Java Generics - inserting inner type parameter

Question

I am new to java. I am just trying to pass Comparable<String> into a method parameter of generic type <E extends Comparable<E>> . I believe the meaning of <E extends Comparable<E>> is any object that extends Comparable. Please let me know how to pass Comparable<String> or any object that extends Comparable<String> and has an other object in it.

Compiler is giving me error The inferred type Compare<String> is not a valid substitute for the bounded parameter <E extends Comparable<E>>

Code:

public class Compare<T> implements Comparable<T>{

    public int compareTo(T o) {

        return 0; // Not worried about logic

    }

}

class CompareTest{

    public <E extends Comparable<E>>void testGeneric(E e){

        System.out.println("Executed");
    }

    public static void main(String args[]){
        Compare<String> compare = new Compare<String>();

        CompareTest test = new CompareTest();
        test.testGeneric(compare);
        //The inferred type Compare<String> is not a valid substitute for the bounded
        //parameter <E extends Comparable<E>>
    }
}

JB Nizet · Accepted Answer · 2013-11-25 22:33:04Z

3

E extends Comparable<E> means: a type E that is able to compare to other objects of the same type E.

But your Compare type doesn't qualify. It can't compare with another Compare. A Compare<T> can only compare itself to a T, and not to a Compare<T>, since it's declared as

public class Compare<T> implements Comparable<T>

It's quite hard to understand what you want to achieve with this Compare type.

answered Nov 25, 2013 at 22:33

JB Nizet

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1 Comment

Mr A Over a year ago

I was just practicing the notion < E extends Comparable< E > > . I though If I pass an object like new Comparable<String>() it would accept as It Comparable< Strinng > extends Comparable. Not sure where am I missing.

Sotirios Delimanolis · Accepted Answer · 2013-11-25 22:35:59Z

2

Your method

public <E extends Comparable<E>> void testGeneric(E e){

expects a type E that is a sub type of Comparable<E>. But you are passing it a Compare<String> which is not a sub type of <Comparable<Compare<String>>, but a sub type of Comparable<String>.

You'll have to clarify what you are trying to do if you need more help.

answered Nov 25, 2013 at 22:35

Sotirios Delimanolis

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Comments

rgettman · Accepted Answer · 2013-11-25 22:34:58Z

1

The error in my IDE says:

Inferred type 'Compare<java.lang.String>' for type parameter 'E' is not within its bound; 
should implement 'java.lang.Comparable<Compare<java.lang.String>>'

It seems that E is inferred as Compare<String> instead of String. To get E to be String, try this:

public <E extends Comparable<E>> void testGeneric(Comparable<E> e){

answered Nov 25, 2013 at 22:34

rgettman

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1 Comment

Mr A Over a year ago

it worked like a heaven. I was expecting < E extends Comparable< E > > to expect Compare< String > and match Compare < String > . I have no clue how come Compare< String > = new Compare < String >(); is being treated Comparable<Compare<String>> instead of Compare< String >

Bohemian · Accepted Answer · 2013-11-27 13:31:07Z

1

To define a class as being comparable, the generic parameter to Comparable must be the class itself:

public class MyComparable implements Comparable<MyComparable> {
    public int compareTo(MyComparable o) {
        return 0;
    }
}

Applying that to your class, you get the following code (which compiles):

public static class Compare<T> implements Comparable<Compare<T>> {
    public int compareTo(Compare<T> o) {
        return 0; // Not worried about logic
    }
}

class CompareTest {
    public <E extends Comparable<E>> void testGeneric(E e) {
        System.out.println("Executed");
    }
}

public static void main(String[] args) {
    Compare<String> compare = new Compare<String>();
    CompareTest test = new CompareTest();
    test.testGeneric(compare);
}

edited Nov 27, 2013 at 13:31

answered Nov 25, 2013 at 23:01

Bohemian♦

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2 Comments

Mr A Over a year ago

you are absolutely right. It was the problem of being able to create an object of type you created in the seconds code block. i.e. MyComparable < MyComparable < MyComparable so on and on ... > > . Can't we create an object of class MyComparable in the second block. If so an example of that would be highly appreciated.

Bohemian Over a year ago

I have simplified my answer for clarity and provided you with a repaired (ie compilable) version of your code.

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