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copied!<p>The concrete class compiles fine because B extends A, and since Java 5 a return type on a subclass can be a subclass of the return type on the superclass.</p> <p>As for the generic, I have hit the same wall, and you have two options that I know of.</p> <p>One is to paramaterize A:</p> <pre><code> public interface A<T extends A> extends Iterable<T> </code></pre> <p>then:</p> <pre><code>public class B implements A<B> </code></pre> <p>However, that has a disadvantage that you will need to give a parameter of A, instead of having it fixed, even if you want A:</p> <pre><code>private class NoOneSees implements A<A> </code></pre> <p>As to the question of if you actually want <code>Iterator<B></code>, in the case of an Iterable, most likely that is preferable, and considering that these are interfaces, the need to redeclare the parameter is probably reasonable. It gets a little more complicated if you start inheriting concrete classes, and for things that have meaning other than an Iterable, where you may want covariance. For example:</p> <pre><code>public interface Blah<T> { void blah(T param); } public class Super implements Blah<Super> { public void blah(Super param) {} } public class Sub extends Super { public void blah(Super param) {} //Here you have to go with Super because Super is not paramaterized //to allow a Sub here and still be overriding the method. } </code></pre> <p>Similarly for covariance, you can't declare a variable of type <code>Iterator<A></code> and then assign an <code>Iterator<B></code> in it, even if B extends A.</p> <p>The other option is live with the fact that further implementation/subclasses will still reference A.</p>

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