title: Is there a minimalist way to get a default ResourceBundle when you do not need localization?
body: <p>I want to inherit to extend <code>java.util.logging.Logger</code>.  The only visible constructor is </p>
  
  <pre><code>protected Logger(String name, String resourceBundleName)
  </code></pre>
  
  <p>This is despite the fact that when you want to get an instance, they offer two ways:</p>
  
  <pre><code>static Logger   getLogger(String name)
  static Logger   getLogger(String name, String resourceBundleName)  
  </code></pre>
  
  <p>Is there a way to obtain some sort of default <code>ResourceBundle</code> or default name of a <code>ResourceBundle</code> so that I can use the constructor without creating an internationalization properties file and associated object?  </p>
  
  <p>Of course I could simply produce a wrapper class and use <code>getLogger(String name)</code> to avoid the need for the <code>ResourceBundle</code> name.  Unfortunately that means I would have to implement a wrapper method for every <code>Logger</code> method I want to use.  (Too much typing).</p>
  
  <p><strong>Edit:</strong> The following attempt to get the default localization data from Logger does not work.  </p>
  
  <pre><code>import java.util.logging.Logger;
  
  class JLogger extends Logger
  {
      static Logger dummyLogger = Logger.getLogger("com.dummy.utilities");
  
      private JLogger(String name)
      {
          super(name, JLogger.dummyLogger.getResourceBundleName());
      }
  
      static JLogger getJLogger(String name)
      {
          return new JLogger(name);
      }
  
      public void severe() // a zero arguments version of severe
      {
          severe("");
      }
  }
  </code></pre>
  
  <p>The <code>getResourceBundleName()</code> just returns a null.  The <code>super(name, resourceName)</code> constructor builds something which does not do any logging if the second argument is <code>null</code>.</p>