How to set constraints on generic types in Java?

Question

I have a generic class:

public class ListObject<T>
{
    // fields
    protected T _Value = null;
      // ..
}

Now I want to do something like the following:

ListObject<MyClass> foo = new ListObject<MyClass>();
ListObject<MyClass> foo2 = new ListObject<MyClass>();
foo.compareTo(foo2);

Question:

How can I define the compareTo() method with resprect to the generic T?

I guess I have to somehow implement a constraint on the generic T, to tell that T implements a specific interface (maybe Comparable, if that one exists).

Can anyone provide me with a small code sample?

Community · Accepted Answer · 2017-05-23 12:17:58Z

62

Read also the discussion here: Generics and sorting in Java

Short answer, the best you can get is:

class ListObject<T extends Comparable<? super T>> {
    ...
}

But there is also reason to just use:

class ListObject<T extends Comparable> {
    ...
}

edited May 23, 2017 at 12:17

CommunityBot

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answered Jan 17, 2010 at 16:43

nanda

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2 Comments

citronas Over a year ago

I'll stick to ListObject<T extends Comparable> for the moment.

sateesh Over a year ago

Also check the related thread: stackoverflow.com/questions/2064169/…, which discusses the need for wildcard of form:<? super T>

JaredPar · Accepted Answer · 2010-01-17 16:34:05Z

5

This depends on exactly what you want the compareTo method to do. Simply defining the compareTo method to take other ListObject<T> values is done by the following

public class ListObject<T> {
  public int compareTo(ListObject<T> other) {
    ...
  }
}

However if you want to actually call methods on that parameter you'll need to add some constraints to give more information about the T value like so

class ListObject<T extends Comparable<T>> {
  ...
}

answered Jan 17, 2010 at 16:34

JaredPar

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Comments

John Feminella · Accepted Answer · 2010-01-17 17:04:54Z

4

Try public class ListObject<T extends U>. Only Ts which implement U (or derive from U) will be allowable substitutions.

edited Jan 17, 2010 at 17:04

answered Jan 17, 2010 at 16:31

John Feminella

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2 Comments

citronas Over a year ago

Gives me a compiler error. Even with such a simple TestClass: public class ConstraintTest<T implements Comparable> { } Syntax error on token "implements",, expected What am I missing?

Esko Over a year ago

@citronas: It's actually extends even for interfaces. Do note that if you use a base class with the extends attribute, you can actually add the base class directly to the generified object anymore, that's the biggest (and worst) side effect of the generics extension.

Steve Emmerson · Accepted Answer · 2010-01-17 16:31:40Z

-3

public class ListObject<T implements Comparable> {...}

answered Jan 17, 2010 at 16:31

Steve Emmerson

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1 Comment

helpermethod Over a year ago

-1 Voted down because Comparable without the type parameter is a raw type which u shouldn't use with Java 1.5+

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