Java generics and varargs

Question

I\'d like to implement a function with both generics and varargs.

public class Question {
    public static  void doNastyThingsToClasses(Class

        

Answer 1

The second argument Class<? extends A>... that must extend the class that the first argument is (ex. argument one is a Question so the second argument be something that extends Question.

The Breakdown:
NastyThingsToClasses(Object.class, Question.class, SomeQuestion.class); // OK
Everything extends Object so the second argument is correct.

NastyThingsToClasses(Question.class, SomeQuestion.class); // OK
SomeQuestion extends Question so thats fair game.

NastyThingsToClasses(Question.class, Object.class, SomeQuestion.class);
Object does not extend Question hence error.

hopefully that cleared things up.

-Brett

Answer 2

My solution to this problem was to

1. create a class Nastier
2. remove ... from doNastyThingsToClasses
3. make doNastyThingsToClasses none static method
4. make the name short, like do
5. return this

6. move repetitive args to class properties

   class Nastier {
     private final Class<A> parent;
   
     public Nastier(Class<A> parent) {
        this.parent = parent;
     }
   
     public <A, C extends A> Nastier do(Class<? extends A> clazz) {
        System.out.println(clazz);
        return this;
     }
   }
   
   public static void main(String[] args) {   
     Nastier nastier = new Nastier(Object.class);
     nastier.do(Question.class).do(SomeQuestion.class).do(NotQuestion.class);
   }
   

I believe the code looks clean and I am happy.... :)

Answer 3

As almost always, Angelika Langer's Java generics FAQ explains it in great detail. (Scroll to "Why does the compiler sometimes issue an unchecked warning when I invoke a "varargs" method?" - the ID doesn't work well.)

Basically, you end up losing information in a worse way than normal. Yet another little pain point in Java generics :(

Answer 4

As an aside, the warning can now be suppressed using Java 7's new @SafeVarargs annotation.

@SafeVarargs
public static <A> void func( Class<A> parent, Class<? extends A>... classes ) {
    // Do func...
}

Answer 5

OK, so finally I end up throwing the varargs away:

public class Question {

    public static <A, C extends A> void doNastyThingsToClasses(Class<A> parent, List<Class<? extends A>> classes) {
        /******/
        for(Class<? extends A> clazz : classes) {
            System.out.println(clazz);
        }
    }

    public static class NotQuestion {
    }
    public static class SomeQuestion extends Question {
    }

    public static void main(String[] args) {

        ArrayList<Class<? extends Object>> classes = new ArrayList<Class<? extends Object>>();
        classes.add(Question.class);
        classes.add(SomeQuestion.class);
        classes.add(NotQuestion.class);
        doNastyThingsToClasses(Object.class, classes);

        ArrayList<Class<? extends Question>> clazzes = new ArrayList<Class<? extends Question>>();
        clazzes.add(Question.class);
        clazzes.add(SomeQuestion.class);
        clazzes.add(NotQuestion.class); // yes, this will _not_ compile
        doNastyThingsToClasses(Question.class, clazzes);

    }

}

The only flaw is the long code for populating the collection used to carry function's arguments.

Answer 6

Jon Skeet's answer is (of course) correct; I'll expand on it a little by pointing out that you CAN get rid of this warning, with a big 'if'. You can avoid this warning IF you're willing to commit to having your project build using Java 7.

Bob Lee wrote a proposal to let this warning be suppressed at method-declaration site, rather than usage site, as part of Project Coin.

This proposal was accepted for JDK7 (though the syntax changed slightly, to @SuppressWarnings("varargs")); you can, if you're curious, look at the commit that added this support to the JDK.

Not necessarily helpful for you, but I thought I'd make this a separate answer so it lives on for future readers, who may be lucky enough to live in a post-Java-7 world.