Suppose I have created StringUtil class by extending String. Now any third party calling StringUtil can use all the methods defined for String class. How can I restrict usage of some methods?
String class is final, and you can't extend a final class in Java. You have to read more about Java core. For example, begin by reading https://www.w3schools.in/java-tutorial/super-final-keywords/
Also, when you want a string utility, you have several options like Apache-Commons-Lang or Spring-Core that makes your code readable and has more functionalities.
In case you have multiple classes such that you want to give each of them access to a group of methods inside StringUtil, you can do the following:
If Class1, for example, needs only methods: f1, f2, f3 then you could create an interface interface1 such that exposes only f1, f2, f3 and Class will receive an object of type interface1 as a param to its constructor. StringUtil implements interface1.
Do the same for other classes so each class will be able to use only what you expose to them.
You should read also on dependency injection.
The interfaces:
interface Interface1 {
void f1();
}
interface Interface2 {
void f2();
void f3();
}
StringUtils class:
class StringUtils implements Interface1, Interface2 {
#Override
public void f1() {
}
#Override
public void f2() {
}
#Override
public void f3() {
}
}
The classes who use StringUtils class:
class A {
private Interface1 interface1; // you can mark it as final if it's permanent.
public A(Interface1 interface1) {
this.interface1 = interface1;
}
}
class B {
private Interface2 interface2; // you can mark it as final if it's permanent.
public B(Interface2 interface2) {
this.interface2 = interface2;
}
}
In such scenarios, however, there may be some workaround like: override the method in the base throw some expectation from it, but it will not be a real solution. My recommendation will be to use composition over inheritance. That means your base should has a relationship with superclasses. For example, a let assume 2 classes A (same as your String class) and B (same as your StringUtil class).
class A {
public int add(int a, int b) {
return a + b;
}
public int square(int a) {
return a * a;
}
}
class B {
A obj;
B(A obj) {
this.obj = obj;
}
public int add(int a, int b) {
return obj.add(a, b);
}
}
Now your class B only expose only one method add which delegates its behavior to class A's add method. Also square is not available on B.
Related
I want to call the constructor of a class inside the method of an interface.
For example, if I have two classes B and C and they both implement SomeInterface, so that each of them has method foo().
interface SomeInterface {
public SomeInterface foo();
}
class B implements SomeInterface {
public B(int fst, int snd) {}
#Override
public SomeInterface foo() {
return new B(1, 1);
}
}
class C implements SomeInterface {
public C(int fst, int snd) {}
#Override
public SomeInterface foo() {
return new C(1, 1);
}
}
And let's say, for the sake of this question, that I have a lot more classes that implement SomeInterface and they all do the same thing, that is return new <nameoftheclass>(1,1)
and all these classes extend the parent class A.
Is there a way for me to create only one method in A such that if any of these classes use the foo method that is found in A it will call their constructor and just like that save me lines of code?
You can do something like this with reflection, although it will be prone to failure.
public SomeInterface foo() {
Constructor<? extends SomeInterface> c = getClass().getConstructor(int.class, int.class);
return c.newInstance( 1, 1);
}
You'll have to manage some exceptions, but is this what you're after?
The question would then be, where can this be used? Interfaces don't have a common constructor.
public interface SomeInterface{
default SomeInterface another(){
Constructor<? extends SomeInterface> c = getClass().getConstructor(int.class, int.class);
return c.newInstance( 1, 1);
}
}
That would work provided whatever the implementations try to use it have that constructor. There is no guarantee that constructor exists though. Maybe you would want it on an abstract class?
use the foo method that is found in A it will call their constructor and just like that save me lines of code?
You are getting it wrong. Class design decisions must be based on use cases and relationships of the classes in your domain. If your main criteria will be to spare some lines of code, you can end up with a coffee machine extending combine harvester because both of them have tree dimensions. Don't take a pill if you have no headache.
Parent class A that you've mentioned doesn't make any sense because method foo() returns an instance of SomeInterface interface which A doesn't implement (because if it does, its subclasses don't need to declare to implement it). I.e. A and SomeInterface are not compatible and compiler will not allow to type cast between them. Therefore, I'll omit the parent class.
As an example, the "template" you've provided might be useful, will be a situation when classes with similar functionality need to grouped together.
The interface can serve as a single entry point for the user of the code. Every class will implement the behavior defined by the interface, and only through the interface it'll be possible to get an instance of the class with a particular flavor of functionality. The actual classes will be hidden from the user.
Similarly, abstract class NumberFormat from the JDK provides a way to obtain different kinds of formatters, but actual implementations are hidden are not exposed (the approach shown below is far more simple than the actual way of how factory methods of the NumberFormat are implemented).
Note, interface and its implementations must reside in the same package.
public interface BaseInterface {
public static BaseInterface getInstance(Classifier classifier) { // factory
return switch(classifier) {
case A -> new A();
case B -> new B();
};
}
void doSomeThingUseful(); // behaviour that every class should implement
}
enum Classifier { A, B }
class A implements BaseInterface {
A() {}
#Override
public void doSomeThingUseful() {
System.out.println("Class A");
}
}
class B implements BaseInterface {
B() {}
#Override
public void doSomeThingUseful() {
System.out.println("Class B");
}
}
main() - demo
public static void main(String[] args) {
List<BaseInterface> items = List.of(BaseInterface.getInstance(Classifier.A),
BaseInterface.getInstance(Classifier.B));
for (BaseInterface item: items) {
item.doSomeThingUseful();
}
}
Output
Class A
Class B
In my project I have a superclass and two subclasses extending from it. There is a method in the superclass that is overriden differently in each subclass.
I want to know if it's possible to introduce a method (in another class) that takes object of either subclass as a parameter and calls a method overriden in one of subclasses (depending on to which subclass does the object belong).
public class Superclass{
public int method(){return 0;}
}
public class Subclass1 extends Superclass{
public int method(){return 1;}
}
public class Subclass2 extends Superclass{
public int method(){return 2;}
}
public class CallingClass{
public static int dependantCall(Superclass parameter){return parameter.method}
I want to be able to do something like
Subclass1 subclassObject = new Subclass1;
System.out.println(CallingClass.dependantCall(subclassObject));
and get output
1
That is what Polymorphism is for! Defining the Superclass as a parameter type will allow you to pass either subclass in.
For example in your other class you can define it like this:
// classes Dog and Cat extend Animal and override makeNoise()
class Owner{
playWith(Animal a){
a.makeNoise();
}
}
Now the Owner can accept owner.makeNoise(cat) and owner.makeNoise(dog)
More reading: https://docs.oracle.com/javase/tutorial/java/IandI/polymorphism.html
Yes, it is entirely possible. Here's how that method would look like:
public <T extends Superclass> void foo(T subclassObject) {
...
}
Or:
public void foo(Superclass obj) {
...
}
Note that in the above method, you can pass subclasses' objects as well (they are covariant data types).
This is what Java does by default when you create subclases, so no need to do anything special. Each object carries it's type information at run time, and the method invoked would always be the most specific one for the object. Example:
public class Doer {
public void doSomething() {
// Body presence
};
}
public class Painter extends Doer {
#Override
public void doSomething() {
// Paint here
}
}
public class Manager extends Doer {
#Override
public void doSomething() {
// Micromanage here
}
}
// Elsewhere in your code:
public void busyness(Doer doer) {
doer.doSomething();
}
A style note: if it is possible, one should prefer using interfaces instead of base classes (base classes those should be used only if you want to share implementation between subclasses). Example with interfaces:
public interface Doer {
void doSomething();
}
public class JackOfAllTrades implements Does {
#Override
public void doSomething() {
// Do whatever necessary
}
}
// Client code stays exactly the same as above:
public void busyness(Doer doer) {
doer.doSomething();
}
Note that in Java a class can have only one base class but can implement multiple interfaces.
#Override annotations are not strictly required, but they help Java compiler to spot some errors for you (e.g. if you misprint method name).
In your example it would look like
public class CallingClass {
public static int dependantCall(Superclass parameter) {
return parameter.method();
}
}
Subclass1 subclassObject = new Subclass1();
System.out.println(CallingClass.dependantCall(subclassObject));
I have created some interface such that:
public interface A{
}
and i would like to call the method a that I have already implemented in class B in interface A such that:
public class B{
public boolean a(){
return true;
}
}
public interface A{
public void call {
a();
}
}
without any errors, any help please?
What you want to do is strictly speaking impossible, as you cannot define method implementations in an interface. You can get something similar by defining an implementation of the interface that extends B. Hopefully that is close enough.
public class AImplementation extends B implements A{
public void call(){
a();
}
}
If you are using any java version before 8, then stick with the answers of #tinker and #Davis Broda. They provide better design since they do not couple your interface to the B class. If you insist however, in java 8 you can have default method implementations as well as static methods in an interface.
If your method is for inheritance then you have to use a default method. Add the default keyword:
default void call() {
...
}
Now the problem is how to get a reference to the class in order to call the method since you cannot have instance fields in interfaces. You have two choices:
Pass the object of B as a method parameter:
public interface A{
default void call(B b) {
b.a();
}
}
or make the method in B static
public interface A{
default void call() {
B.a();
}
}
If your method is not for inheritance but just a utility than you can make it static as :
public interface A{
public static void call() {
B.a();
}
}
I agree with #Davis Broda's answer, there is no way to have a method definition in an interface. But I have another way to address this.
You can have the interface and then have an abstract class implement this interface, and then have all other classes extend the abstract class. The abstract class doesn't have to extend the class from where you want to call the method, you could call it from an instance of that class too.
public interface A {
void caller();
}
public class B {
public void callMe() {
}
}
public class AbstractA implements A {
private B b;
public AbstractA(B b) {
this.b = b;
}
#Override
public void caller() {
b.callMe();
}
}
This way, all implementations of AbstractA will be able to call B's callMe method. And you can access this directly from the interface using this code:
A anInstance = someInstance;
anInstance.caller();
Your question is not very clear, but if I'm guessing right, you want interface A to be kind of a generic caller.
If you're using Java 8, you can achive that using a method reference:
public class B {
public boolean a() {
return true;
}
}
public interface A<T> {
default T call(Supplier<T> s) {
return s.get();
}
}
public class AImpl
implements A<Boolean> {
}
public class Sample {
public static void main(String[] args) {
AImpl a = new AImpl();
B b = new B();
boolean result = a.call(b::a);
System.out.println(result); // true
}
}
This uses Supplier<T> because your method a() in class B returns a boolean and does not receive any arguments.
I have the following setup:
class Base {};
class ImplA extends Base {};
class ImplB extends Base {};
class ImplC extends Base {};
Base baseFactory(int type) {
switch(type) {
case 0:
return new ImplA();
case 1:
return new ImplB();
case 2:
return new ImplC();
}
Base a = baseFactory(0);
Base b = baseFactory(1);
Base c = baseFactory(2);
List<Base> list = new ArrayList<Base>();
list.add(a);
list.add(b);
list.add(c);
// Somewhere else I have:
interface BaseHandler {
process(ImplA a);
process(ImplB b);
process(ImplC c);
};
Now, what I would like to be able to do is something along the lines of:
class Processor {
BaseHandler bh;
Processor(BaseHandler bh) {
this.bh = b;
}
void processList(List<Base> list) {
for (Base x : list) {
bh.process(x);
}
}
And then have a user implement BaseHandler and be able to construct a Processor to operate on each element in the Base list.
But, this does not work as process(Base) is not defined. It may seem simple to just add 3 if statements, but I already have a switch like structure in building instances of classes extending the Base. It seems unnecessary to repeat this over and over. Is there a way to achieve this idea without writing an intermediate step that determines the runtime class of each Base in the list and calls the appropriate method (in effect another switch case -- but it would be if's)?
I think one work around idea would be to make each Base have an abstract process method which needs to be implemented by the Impl classes. However, this is not acceptable in my situation since the user will not be implementing the Impl classes. Basically, I need process to be a user-defined callback. Further, it does not make sense for process to be a member of the Impl or Base classes since it is in no way related. It's a separate callback that needs to respond dynamically to the type it is called with. And the type is always guaranteed to be a subclass of Base.
You do need the "intermediate step" that you describe, but it need not be if statements. What you're looking for is double dispatch using the visitor pattern. Basically your Base class would have a method:
void accept(BaseHandler handler);
and each subclass would implement it as:
handler.process(this);
where this would resolve to the correct type in each subclass at compile-time.
What you're looking for is the Visitor pattern. You put an abstract method on Base, but all it does is call the appropriate method in BaseHandler:
public interface Base {
void acceptHandler(BaseHandler handler);
}
Then your concrete implementations override acceptHandler and call the correct overload.
public class ImplA implements Base {
public void acceptHandler(BaseHandler handler) {
handler.process(this);
}
}
At this point there's not much value in the overloading, and you'd be better off just giving your methods descriptive names.
It sounds like what you want is the Visitor pattern here:
public interface BaseVisitor {
void caseA(ImplA a);
void caseB(ImplB b);
void caseC(ImplC c);
}
public class MyVisitor implements BaseVisitor {
void visit(List<Base> bases) {
for (Base b : bases) {
b.accept(this);
}
}
public void caseA(ImplA a) { // ... }
public void caseB(ImplB b) { // ... }
public void caseC(ImplC c) { // ... }
}
public abstract class Base {
abstract void accept(BaseVisitor visitor);
}
public class ImplA {
public void accept(BaseVisitor visitor) {
visitor.caseA(this);
}
}
public class ImplB {
public void accept(BaseVisitor visitor) {
visitor.caseB(this);
}
}
public class ImplC {
public void accept(BaseVisitor visitor) {
visitor.caseC(this);
}
}
Let's say I have three classes A, B and C.
B extends A
C extends B
All have a public void foo() method defined.
Now from C's foo() method I want to invoke A's foo() method (NOT its parent B's method but the super super class A's method).
I tried super.super.foo();, but it's invalid syntax.
How can I achieve this?
You can't even use reflection. Something like
Class superSuperClass = this.getClass().getSuperclass().getSuperclass();
superSuperClass.getMethod("foo").invoke(this);
would lead to an InvocationTargetException, because even if you call the foo-Method on the superSuperClass, it will still use C.foo() when you specify "this" in invoke. This is a consequence from the fact that all Java methods are virtual methods.
It seems you need help from the B class (e.g. by defining a superFoo(){ super.foo(); } method).
That said, it looks like a design problem if you try something like this, so it would be helpful to give us some background: Why you need to do this?
You can't - because it would break encapsulation.
You're able to call your superclass's method because it's assumed that you know what breaks encapsulation in your own class, and avoid that... but you don't know what rules your superclass is enforcing - so you can't just bypass an implementation there.
You can't do it in a simple manner.
This is what I think you can do:
Have a bool in your class B. Now you must call B's foo from C like [super foo] but before doing this set the bool to true. Now in B's foo check if the bool is true then do not execute any steps in that and just call A's foo.
Hope this helps.
To quote a previous answer "You can't - because it would break encapsulation." to which I would like to add that:
However there is a corner case where you can,namely if the method is static (public or protected). You can not overwrite the static method.
Having a public static method is trivial to prove that you can indeed do this.
For protected however, you need from inside one of your methods to perform a cast to any superclass in the inheritance path and that superclass method would be called.
This is the corner case I am exploring in my answer:
public class A {
static protected callMe(){
System.out.println("A");
}
}
public class B extends A {
static protected callMe(){
System.out.println("B");
}
}
public class C extends B {
static protected callMe(){
System.out.println("C");
C.callMe();
}
public void accessMyParents(){
A a = (A) this;
a.callMe(); //calling beyond super class
}
}
The answer remains still No, but just wanted to show a case where you can, although it probably wouldn't make any sense and is just an exercise.
Yes you can do it. This is a hack. Try not to design your program like this.
class A
{
public void method()
{ /* Code specific to A */ }
}
class B extends A
{
#Override
public void method()
{
//compares if the calling object is of type C, if yes push the call to the A's method.
if(this.getClass().getName().compareTo("C")==0)
{
super.method();
}
else{ /*Code specific to B*/ }
}
}
class C extends B
{
#Override
public void method()
{
/* I want to use the code specific to A without using B */
super.method();
}
}
There is a workaround that solved my similar problem:
Using the class A, B, and C scenario, there is a method that will not break encapsulation nor does it require to declare class C inside of class B. The workaround is to move class B's methods into a separate but protected method.
Then, if those class B's methods are not required simply override that method but don't use 'super' within that method. Overriding and doing nothing effectively neutralises that class B method.
public class A {
protected void callMe() {
System.out.println("callMe for A");
}
}
public class B extends A {
protected void callMe() {
super.callMe();
methodsForB(); // Class B methods moved out and into it's own method
}
protected void methodsForB() {
System.out.println("methods for B");
}
}
public class C extends B {
public static void main(String[] args) {
new C().callMe();
}
protected void callMe() {
super.callMe();
System.out.println("callMe for C");
}
protected void methodsForB() {
// Do nothing thereby neutralising class B methods
}
}
The result will be:
callMe for A
callMe for C
It's not possible, we're limited to call the superclass implementations only.
I smell something fishy here.
Are you sure you are not just pushing the envelope too far "just because you should be able to do it"? Are you sure this is the best design pattern you can get? Have you tried refactoring it?
I had a problem where a superclass would call an top class method that was overridden.
This was my workaround...
//THIS WOULD FAIL CALLING SUPERCLASS METHODS AS a1() would invoke top class METHOD
class foo1{
public void a1(){
a2();
}
public void a2(){}
}
class foo2 extends foo1{
{
public void a1(){
//some other stuff
super.a1();
}
public void a2(){
//some other stuff
super.a2();
}
//THIS ENSURES THE RIGHT SUPERCLASS METHODS ARE CALLED
//the public methods only call private methods so all public methods can be overridden without effecting the superclass's functionality.
class foo1{
public void a1(){
a3();}
public void a2(){
a3();}
private void a3(){
//super class routine
}
class foo2 extends foo1{
{
public void a1(){
//some other stuff
super.a1();
}
public void a2(){
//some other stuff
super.a2();
}
I hope this helps.
:)
Before using reflection API think about the cost of it.
It is simply easy to do. For instance:
C subclass of B and B subclass of A. Both of three have method methodName() for example.
public abstract class A {
public void methodName() {
System.out.println("Class A");
}
}
public class B extends A {
public void methodName() {
super.methodName();
System.out.println("Class B");
}
// Will call the super methodName
public void hackSuper() {
super.methodName();
}
}
public class C extends B {
public static void main(String[] args) {
A a = new C();
a.methodName();
}
#Override
public void methodName() {
/*super.methodName();*/
hackSuper();
System.out.println("Class C");
}
}
Run class C Output will be:
Class A
Class C
Instead of output:
Class A
Class B
Class C
In my simple case I had to inherit B and C from abstract class, that incapsulates equal methods of B and C. So that
A
|
Abstr
/ \
B C
While it doesn't solve the problem, it can be used in simple cases, when C is similar to B. For instance, when C is initialized, but doesn't want to use initializers of B. Then it simply calls Abstr methods.
This is a common part of B and C:
public abstract class Abstr extends AppCompatActivity {
public void showProgress() {
}
public void hideProgress() {
}
}
This is B, that has it's own method onCreate(), which exists in AppCompatActivity:
public class B extends Abstr {
#Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState); // Call from AppCompatActivity.
setContentView(R.layout.activity_B); // B shows "activity_B" resource.
showProgress();
}
}
C shows its own layout:
public class C extends Abstr {
#Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState); // Call from AppCompatActivity.
setContentView(R.layout.activity_C); // C shows "activity_C" resource.
showProgress();
}
}
This is not something that you should do normally but, in special cases where you have to workaround some bug from a third party library (if it allow to do so), you can achieve calling a super super class method that has already been overwritten using the delegation pattern and an inner class that extends the super super class to use as a bridge:
class A() {
public void foo() {
System.out.println("calling A");
}
}
class B extends A() {
#Overwrite
public void foo() {
System.out.println("calling B");
}
}
class C extends B() {
private final a;
public C() {
this.a = new AExtension();
}
#Overwrite
public void foo() {
a.foo();
}
private class AExtension extends A {
}
}
This way you will be able to not only call the super super method but also combine calls to other super super class methods with calls to methods of the super class or the class itself by using `C.super` or `C.this`.