Difference between generics > and T extends Comparable

Question

Can someone explain the difference between <T extends Number & Comparable<T>> and T extends Comparable<? extends Number>?

These look similar to me and both of them compiles fine for sub classes type. The invalid type args shows below error

Type parameter is not within bound parameter;should implement 'java.lang.Number'

and

Type parameter is not within bound parameter; should extend 'java.lang.Comparable>'

respectively.

Show your code. What error are you talking about? When do you get it? — Erwin Bolwidt, May 26 '16 at 15:31

score 2 · Accepted Answer · edited May 23 '17 at 11:46

2

You won't be able to use:

Comparable<? extends Number>

because the only methods defined by Comparable are consumers (in the sense of PECS): it needs to accept instances of type ? extends Number into its compareTo method - and there is no type which satisfies that bound safely.

Integer is a Comparable<? extends Number>, but so is Double. Thus, you can't safely call instance1.compareTo(instance2) because it would fail if these instances are concretely Integer and Double respectively, since Integer.compareTo can only accept Integer parameters.

As such, the compiler prevents you from calling this method in the first place.

edited May 23 '17 at 11:46

Community

1
1

answered May 26 '16 at 15:39

Andy Turner

137,514
11
162
243

when you say there are no types what do you mean ? Is Integer class not a type that will satisfy the bound ? – s7vr May 26 '16 at 18:05
@ShivV: He means that a value of type `T` won't be able to `.compareTo()` anything. – newacct Jun 03 '16 at 02:43
@ShivV Edited my answer to add more explanation. – Andy Turner Jun 03 '16 at 06:58
Thank you for taking time to answer. Accepted. – s7vr Jun 03 '16 at 07:04

score 2 · Answer 2 · answered May 26 '16 at 16:01

Option 1:

public class A <T extends Number & Comparable<T>>{}

Your Generic Parameter should extend Number and implements Comparable, which means class A is a Number and Comparable.

Option 2:

public class B <T extends Comparable<? extends Number>>{}

T is Comparable on Numbers(can compare Number only) but doesn't have to be a Number, unlike option 1

I will explain by example:

A

public class A <T extends Number & Comparable<T>>{}

B

public class B <T extends Comparable<? extends Number>>{}

IntegerWrapper(Option 2)

    public class IntegerWrapper implements Comparable<Integer>  {

      Integer number;

      public IntegerWrapper(int number) {
        this.number = number;
      }

      @Override
      public int compareTo(Integer o) {
        return number.compareTo(o);
      }
    }

GenericsTest public class GenericsTest {

  public static void main(String args[]){
    A myA = new A<Integer>();

    B myB = new B<IntegerWrapper>();
  }
}

I think option 1 is what you are looking for, because i can't think of many useful scenarios for Option 2(Maybe there is...)

"T is Comparable on Numbers(can compare Number only) but doesn't have to be a Number" Can you elaborate a little more ? — s7vr, May 26 '16 at 18:07
T is Comparable for Number types, but does not have to extend Number class like option 1 -> T extends Number, its only Comparable. and was defined as T extends Number. from my example IntegerWrapper does not extend Number class, but implements Comparable — Elia Rohana, May 26 '16 at 22:18

Difference between generics > and T extends Comparable

2 Answers2