - you
can declare the class in the usual way, but use the object keyword
instead of class
Companion objects
If you need a function or a property to be tied to a class rather than
to instances of it (similar to #staticmethod in Python), you can
declare it inside a companion object
I have two classes Class A and Class SRD (Sample classes for understanding the problem. Real classes are different). Both classes have same Function(method1) with same arguments. Both are not derived from different Classes.
Class SRD is the member of Class A. a function in Class A creates a new object for SRD and calls method1(). It should call the mock function. but it calls the actual implementation
I have Written mock classes for both the classes and defined the mock method in both the classes and Wrote EXPECT_CALL in TEST function
class A{
private:
SRD* srd;
public :
bool Method1();
bool MethodX();
SRD* getsrd() {return srd;}
};
bool A :: MethodX()
{
srd.Method1(); // Not Calling Mock Method - Calling Actual
//Implementation
}
bool A::Method1()
{
....
}
class SRD{
public:
bool Method1();
};
class MockSRD : public SRD{
MOCK_METHOD0(Method1, bool())
};
class MockA : public MockA{
MOCK_METHOD0(Method1, bool())
};
bool SRD::Method1()
{
....
}
class TestA : public A {};
TEST_F(TestA, test1)
{
MockSRD srd;
EXPECT_CALL(srd, Method1())
.Times(testing::AnyNumber())
.WillRepeatedly(Return(true));
srd.Method1() //Working fine - Caling mock Method;
MethodX()
}
When i call s1.Method1(), It should call the mock method. how should i do that ?
I don't want to change the production code.
Thanks for taking time to respond the Question . #Chris Oslen & #sklott
I forgot to make the base class method to Virtual. Its worked fine when i change the base class methods