Re: Locating all classes implementing a given interface?

Bas <basschulte@gmail.com> writes:


  I have written a library routine for this, based on code by
  Ralf Ullrich.

  For example, to find all classes implementing ?java.util.Map?,
  one sets a filter accepting only classes which
  ?java.util.Map? is assignable from:

public boolean accepts( final java.lang.Class class_ )
{ return java.util.Map.class.isAssignableFrom( class_ ); }

  One also needs to provide a starting class to find the jar, which
  is done as follows.

public java.lang.String entryPath(){ return "java.lang.Object"; }

  The jar containing this type will be chosen for exploration.

  In the following example, the client does not need to specify
  this ?entryPath?, as "java.lang.Object" is the default
  obtained by extending
  ?de.dclj.ram.java.lang.reflect.Finder.DefaultSpecification?.

  The example client is:

public class Main
{ /* based on an idea and on code by Ralf Ullrich from 2006 */

  public static void main( final java.lang.String[] args )
  { new de.dclj.ram.java.lang.reflect.Finder
    ( new de.dclj.ram.java.lang.reflect.Finder.DefaultSpecification()
      {
        public boolean isClassFinder(){ return true; }

        public boolean accepts( final java.lang.Class class_ )
        { return java.util.Map.class.isAssignableFrom( class_ ); }

        }).inspectJar(); }}

class java.lang.ProcessEnvironment
class java.rmi.server.RemoteObjectInvocationHandler$MethodToHash_Maps$1
class java.security.AuthProvider
(...)
class java.util.Properties
class java.util.Hashtable
interface java.util.Map

  The library ?ram.jar? is an early publication in alpha state,
  it is experimental, changing, and mostly undocumented. See:

http://www.purl.org/stefan_ram/pub/ram-jar

  Known issues:

    - I have ideas to improve the interface, which are not yet
      implemented.

    - To compile the library from the sources, one might need
      to take care of some minor bugs: If I remember
      correctly, some ?private?s need to be replaced by
      ?public?.