RxJava Single/Observable return implementation of interface

Question

I have faced with interesting problem

I have the following method

@Override
public Single<IResults> getResults() {
    return Single.just(new ConcreteResults());
}

IResults is an interface and ConcreteResults is the implementation.

However I am getting an error

Error:(149, 27) error: incompatible types: Single<ConcreteResults> cannot be converted to Single<IResults>

How to solve this issue ? I need to have an interface to abstract return type.

However, this dirty trick seems to work :)

Single.just(new ConcreteResults())
                .map(new Function<ConcreteResults, IResults>() {
                    @Override
                    public IResults apply(@NonNull ConcreteResults results) throws Exception {
                        return results;
                    }
                });

Answer 1

Use:

@Override public Single<? extends IResults> getResults() { 
    return Single.just(new ConcreteResults()); 
}

Please remember that hieritance with genetics is a bit different.

Although IResults is a sub type of ConcreteResults, Single of ConcreteResults is NOT a sub type of Single of IResults . You need to use a wildcard to create a relationship between then

You can search for more information about it here: https://docs.oracle.com/javase/tutorial/java/generics/subtyping.html

Answer 2

You need to create a variable before returning the value. The compiler will figure out, that IResult is the over-type of ConcretResult. If you return it immediately the compiler does not get it.

@Test
void name() {
    Single<IResult> result = getResult();
}

private Single<IResult> getResult() {
    Single<IResult> just = Single.just(new Result());

    return just;
}

interface IResult {

}

class Result implements IResult {

}

For 1_7 Settings you must use:

compileOptions {
    sourceCompatibility JavaVersion.VERSION_1_7
    targetCompatibility JavaVersion.VERSION_1_7
}

You have to give the compiler pointers, because in 1_7 the compiler does not inference the type from definition.

private Single<IResult> getResult() {
    Single<IResult> just = Single.<IResult>just(new Result());

    return just;
}