Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

Java generics: Illegal forward reference

Given a generic interface

interface Foo<A, B> { }

I want to write an implementation that requires A to be a subclass of B. So I want to do

class Bar<A, B super A> implements Foo<A, B> { }
// --> Syntax error

or

class Bar<A extends B, B> implements Foo<A, B> { }
// --> illegal forward reference

But the only solution that seems to work is this:

class Bar<B, A extends B> implements Foo<A, B> { }

which is kind of ugly, because it reverses the order of the generic parameters.
Are there any solutions or workarounds to this problem?

like image 634
Cephalopod Avatar asked Jan 07 '11 17:01

Cephalopod


3 Answers

Since this isn't possible in Java, try to think of Bar<B, A extends B> differently.

When you declare a variable for Bar, you're specifying the parent class first and then the child class. That's how Bar works. Don't think of it as being backwards - think of it as being forwards. The parent should naturally be specified before the child. This additional relationship you added is what drives the parameter order, not the underlying interface.

like image 166
Erick Robertson Avatar answered Nov 04 '22 01:11

Erick Robertson


After seeing this question, I spent a little bit trying some different techniques that I thought might work. For example, building a generic interface ISuper<B,A extends B> and then having Bar<A,B> implements ISuper<B,A> (and a similar technique with a sub-class and extends rather than implements) but this just results in a type error, Bar.java:1: type parameter A is not within its bound. Likewise, I tried creating a method private <A extends B> Bar<A,B> foo() { return this; }; and calling it from the constructor, but this just results in the fun type error message Bar.java:2: incompatible types found : Bar<A,B> required: Bar<A,B>

So, I think that, unfortunately, the answer is no. Obviously, it's not the answer you were hoping for, but it seems that the correct answer is that this just isn't possible.

like image 31
Keith Irwin Avatar answered Nov 04 '22 01:11

Keith Irwin


It was pointed out already that there is neither a solution nor a nice workaround. Here is what I finally did. It only works for my special case, but you can take it as an inspiration if you run into similar problems. (It also explains why I ran into this problem)

First of all, there is this class (showing only the relevant interface):

class Pipe<Input, Output> {

    boolean hasNext();

    Input getNext();

    void setNext(Output o);

}

The Foo interface is actually

interface Processor<Input, Output> {

    process(Pipe<Input, Output> p);

}

and the class Bar should work like this

class JustCopyIt<Input, Output> implements Processor<Input, Output> {

    process(Pipe<Input, Output> p) {
       while (p.hasNext()) p.setNext(p.getNext());
    }

}

The easiest way would be to cast the values like this: p.setNext((Output) p.getNext()). But this is bad as it would allow to create an instance of JustCopyIt<Integer, String>. Calling this object would mysteriously fail at some point, but not at the point where the actual error is made.

Doing class JustCopyIt<Type> implements Processor<Type, Type> would also not work here, because then I am not able to process a Pipe<String, Object>.

So what I finally did was to change the interface to this:

interface Processor<Input, Output> {

    process(Pipe<? extends Input, ? super Output> p);

}

This way, a JustCopyIt<List> is able to process a Pipe<ArrayList, Collection>.

While this technically seems to be the only valid solution, it is still bad because it 1) only works for this special case, 2) required me to change the interface (which is not always possible) and 3) made the code of the other processors ugly.

Edit:
Reading Keiths answer again inspired me for another solution:

public abstract class Bar<A, B> implements Foo<A, B> {

    public static <B, A extends B> Bar<A, B> newInstance() {
        return new BarImpl<B, A>();
    }

    private static class BarImpl<B, A extends B> extends Bar<A, B> {
        // code goes here
    }

}

// clean code without visible reversed parameters
Bar<Integer, Object> bar1 = Bar.newInstance();
Bar<Object, Integer> bar2 = Bar.newInstance(); // <- compile error
like image 1
Cephalopod Avatar answered Nov 03 '22 23:11

Cephalopod