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).
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 ofBasedList<T>. The interface could just be: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:That seems like kind of a waste though, and I don’t really think the original declaration looks that bad.