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Home/ Questions/Q 868403
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T10:08:04+00:00 2026-05-15T10:08:04+00:00

Given the interface: public interface BasedOnOther<T, U extends BasedList<T>> { public T getOther(); public

  • 0

Given the interface:

public interface BasedOnOther<T, U extends BasedList<T>> {

    public T getOther();

    public void staticStatisfied(final U list);

}

The BasedOnOther<T, U extends BasedList<T>> looks very ugly in my use-cases. It is because the T type parameter is already defined in the BasedList<T> part, so the “uglyness” comes from that T needs to be typed twice.

Problem: is it possible to let the Java compiler infer the generic T type from BasedList<T> in a generic class/interface definition?

Ultimately, I’d like to use the interface like:

class X implements BasedOnOther<Y> {
    public SomeType getOther() { ... }
    public void staticStatisfied(final Y list) { ... }
} // Does not compile, due to invalid parameter count.

Where Y extends BasedList<SomeType>.

Instead:

class X implements BasedOnOther<SomeType, Y> {
    public SomeType getOther() { ... }
    public void staticStatisfied(final Y list) { ... }
}

Where Y extends BasedList<SomeType>.

Update: ColinD suggested

public interface BasedOnOther<T> {
    public T getOther();
    public void staticSatisfied(BasedList<T> list);
}

It is impossible to create an implementation such as:

public class X implements BasedOnOther<SomeType> {
    public SomeType getOther() { ... }
    public void staticStatisfied(MemoryModel list);
} // Does not compile, as it does not implement the interface.

Where MemoryModel extends BasedList<SomeType>, which is needed (as it provides other methods).

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-15T10:08:05+00:00Added an answer on May 15, 2026 at 10:08 am

    It looks as if you don’t actually need the type parameter U extends BasedList<T>, if you don’t actually need to do anything in the class that requires some specific subclass/implementation of BasedList<T>. The interface could just be:

    public interface BasedOnOther<T> {
      public T getOther();
      public void staticSatisfied(BasedList<T> list);
    }
    

    Edit: Based on your update, I don’t think there’s any way you can do this. I think you’ll have to either just go with your original declaration or make some intermediate type that specifies T, like:

    public interface BasedOnSomeType<U extends BasedList<SomeType>>
             extends BasedOnOther<SomeType, U>
    {
    }
    
    public class X implements BasedOnSomeType<MemoryModel> { ... }
    

    That seems like kind of a waste though, and I don’t really think the original declaration looks that bad.

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