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Asked: 2026-06-07T23:54:33+00:00

I have public class First<T> {} public class Second<T extends SomeConcreteClass> extends First<T> {}

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I have

public class First<T> {}

public class Second<T extends SomeConcreteClass> extends First<T> {}

public class Third<T> extends Second<T> {} //Compile-time error

I get the compile-time error

Type argument T is not with bounds of type-variable T.

When I contruct a Third, I want to be able to give the generic parameter as SomeConcreteClass (or derived class thereof), and for a run-time error to be thrown if I’ve offered up a type that is not part of SomeConcreteClass‘s inheritance hierarchy.

I would think that the specification in Second‘s declaration would simply propagate downward, i.e. it should be implicit in the declaration (and any instantiations) of Third.

What’s with the error?

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1 Answer

  1. Editorial Team

    All you need is

    public class Third<T extends SomeConcreteClass> extends Second<T>

    You just need to respecify the bound. It doesn’t propagate like you think it does.

    (I’m not positive of the reason for this, but I have some guesses — what if it was Third<T> extends Second<Foo<T>>? The appropriate bound on T isn’t obvious, if there even is one. So instead, it just doesn’t propagate automatically; you have to specify it.)

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