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copied!<p>From Java Language Specification:</p> <p><a href="http://docs.oracle.com/javase/specs/jls/se7/html/jls-8.html#jls-8.4.8.3" rel="nofollow noreferrer">jls-8.4.8.3</a></p> <blockquote> <p>The access modifier (§6.6) of an overriding or hiding method must provide at least as much access as the overridden or hidden method, as follows:</p> <ul> <li>If the overridden or hidden method is public, then the overriding or hiding method must be public; otherwise, a compile-time error occurs.</li> <li>...</li> </ul> </blockquote> <p>Notice that you are trying to override <code>public</code> method <code>getNum()</code> inherited from class <code>B</code> (and also from interface <code>A</code>) with new one that has <code>protected</code> access modifier. It means that you are trying to reduce visibility of this method which according to specification is incorrect.<br> To be able to override this method you need to use <code>public</code> access modifier with your new version of that method.</p> <hr> <p>Why you cant reduce visibility? Take a look at below code which uses your classes but is placed inside some other package and ask yourself "how should this code behave?".</p> <pre><code>package my.pckage; import your.pckage.A; import your.pckage.C; public class Test{ public static void main (String[] args){ C C = new C(); c.getNum();// ERROR: Test class doesn't have access to `c`s protected method. // Why should it have, Test doesn't extend C. A a = (A)c;// Lets try using other reference a.getNum();// Should `a` have access to method that is protected in `C`? // If yes, then what is the point of declaring this method // protected if all I would have to do to get access to it is // casting instance of C to A interface? } } </code></pre>

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