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    <p>This is one situation where the java syntax hides what's really going on, and C# makes it a bit clearer.</p> <p>In C# your B would look like</p> <pre><code>class B : A { public B() : base(method()) { } private static String method() { return "method invoker"; } } </code></pre> <p>Although the java syntax places <code>super(method)</code> within the constructor it's not really called there: All the parent initialization is run <strong>before</strong> your subclass constructor. The C# code shows this a little more clearly; by placing <code>super(method())</code> at the first line of the java constructor you're simply telling java to use the parameterized constructor of the super class rather than the parameterless version; this way you can pass variables to the parent constructor and they'll be used in the initialization of the parent level fields <strong>before</strong> your child's constructor code runs. </p> <p>The reason that <code>super(method())</code> is valid (as the first line in a java constructor) is because <code>method()</code> is being loaded with the static elements--before the non-static ones, including the constructors--which allows it to be called not only before <code>B()</code>, but before <code>A(String)</code> as well. By saying</p> <pre><code>public B() { String s = method(); super(s); } </code></pre> <p>you're telling the java compiler to initialize the super object with the default constructor (because the call to <code>super()</code> isn't the first line) and you're ready to initialize the subclass, but the compiler then becomes confused when it sees that you're trying to initialize with <code>super(String)</code> after <code>super()</code> has already run.</p>
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