now i want to implement this interface with the class.
so how should i do it?
public class TMark<E> implements ITMark{}
is this the way but throwing errors
I am getting the following:
ITMark is a raw type. References to generate type ITMark<E> should be parametrized
I am implementing this code in Eclipse IDE
ITMark is a raw type because it has no declared generic parameters.
If you declared TMark as TMark<E extends Comparable<E>> implements ITMark<E>, it would not longer be a raw type because you declared its generic parameter.
You left out the generic parameter, that is, the part that goes in the angle brackets. You need something like:
public class TMark <E extends Comparable <E> implements ITMark<E>
{
...
}
For a particular generic type you put a suitable 'Comparable' type inside the angle brackets, something like:
public class IntegerTMark extends TMark <Integer>
{
...
}
For a good introduction to generics, read the Java tutorials, the free chapter from Joshua Bloch's Effective Java at http://java.sun.com/docs/books/effective/generics.pdf and the many articles about generics at https://www.ibm.com/developerworks/java/.
Do this:
public class TMark<SomeComparableClass> implements ITMark<SomeComparableClass> {
// implement the methods of ITMark for type SomeComparableClass
}
You must specify which Comparable class you are implementing for this class. FYI, most common java types (eg Integer, String, Date, etc) are Comparable.
Related
In java, can I do something -more or less- like this? and how?
public class SomeGenericClass<T> extends T{
}
I think this is the relevant quote from the JLS (§8.1.4):
The ClassType [provided in the extends clause] must name an accessible (§6.6) class type, or a compile-time error occurs.
(The bit about accessibility isn't relevant).
A class type is not the same thing as a type variable (§4.3) (which T is), so attempting to do this would be a compile-time error.
you can't do public class SomeGenericClass<T> extends T for multiple reasons:
Type erasure will erase T to Object., and it would be pointless to say public class SomeGenericClass<T> extends Object ,as Object is already the base class of all types.
Imagine if it were allowed and i did something like public class SomeGenericClass<T extends Comparable> extends T , type erasure will turn this into public class SomeGenericClass<Comparable> extends Comparable but interface is implemented not extended .
public class SomeGenericClass<T> extends T{
}
You must understand Generics better. SomeGenericClass is a template (C++ terminology). As it is, it means nothing to compiler. You cannot write
SomeGenericClass<T> obj = new SomeGenericClass<T>() // this is invalid
Instead you need to supply the type attribute. This information complements the above answer provided by #Andy Turner
I have
1) a basic interface,
2) a few classes that implement this interface,
3) and a generic class that I want to accept, as a parameter, any of the implementing classes
I have tried the following:
public class Foo extends Bar<? extends SomeInterface> {
public Foo(List<? extends SomeInterface> someInterfaceList) {
super(someInterfaceList);
}
...
}
I receive the error No Wildcard Expected. Elsewhere in my code I have statements such as List<? extends SomeInterface> and I receive no errors, so why am I running into problems here? How can I fix this problem and still get the desired results?
I have tried search 'No Wildcard Expected' and 'wildcard in class declaration' to no avail. Thanks in advance!
It sounds like you want to declare a generic type argument that you will reference elsewhere. Wildcards only make sense when the type is used only once, and when declaring a generic type parameter for a class this doesn't make any sense.
Try this instead:
public class Foo<T extends SomeInterface> extends Bar<T> {
public Foo(List<T> someInterfaceList) {
super(someInterfaceList);
}
...
}
As your code was written, there was no way for the user of your class to specify the generic type argument for Bar<>, since Foo wasn't itself a generic type.
Further, if this were possible, it would have been possible for the generic argument to Bar<> to be different than the generic argument to List<> -- as long as both types implemented SomeInterface there would not be a compile-time issue with these definitions, but there could have been a much more confusing error message later when you incorrectly assumed that both types must be the same.
So, declare the generic type once as a generic argument to the Foo class, and then use that type (T in my example) elsewhere to refer to that type instead of accepting some new generic type argument that may not refer to the same type.
I'm not exactly sure what you're looking for, so it might help if you could provide a little more detail. Perhaps you could be a little more specific about how you're planning to instantiate and use these objects?
Anyways, I think you might be looking for something like this:
import java.util.List;
public class Foo<T extends SomeInterface> {
public Foo(List<T> someInterfaceList) {
for (T item : someInterfaceList) {
// do something with each item
}
}
}
class Bar<T> {}
interface SomeInterface<T> {
T x(T y);
}
Or, alternatively, you could just use the following for the constructor:
public Foo(List someInterfaceList) {
but you wouldn't have an easy way of getting the type T of the items in the list.
As per my Previous Question, I am reading the article from Angelika Dissecting Enum. Except for the points that a type can only be instantiated for its subtypes and the subtypes do inherit some common methods, I am not able to understand the article.
What is the meaning of abstract Enum class declared in this way? How is it helpful?
The document in the last part has described three aspects, can someone explain them in easier terms to me?
I do see in the code sketch the Enum class is declaring the compareTo method. When Enum is implicitly implementing Comparable interface. Why do it needs to define its own compareTo method?
Seems like it is a concept of recursive generics. What does recursive generics exactly mean? After doing a bit of R&D and understanding my last question answer, I understand that it forces the class to be parameterized on itself.
Still, a detailed explanation would be useful.
I think the main benefit of declaring generic types as Type<E extends Type<E>> is that such generic classes will make subclasses to inherit methods which return or accept arguments with subtype's type. Such methods in java.lang.Enum are:
public final int compareTo( E o) { ... }
public final Class< E > getDeclaringClass() { ... }
So, if we declare the enum Color, that implicitly means:
public class Color extends Enum<Color>
so in this instantiation of Enum the type paramater E is assigned the type argument Color, so the above methods will look like these:
public final int compareTo(Color o) { ... }
public final Class<Color> getDeclaringClass() { ... }
When saying something like Enum<Color extends Enum<Color>>, that sounds like you are declaring a generic type parameter Color that makes sure that it extends Enum with a type parameter matching Color.
But that isn't where generic type parameters for a class are declared. You must declare them next to the class name; you can only use them later in the extends clause. E.g.
// Use "extends" here ... not here.
public class MyClass<E extends MyClass<E>> extends MySuperClass<E>
In this example, you are declaring the class Color to be the value of the generic type parameter that is already defined on Enum.
I have a class defined like so:
public class AddRecordsToolbar<D extends IDataSource<T>, T extends Serializable>
extends AbstractToolbar<D, T>
which my IDE IntelliJ IDEA declares as legal. It looks and feels wrong to me.
I want to declare it like this:
public class AddRecordsToolbar<D extends IDataSource<T extends Serializable>, T>
extends AbstractToolbar<D, T>
however that syntax is illegal thanks to something to do with Javas type erasure.
D extends IDataSource<T> is required by the superclass.
My Class is using Serializable to do a deep copy. Hence the T extends Serializable.
So now on to the Question: If I specify T extends Serializable as the second type parameter for my class will it still enforce T extends Serializable for D as well?
Answering to your question, yes its do.
The order of generic parameter it only in your mind.
If we would rephrase that implementation to:
public class AddRecordsToolbar<T extends Serializable, D extends IDataSource<T>> extends AbstractToolbar<D, T>
you will be not so surprised, and looks that the way it should be.
I will try to find the explanation for this in Java Language Specification (when it will work) but for now that the way it is.
I believe that this is addressed in section 8.1.2 of the java language spec:
The scope of a class' type parameter is the entire declaration of the
class including the type parameter section itself. Therefore, type
parameters can appear as parts of their own bounds, or as bounds of
other type parameters declared in the same section.
Yes. The "T" in the second parameter is the same T as in the first. Nothing world work, otherwise.
This means that
class IFoo extends IDataSource<String>{};
AddRecordsToolbar<IFoo, Integer> x;
Is illegal. Integer and String are both serialisable, but the declaration of AddRecordsToolbar says that the data source has to be a source of the data type in your second parameter. And that second parameter says that it has to be serializable.
I'd like to define a generic type, whose actual type parameter can only be
One of the numeric primitive wrapper classes (Long, Integer, Float, Double)
String
I can meet the first requirement with a definition like this
public final class MyClass<T extends Number> {
// Implementation omitted
}
But I can't figure out how to meet both of them. I suspect this is not actually possible, because AFAIK there's no way to specify "or" semantics when defining a formal type parameter, though you can specify "and" semantics using a definition such as
public final class MyClass<T extends Runnable & Serializable > {
// Implementation omitted
}
Cheers,
Don
Java generics does not support union types (this parameter can be A OR B).
On a related note that may be of interest to some, it does support multiple bounds, if you want to enforce multiple restrictions. Here's an example from the JDK mentioned in the Java generics tutorial:
public static <T extends Object & Comparable<? super T>> T max(Collection<? extends T> coll)
You could use factory methods for all supported types and make the constructor private/protected. You have to fix the generic type in the constructor anyway so that it makes sense, so you could probably code it like this:
public final class MyClass<T> {
public static MyClass<Integer> newInstance(int i) {
return new MyClass<Integer>(i);
}
public static MyClass<String> newInstance(String s) {
return new MyClass<String>(s);
}
//More factory methods...
protected MyClass(T obj) {
//...
}
}
Or if you do not want the constructor parameter, something like this:
public final class MyClass {
public static MyClass newIntegerInstance() {
return new MyClass();
}
//...
}
As erickson stated, the common implementation can rely only on Object anyway, so the only restriction is, that you can create other implementations for other types besides the primitive and String.
While generics won't work here, a base type with derived types for Number and String will. Since a generic type would have erased to Object anyway, any functionality you would have put there can go in an abstract base class. You're likely to need only a type-specific accessor on the subclass to get the value.
Also, be careful with the Number class. It's not limited to boxing the primitive types, as anyone can extend it—e.g., BigInteger.
Interesting question, it boggled me a bit. However apparently this is impossible. I tried several different hacks, none really work.
Maybe you could do what follows:
Make MyClass<T> a package-default class, invisible to other components, or at least with a package-default ctors only, so that it cannot extended or instantiated outside the package.
Create two public classes in the package of MyClass<T>:
MyNumericClass<T extends Number> extends MyClass<T>
MyStringClass extends MyClass<String>
This way all subclasses of MyClass will be limited to those parametrized with a Number subclass or String.