Consider the code:
public class A<T extends X> {
public static interface Delegate {
void doMagic(T t); // why can't I access "T" here?
}
public A(Delegate delegate) { ... }
}
...
public class TheDelegate implements A<Y> { ... }
...
A<Y> a = new A<Y>(new A<Y>.Delegate() {
#Override
public void doMagic(Y y) {
...
}
});
Why can't I access T from Delegate interface?
It's because your inner interface is static. The generic parameter only applies to an instance of A as opposed to applying to the class, so the scope of T is the non-static scope of A.
In case you didn't know, all interfaces and enumerations are static in Java, even if they are not declared as static and are inside another class. Therefore there is no way to work around this with an interface.
See this answer also.
EDIT: Steven's answer is correct. However, your user code will look like this:
// Note the extra declaration of the generic type on the Delegate.
A<Integer> a = new A<Integer>(new A.Delegate<Integer>() {
#Override
public Integer myMethod() {
return null;
}
});
Your inner interface can have its own generic bounds. Try declaring and using it as Delegate<T> and it should work fine.
Related
I am trying to create a generic method, but it requires data from the sub class. Is there a way to do this, or is there a better implementation?
Example of my situation:
public class Super {
public static Object method() {
return doSomethingWith(specificToSubClassValue);
}
}
public class Sub1 extends Super {
public static String specificToSubClassValue = "123";
}
public class Sub2 extends Super {
public static String specificToSubClassValue = "456";
}
I obviously cannot do this. What is a better approach?
One alternative I can think of is to override the #method method in each sub class, but it will be the same code in each instance so I wanted to throw it in the parent class (and it won't be truly overridden since it is static), but I am not sure how to approach it since it is dependent on the sub class value.
Static methods in Java can't be overwritten, and can't access children-specific information: they know nothing about inheritance.
What you need here is instance method, which you can overwrite. An you also may use generics.
public class Super<T> {
public Object method() {
final T specificToSubClassValue = getSpecificToSubClassValue();
if (specificToSubClassValue != null) {
return specificToSubClassValue.hashCode();
} else {
return null;
}
}
protected T getSpecificToSubClassValue() {
return null;
}
}
class Sub1 extends Super<String> {
#Override
protected String getSpecificToSubClassValue() {
return "abc";
}
}
class Sub2 extends Super<Integer> {
#Override
protected Integer getSpecificToSubClassValue() {
return 123;
}
}
Declare an abstract method in Super, which will be used to return the value from the implementing classes. Note that this can only be achieved with non-static methods. As per #JB Nizet's comment, static methods cannot be overriden by subclasses. I've removed the static modifier from the code below to shown you how it would work.
public class Super {
public static Object method() {
return doSomethingWith(specificToSubClassValue);
}
protected abstract Object getValue ();
}
public class Sub1 extends Super {
public static String specificToSubClassValue = "123";
#Override
protected Object getValue () {
return specificToSubClassValue;
}
}
public class Sub2 extends Super {
public static String specificToSubClassValue = "456";
#Override
protected Object getValue () {
return specificToSubClassValue;
}
}
Well, te whole idea of inheritance is that the superclass should not be able to do that much without the sub-class. Otherwise the whole inheritance would be pointless exercise and spaghetti code. But you are tackling the problem the wrong way. Make sub-class "spill" the data you need (through getter) and use the generic method from superclass on the data in the sub-class.
Also the overriding of super class methods is highly overrated. You should strive for your super method to be as flexible and re-usable as possible, and even then strive rather for overloading, instead of overriding.
You could have:
public class Super {
public Object method(String specificValue) {
return doSomethingWith(specificToSubClassValue);
}
And then have your sub do this
public class Sub1 extends Super {
public static String specificToSubClassValue = "123";
Object method(specificToSubClassValue);
}
This way you accomplish exactly what you want, operate on the class specific value using the super method.
I went to an interview. Interviewer asked me if one can instantiate an interface and abstract class? As per my knowledge I said "No". But he said "Yes, we can with the help of an anonymous class".
Can you please explain to me how?
This was a trick questions.
No you can not instantiate an interface or abstract class.
But you can instantiate an anonymous class that implements/extends the interface or abstract class without defining a class object. But it is just a shortcut to defining a fully named class.
So I would say technically your answer was correct.
I don't know what is "instantiation of interface and abstract class".
I think it's an inaccurate, improper expression of something,
we can only guess at the intended meaning.
You cannot create an instance of an interface or an abstract class in Java.
But you can create anonymous classes that implement an interface or an abstract class.
These won't be instances of the interface or the abstract class.
They will be instance of the anonymous class.
Here's an example iterator from the Iterator interface that gives you an infinity of "not really":
new Iterator<String>() {
#Override
public boolean hasNext() {
return true;
}
#Override
public String next() {
return "not really";
}
};
Or a funky AbstractList that contains 5 "not really":
List<String> list = new AbstractList<String>() {
#Override
public int size() {
return 5;
}
#Override
public String get(int index) {
return "yes";
}
};
Assume you have an abstract class: MyAbstractClass with abstract void method myAbstractMethod. Then you can make an "instance" of this class via this code:
MyAbstractClass myAbstractClassInstance = new MyAbstractClass() {
public void myAbstractMethod() {
// add abstract method implementation here
}
};
myAbstractClassInstance extends your MyAbstractClass in this case. When you instantiate this class you have to implement all abstract methods as you can see from the code above.
The same way works for interfaces, assume you have an interface MyInterface with a void method myInterfaceMethod inside, then you can create an "instance" (implementation of this instance) via this code:
MyInterface myInterfaceImpl = new MyInterface() {
public void myInterfaceMethod() {
// add method implementation here
}
}
myInterfaceImpl is an implemetation of MyInterface in this case. When you create an object using interface, you have to implement interface methods as it is shown above.
Interface :
interface Interface1 {
public void m1();
}
When you right
new Interface1() {
public void m1() {
}
}
Its not actually creating the instance of Interface. Its creating an instance of its subtype which doesnt have any name/reference. Hence we cannot create an instance of interface or abstract class
You cannot create instances of abstract classes or interfaces using the new operator. For example,
new AbstractSet(); // That's wrong.
You can, however, use them to declare reference variables. For example, You can do this:
AbstractSet set;
You can instantiate anonymous as well as declared implementing classes or subclass.
For example, Set extends AbstractSet, so you can instantiate Set.
Yes, we can create by having defining the abstract methods or the interface methods on the fly during instantiation. That's like a Named anonymous class.
//interface
Runnable r = new Runnable(){
public void run() {
System.out.println("Here we go");
}
};
//Abstract class
abstract class MyAbstract {
abstract void run();
}
MyAbstract ab = new MyAbstract(){
#Override
void run() {
System.out.println("Here we go");
}};
I wanted to implement a method in a abstract class that is called by the inherited classes and uses their values.
For instance:
abstract class MyClass{
String value = "myClass";
void foo(){System.out.println(this.value);}
}
public class childClass{
String value="childClass";
void foo(){super.foo();}
}
public static void main(String[] args){
new childClass.foo();
}
This will output "myClass" but what I really want is to output "childClass". This is so I can implement a "general" method in a class that when extended by other classes it will use the values from those classes.
I could pass the values as function arguments but I wanted to know if it would be possible to implement the "architecture" I've described.
A super method called by the inherited class which uses the values from the caller not itself, this without passing the values by arguments.
You could do something like this:
abstract class MyClass {
protected String myValue() {
return "MyClass";
}
final void foo() {
System.out.println(myValue());
}
}
public class ChildClass extends MyClass {
#Override
protected String myValue() {
return "ChildClass";
}
}
and so on
This is a place where composition is better than inheritance
public class Doer{
private Doee doee;
public Doer(Doee doee){
this.doee = doee;
}
public void foo(){
System.out.println(doee.value);
}
}
public abstract class Doee{
public String value="myClass"
}
public ChildDoee extends Doee{
public String= "childClass"
}
...
//Excerpt from factory
new Doer(new ChildDoee);
I believe you are asking whether this is possible:
public class MyClass {
void foo() {
if (this instanceof childClass) // do stuff for childClass
else if (this intanceof anotherChildClass) // do stuff for that one
}
}
So the answer is "yes, it's doable", but very much advised against as it a) tries to reimplement polymorphism instead of using it and b) violates the separation between abstract and concrete classes.
You simply want value in MyClass to be different for an instance of childClass.
To do this, change the value in the childClass constructor:
public class childClass {
public childClass() {
value = "childClass";
}
}
Edited:
If you can't override/replace the constructor(s), add an instance block (which gets executed after the constructor, even an undeclared "default" constructor):
public class childClass {
{
value = "childClass";
}
}
I have the following situation:
public abstract class A {
private Object superMember;
public A() {
superMember = initializeSuperMember();
// some additional checks and stuff based on the initialization of superMember (***)
}
protected abstract Object initializeSuperMember();
}
class B extends A {
private Object subMember;
public B(Object subMember) {
super();
subMember = subMember;
}
protected Object initializeSuperMember() {
// doesn't matter what method is called on subMember, just that there is an access on it
return subMember.get(); // => NPE
}
}
The problem is that I get a NPE on a new object B creation.
I know I can avoid this by calling an initializeSuperMember() after I assign the subMember content in the subclass constructor but it would mean I have to do this for each of the subclasses(marked * in the code).
And since I have to call super() as the first thing in the subclass constructor I can't initialize subMember before the call to super().
Anyone care to tell me if there's a better way to do this or if I am trying to do something alltogether wrong?
Two problems:
First, you should never call an overrideable member function from a constructor, for just the reason you discovered. See this thread for a nice discussion of the issue, including alternative approaches.
Second, in the constructor for B, you need:
this.subMember = subMember;
The constructor parameter name masks the field name, so you need this. to refer to the field.
Follow the chain of invocation:
You invoke the B() constructor.
It invokes the A() constructor.
The A() constructor invokes the overridden abstract methot
The method B#initializeSuperMember() references subMember, which has not yet been initialized. NPE.
It is never valid to do what you have done.
Also, it is not clear what you are trying to accomplish. You should ask a separate question explaining what your goal is.
Hum, this code does not look good and in all likelyhood this is a sign of a bad situation. But there are some tricks that can help you do what you want, using a factory method like this:
public static abstract class A {
public abstract Object createObject();
}
public static abstract class B extends A {
private Object member;
public B(Object member) {
super();
this.member = member;
}
}
public static B createB(final Object member) {
return new B(member) {
#Override
public Object createObject() {
return member.getClass();
}
};
}
The problem is when you call super(), the subMember is not initialized yet. You need to pass subMemeber as a parameter.
public abstract class A {
public A (Object subMember) {
// initialize here
}
}
class B extends A {
public B (Object subMember) {
super(subMember);
// do your other things
}
}
Since you don't want to have subMember in the abstract class, another approach is to override the getter.
public abstract class A {
public abstract Object getSuperMember();
protected void checkSuperMember() {
// check if the supberMember is fine
}
}
public class B extends A {
private Object subMember;
public B(Object subMember) {
super();
this.subMember = subMember;
checkSuperMemeber();
}
#Override
public Object getSuperMember() {
return subMember.get();
}
}
I hope this can remove your duplicate code as well.
In Java, I'd like to have something as:
class Clazz<T> {
static void doIt(T object) {
// ...
}
}
But I get
Cannot make a static reference to the non-static type T
I don't understand generics beyond the basic uses and thus can't make much sense of that. It doesn't help that I wasn't able to find much info on the internet about the subject.
Could someone clarify if such use is possible, by a similar manner? Also, why was my original attempt unsuccessful?
You can't use a class's generic type parameters in static methods or static fields. The class's type parameters are only in scope for instance methods and instance fields. For static fields and static methods, they are shared among all instances of the class, even instances of different type parameters, so obviously they cannot depend on a particular type parameter.
It doesn't seem like your problem should require using the class's type parameter. If you describe what you are trying to do in more detail, maybe we can help you find a better way to do it.
Java doesn't know what T is until you instantiate a type.
Maybe you can execute static methods by calling Clazz<T>.doit(something) but it sounds like you can't.
The other way to handle things is to put the type parameter in the method itself:
static <U> void doIt(U object)
which doesn't get you the right restriction on U, but it's better than nothing....
I ran into this same problem. I found my answer by downloading the source code for Collections.sort in the java framework. The answer I used was to put the <T> generic in the method, not in the class definition.
So this worked:
public class QuickSortArray {
public static <T extends Comparable> void quickSort(T[] array, int bottom, int top){
//do it
}
}
Of course, after reading the answers above I realized that this would be an acceptable alternative without using a generic class:
public static void quickSort(Comparable[] array, int bottom, int top){
//do it
}
I think this syntax has not been mentionned yet (in the case you want a method without arguments) :
class Clazz {
static <T> T doIt() {
// shake that booty
}
}
And the call :
String str = Clazz.<String>doIt();
Hope this help someone.
It is possible to do what you want by using the syntax for generic methods when declaring your doIt() method (notice the addition of <T> between static and void in the method signature of doIt()):
class Clazz<T> {
static <T> void doIt(T object) {
// shake that booty
}
}
I got Eclipse editor to accept the above code without the Cannot make a static reference to the non-static type T error and then expanded it to the following working program (complete with somewhat age-appropriate cultural reference):
public class Clazz<T> {
static <T> void doIt(T object) {
System.out.println("shake that booty '" + object.getClass().toString()
+ "' !!!");
}
private static class KC {
}
private static class SunshineBand {
}
public static void main(String args[]) {
KC kc = new KC();
SunshineBand sunshineBand = new SunshineBand();
Clazz.doIt(kc);
Clazz.doIt(sunshineBand);
}
}
Which prints these lines to the console when I run it:
shake that booty 'class com.eclipseoptions.datamanager.Clazz$KC' !!!
shake that booty 'class com.eclipseoptions.datamanager.Clazz$SunshineBand' !!!
It is correctly mentioned in the error: you cannot make a static reference to non-static type T. The reason is the type parameter T can be replaced by any of the type argument e.g. Clazz<String> or Clazz<integer> etc. But static fields/methods are shared by all non-static objects of the class.
The following excerpt is taken from the doc:
A class's static field is a class-level variable shared by all
non-static objects of the class. Hence, static fields of type
parameters are not allowed. Consider the following class:
public class MobileDevice<T> {
private static T os;
// ...
}
If static fields of type parameters were allowed, then the following code would be confused:
MobileDevice<Smartphone> phone = new MobileDevice<>();
MobileDevice<Pager> pager = new MobileDevice<>();
MobileDevice<TabletPC> pc = new MobileDevice<>();
Because the static field os is shared by phone, pager, and pc, what is the actual type of os? It cannot be Smartphone, Pager, and
TabletPC at the same time. You cannot, therefore, create static fields
of type parameters.
As rightly pointed out by chris in his answer you need to use type parameter with the method and not with the class in this case. You can write it like:
static <E> void doIt(E object)
Something like the following would get you closer
class Clazz
{
public static <U extends Clazz> void doIt(U thing)
{
}
}
EDIT: Updated example with more detail
public abstract class Thingo
{
public static <U extends Thingo> void doIt(U p_thingo)
{
p_thingo.thing();
}
protected abstract void thing();
}
class SubThingoOne extends Thingo
{
#Override
protected void thing()
{
System.out.println("SubThingoOne");
}
}
class SubThingoTwo extends Thingo
{
#Override
protected void thing()
{
System.out.println("SuThingoTwo");
}
}
public class ThingoTest
{
#Test
public void test()
{
Thingo t1 = new SubThingoOne();
Thingo t2 = new SubThingoTwo();
Thingo.doIt(t1);
Thingo.doIt(t2);
// compile error --> Thingo.doIt(new Object());
}
}
Since static variables are shared by all instances of the class. For example if you are having following code
class Class<T> {
static void doIt(T object) {
// using T here
}
}
T is available only after an instance is created. But static methods can be used even before instances are available. So, Generic type parameters cannot be referenced inside static methods and variables
When you specify a generic type for your class, JVM know about it only having an instance of your class, not definition. Each definition has only parametrized type.
Generics work like templates in C++, so you should first instantiate your class, then use the function with the type being specified.
Also to put it in simple terms, it happens because of the "Erasure" property of the generics.Which means that although we define ArrayList<Integer> and ArrayList<String> , at the compile time it stays as two different concrete types but at the runtime the JVM erases generic types and creates only one ArrayList class instead of two classes. So when we define a static type method or anything for a generic, it is shared by all instances of that generic, in my example it is shared by both ArrayList<Integer> and ArrayList<String> .That's why you get the error.A Generic Type Parameter of a Class Is Not Allowed in a Static Context!
#BD at Rivenhill: Since this old question has gotten renewed attention last year, let us go on a bit, just for the sake of discussion.
The body of your doIt method does not do anything T-specific at all. Here it is:
public class Clazz<T> {
static <T> void doIt(T object) {
System.out.println("shake that booty '" + object.getClass().toString()
+ "' !!!");
}
// ...
}
So you can entirely drop all type variables and just code
public class Clazz {
static void doIt(Object object) {
System.out.println("shake that booty '" + object.getClass().toString()
+ "' !!!");
}
// ...
}
Ok. But let's get back closer to the original problem. The first type variable on the class declaration is redundant. Only the second one on the method is needed. Here we go again, but it is not the final answer, yet:
public class Clazz {
static <T extends Saying> void doIt(T object) {
System.out.println("shake that booty "+ object.say());
}
public static void main(String args[]) {
Clazz.doIt(new KC());
Clazz.doIt(new SunshineBand());
}
}
// Output:
// KC
// Sunshine
interface Saying {
public String say();
}
class KC implements Saying {
public String say() {
return "KC";
}
}
class SunshineBand implements Saying {
public String say() {
return "Sunshine";
}
}
However, it's all too much fuss about nothing, since the following version works just the same way. All it needs is the interface type on the method parameter. No type variables in sight anywhere. Was that really the original problem?
public class Clazz {
static void doIt(Saying object) {
System.out.println("shake that booty "+ object.say());
}
public static void main(String args[]) {
Clazz.doIt(new KC());
Clazz.doIt(new SunshineBand());
}
}
interface Saying {
public String say();
}
class KC implements Saying {
public String say() {
return "KC";
}
}
class SunshineBand implements Saying {
public String say() {
return "Sunshine";
}
}
T is not in the scope of the static methods and so you can't use T in the static method. You would need to define a different type parameter for the static method. I would write it like this:
class Clazz<T> {
static <U> void doIt(U object) {
// ...
}
}
For example:
public class Tuple<T> {
private T[] elements;
public static <E> Tuple<E> of(E ...args){
if (args.length == 0)
return new Tuple<E>();
return new Tuple<E>(args);
}
//other methods
}