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copied!<p>The "pattern" (idiom) of passing an instance of <code>Class<T></code> (typically to the constructor) is using <a href="http://docs.oracle.com/javase/tutorial/extra/generics/literals.html" rel="noreferrer">Class Literals as Runtime-Type Tokens</a>, and is used to keep a runtime reference to the generic type, which is otherwise erased.</p> <p>The solution is firstly to change the token class bound to:</p> <pre><code>Class<? extends T> </code></pre> <p>and then to put a similar requirement on your generic subclass as you did with your super class; have the concrete class pass a type token, but you can type it properly as a parameter:</p> <p>These classes compile without casts or warnings:</p> <pre><code>public abstract class Abstract<T extends Abstract<T>> { private final Class<? extends T> subClass; protected Abstract(Class<? extends T> subClass) { this.subClass = subClass; } } public class NonGeneric extends Abstract<NonGeneric> { public NonGeneric() { super(NonGeneric.class); } } public class Generic<T> extends Abstract<Generic<T>> { public Generic(Class<? extends Generic<T>> clazz) { super(clazz); } } </code></pre> <p>And finally at the concrete class, if you declare the usage as its own class, it doesn't require a cast anywhere:</p> <pre><code>public class IntegerGeneric extends Generic<Integer> { public IntegerGeneric() { super(IntegerGeneric.class); } } </code></pre> <hr> <p>I haven't figured out how to create an instance of <code>Generic</code> (anonymous or not) without a cast:</p> <pre><code>// can someone fill in the parameters without a cast? new Generic<Integer>(???); // typed direct instance new Generic<Integer>(???) { }; // anonymous </code></pre> <p>I don't think it's possible, but I welcome being shown otherwise.</p>

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