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copied!<blockquote> <p>If all objects directly or indirectly inherit from Object how is it possible to elicit the compiler rejection spoken of? </p> </blockquote> <p>I think the basis is of your misunderstanding is that you are <em>thinking</em> that constructors are inherited. In fact, constructors are NOT inherited in Java. So consider the following example:</p> <pre><code>public class A { public A(int i) { super(); ... } } public class B extends A { public B() { super(); ... } } </code></pre> <p>The class <code>A</code>:</p> <ul> <li>does not inherit any constructors from <code>Object</code>, </li> <li>does not explicitly declare a no-args constructor (i.e. <code>public A() {...}</code>), and</li> <li>does not have a default constructor (since it <em>does</em> declare another constructor).</li> </ul> <p>Hence, it has one and only one constructor: <code>public A(int)</code>.</p> <p>The call to <code>super()</code> in the <code>B</code> class tries to use a non-existent no-args constructor in <code>A</code> and gives a compilation error. To fix this, you either need to change the <code>B</code> constructor to use the <code>A(int)</code> constructor, or declare an explicit no-args constructor in <code>A</code>.</p> <p>(Incidentally, it is not necessary for a constructor to <em>explicitly</em> call a superclass constructor ... as I've done. But a lot of people think it is good style to include an explicit call. If you leave it out, the Java compiler inserts an implicit call to the superclasses no-args constructor ... and that results in a compilation error if the no-args constructor does not exist or is not visible to the subclass.)</p> <blockquote> <p>Does it have to do with the constructor being private?</p> </blockquote> <p>Not directly. However, declaring a constructor private will prevent that constructor being called from a child class.</p>

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