Consider that I have following interface:
public interface MyInterface<T extends Number>
In another class I want to declare a method like this:
public <T extends MyInterface<?>> void abc(T a);
Is this a correct way? Or maybe should I write:
public <T extends MyInterface<T>> void abc(T a);
What is the difference between these two declarations?
Some interfaces are intended to take the implementing class as the generic type.
For instance, the Comparable interface is usually implemented as:
class MyClass implements Comparable<MyClass>
because then the method
int compareTo(T o);
declared in Comparable can be used to compare an instance of MyClass to another instance of the same class.
Other interfaces have a generic parameter of a different type. For instance, List<T>, where T indicates the type of object contained within the list.
If you declare a method with:
public <T extends MyInterface<T>> void abc(T a);
then you're saying that the class T implements the interface MyInterface in the way that classes implement the Comparable interface - with themselves as the generic type. That might or might not be appropriate depending on what MyInterface actually is.
If you declare a method with:
public <T extends MyInterface<?>> void abc(T a):
then you not placing any constraint on the generic type in T's implementation of MyInterface.
The other option is
public void abc(MyInterface<?> a);
which is the simplest way to write a method that will accept any implementation of MyInterface.
Related
The full context being:
public class RClass<T extends Comparable<T>>
Would I be right in saying that the statement in the title means that the arguments plugged into the method must either be an object of a class which implements Comparable OR one of its derived classes?
Thanks.
This means that the type parameter must support comparison with other instances of its own type, via the Comparable interface.
An example of such a class is provided in the Oracle tutorial Object Ordering. Note the similar pattern to T extends Comparable<T> in the excerpt below:
public class Name implements Comparable<Name> {
...
public int compareTo(Name n) { ... }
}
Java- The meaning of <T extends Comparable<T>>?
a) Comparable <T>
is a generic interface (remember it's an "interface" i.e not a "class")
b) extends means inheritance from a class or an interface.
From above-said point#a, it is an interface..(Remember it is an inheritance from an "interface" i.e not from a "class")
c)From above-said both points #a & #b,
here "one interface" extends "another interface".
There should be an interface defined for this class..
just an example here is
interface MinMax<T extends Comparable<T>> {
T min();
T max();
}
d) now your class i.e public class RClass {} SHOULD
1# EITHER "implement" this "generic interface" Comparable<T> ..!!!
ex: public class RClass<T> implements Comparable<T>
2# OR create an interface and extend to this "generic interface" Comparable<T>
ex:
interface MinMax<T extends Comparable<T>> {
T min();
T max();
}
class RClass<T extends Comparable<T>> implements MinMax<T> {
.....
.....
}
Here,
Pay special attention to the way that the type parameter T is declared by RClass and
then passed to MinMax. Because MinMax requires a type that implements Comparable,
the implementing class (RClass in this case) must specify the same bound. Furthermore,
once this bound has been established, there is no need to specify it again in the implements clause.
Somewhere in that class, the programmer needs to write something like
if(t.compareTo(othert) < 0) {
...
}
For that to work, the type T must have a compareTo-method which compares it to another object of type T. Extending Comparable guarantees the existence of such a method, among other things.
It means that you can only create an instance of RClass with a type which quite literally extends Comparable<T>. Thus,
RClass<Integer> a;
is acceptable, since Integer extends Comparable<Integer>, while
RClass<Object> b;
is not, since Object is not a class which extends comparable at all.
Yes, and bear in mind that objects of classes derived from Comparable ARE Comparable objects. Inheritance is a is-a relationship.
Simply put, the generic type T must be comparable in order to compareTo. otherwise you cannot do T.compareTo.
In Item 28 Effective java, it suggests: "always use Comparable<? super T> in preference to Comparable<T>. <T extends Comparable<? super T>>"
==>guhanvj, <T extends Comparable<T>> means that <T> is having upper bound of Comparable<T> objects. So <T> can have types of Byte, Character, Double, Float, Long, Short, String, and Integer classes which all implements Comparable<T> interface for natural ordering.
Is it possible to use wildcards as an attribute itself and not as a type parameter?
public void doSomething(<? extends Context> myObject){
}
I want to make sure that my object extends Context and also implements a certain interface.
Using wildcards as a return type is valid but using it as an attribute seems to not work
public <? extends MyClass> validMethod(){
}
Yes it's possible, you have to define it before the return type.
public <T extends MyClass> void myMethod(T param);
You don't explicitly state (for your case) which is the interface and which is the class, but if you wanted to have a generic type bound to both extending a class and an interface, this would be the general form for it.
public <T extends Number & Comparable<T>> T findNumber(List<T> elements) { }
Number is an abstract class; Comparable is an interface. I believe what you have to specify them in this order - concrete before interface.
Given an interface like this
public interface MyInterface1<T extends MyAbstractClass> {
...
}
I want to make another interface MyInterface2 taking a MyInterface1 as a generic type and in MyInterface2 I want to reference the actual types of the actual MyInterface1
public interface MyInterface2<INTERFACE extends MyInterface1<MYCLASS>> {
MYCLASS returnInstanceOfMyClass();
}
So I want to say that method "returnInstanceOfMyClass" returns the actual T of the actual MyInterface1 given to MyInterface2.
The thing is that I am not allowed to write the following
public interface MyInterface2<INTERFACE extends MyInterface1<MYCLASS>> {
I am allowed to write
public interface MyInterface2<INTERFACE extends MyInterface1<?>> {
but then I am not able to reference the actual type of T in MyInterface1 in the method signature in MyInterface2 - because I have given it no name to be used when referencing.
I want to be able to do the following in a type-safe way
class MyClass extends MyAbstractClass {
...
}
MyClass c = new MyInterface2<MyInterface1<MyClass>>.returnInstanceOfMyClass();
No casting to MyClass should be necessary, because it can see that the actual class of MyInterface1 given to MyInterface2 is MyClass, and that is what is returned from returnInstanceOfMyClass.
How to do that?
You need a second generic parameter:
public interface MyInterface2<U estends MyAbstractClass, T extends MyInterface1<U>> {
U returnInstanceOfMyClass();
}
I want to write a method that gets 2 parameters:
an interface.
a class that implements that interface.
I want it to be generic (compile time type safe).
is there a way?
if not, whats the alternative?
is there an option to get generic param that is an interface? ho do I declare it?
?
some thing like this ??
public <I, K extends I> void method(I i, K k){
}
In the above method first parameter would be an interface, and second parameter would be any class that implements that interface.
Interface1 i1;
method(i1, class1); //class1 implements Interface1
In generics, interface implementation and class extending is represented using extends keyword. there is no implements keyword in the world of generics.
public <T extends Interface> T myMethod(Interface I, Class<T> myClass) () {...}
if your method returns T
The full context being:
public class RClass<T extends Comparable<T>>
Would I be right in saying that the statement in the title means that the arguments plugged into the method must either be an object of a class which implements Comparable OR one of its derived classes?
Thanks.
This means that the type parameter must support comparison with other instances of its own type, via the Comparable interface.
An example of such a class is provided in the Oracle tutorial Object Ordering. Note the similar pattern to T extends Comparable<T> in the excerpt below:
public class Name implements Comparable<Name> {
...
public int compareTo(Name n) { ... }
}
Java- The meaning of <T extends Comparable<T>>?
a) Comparable <T>
is a generic interface (remember it's an "interface" i.e not a "class")
b) extends means inheritance from a class or an interface.
From above-said point#a, it is an interface..(Remember it is an inheritance from an "interface" i.e not from a "class")
c)From above-said both points #a & #b,
here "one interface" extends "another interface".
There should be an interface defined for this class..
just an example here is
interface MinMax<T extends Comparable<T>> {
T min();
T max();
}
d) now your class i.e public class RClass {} SHOULD
1# EITHER "implement" this "generic interface" Comparable<T> ..!!!
ex: public class RClass<T> implements Comparable<T>
2# OR create an interface and extend to this "generic interface" Comparable<T>
ex:
interface MinMax<T extends Comparable<T>> {
T min();
T max();
}
class RClass<T extends Comparable<T>> implements MinMax<T> {
.....
.....
}
Here,
Pay special attention to the way that the type parameter T is declared by RClass and
then passed to MinMax. Because MinMax requires a type that implements Comparable,
the implementing class (RClass in this case) must specify the same bound. Furthermore,
once this bound has been established, there is no need to specify it again in the implements clause.
Somewhere in that class, the programmer needs to write something like
if(t.compareTo(othert) < 0) {
...
}
For that to work, the type T must have a compareTo-method which compares it to another object of type T. Extending Comparable guarantees the existence of such a method, among other things.
It means that you can only create an instance of RClass with a type which quite literally extends Comparable<T>. Thus,
RClass<Integer> a;
is acceptable, since Integer extends Comparable<Integer>, while
RClass<Object> b;
is not, since Object is not a class which extends comparable at all.
Yes, and bear in mind that objects of classes derived from Comparable ARE Comparable objects. Inheritance is a is-a relationship.
Simply put, the generic type T must be comparable in order to compareTo. otherwise you cannot do T.compareTo.
In Item 28 Effective java, it suggests: "always use Comparable<? super T> in preference to Comparable<T>. <T extends Comparable<? super T>>"
==>guhanvj, <T extends Comparable<T>> means that <T> is having upper bound of Comparable<T> objects. So <T> can have types of Byte, Character, Double, Float, Long, Short, String, and Integer classes which all implements Comparable<T> interface for natural ordering.