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Can't reference generated-sources class outside of default package

I'm working with Google's Protocol Buffer (in combination with the Protocol Buffers maven plugin) which compiles a .proto file into a class. I can use the generated class in the default package perfectly, but not outside of it. I don't really know how to explain it any better so I'm going to show you some pictures.
I've tried subclassing the Hrp class but that doesn't work (the generated class is final). It is also not an option to move the class every time I re-generate the Hrp class.
I'm not sure if this is relevant, but the generated class is public final. It contains an empty, private constructor.
I have also tried setting the generated sources package prefix for the generated sources folder but that also does not work.
Any help would be greatly appreciated.

Try adding a package id to your Protocol Buffers definition. See Protocol Buffers Package
i.e.
syntax = "proto3";
package MyPackage;
option optimize_for = SPEED;
message Product {
repeated ASale sale = 1;
}
Then when you Generate the Java~Protocol~Buffers code (using protoc), it will be in package MyPackage and you will be able to import it into your java code in the normal way.
In java, you can not import anything from the Default package; which I believe is your problem. See How to access java-classes in the default-package?

Class reference is taken from different jar

There are some classes used in a java program referred from abc package from xyz.jar. The package is imported in the java file.
Also the same class is in other lmn.jar.
So if I delete the jar file from the project, i should be getting the error.
But the class compiles and takes the class from the other lmn.jar.
Eg.
weblogic.jdbc.oci.Blob is a class in weblogic.jar
But if i delete weblogic.jar, it takes it from java.sql.Blob.
I don't want this to happen, the program should display error.

In such a case you can use fully qualified name of class, which will include package name. For example, instead of:
Blob blob = new Blob();
You can write:
weblogic.jdbc.oci.Blob blob = new weblogic.jdbc.oci.Blob();

When importing a class you are also defining the package. So when you import "weblogic.jdbc.oci.Blob" and remove this class from classpath it will not automatically import it from a different package unless you change the import statement as well.
Some IDEs might automatically try to resolve classes and add the missing import statements. Maybe check that.

For one of the case try to use the entire path, ie
For example if you need Blob from weblogic.jar
then try to call weblogic.jdbc.oci.Blob bl = ...

Access classes from package

I'm developing an android test app and i'm going to access all internal class of android.view package. android.view is a package that is present in jar file. I tried by loading package name but it doesn't display the classes if any one tried
this already, please help.
Here's what I tried so far:
public static void main() throws ClassNotFoundException{
Class o =Class.forName("android.view");
Class[] C=o.getDeclaredClasses();
for(int i=0;i<C.length;i++) {
Classname = C[i].getName();
ClassesDisplayActivity.your_array_list3.add(Classname);
Log.i("Ramu","classname "+ C[i].getName());
}
}
}

It is not possible to determine at runtime all of the classes that are in a package using a standard class loader.
You might have some luck with this library though:
https://code.google.com/p/reflections/

Package is not a class. You cannot call Class.forName() for package and access classes that belong to class using getDelcaredClasses().
I do not know what do you really need, so I'd recommend you to explain this in separate question. probably you will receive better solutions.
However if you really need this you have to do the following:
Get your classpath by calling System.getProperty(java.class.path)
split this property to its elements by colon
iterate over the list and read each resource. If resource is jar you can use ZipInputStream, if it is a directory use File class.
filter list of resources you got at #3.
Fortunately you can use 3rd party library named Reflections that helps you to do all this without writing code.

Java type cannot be resolved MIME type class

I know this is a common question asked but I've been searching and I've included the class into eclipse through the buildpath. I start to write the import statement and it autocompletes options for me so I know it's finding the class.
My problem is how come it's giving this error when I'm reading the docs and it says the constructor method is MimeUtil2() ?
http://www.jarvana.com/jarvana/view/eu/medsea/mimeutil/mime-util/2.1/mime-util-2.1-javadoc.jar!/eu/medsea/mimeutil/MimeUtil2.html#MimeUtil2()
package com.jab.app;
import java.io.File;
import eu.medsea.mimeutil.*;
public class CheckFileType {
private void GetMimeType(File filename){
MimeUtil2 test = new MimeUtil2(); //Produces the error saying java type cannot be resolved
}

I think you need to import
import eu.medsea.mimeutil.*;
According to the documentation, the type is eu.medsea.mimeutil.MimeUtil2

I ended up finding out that I was using the test-source.jar not the main jar file itself. The sourceforge page made the default as the source file instead of the main jar file.
It was buried inside of the files page.

How to get current class name including package name in Java?

I'm working on a project and one requirement is if the 2nd argument for the main method starts with “/” (for linux) it should consider it as an absolute path (not a problem), but if it doesn't start with “/”, it should get the current working path of the class and append to it the given argument.
I can get the class name in several ways: System.getProperty("java.class.path"), new File(".") and getCanonicalPath(), and so on...
The problem is, this only gives me the directory in which the packages are stored - i.e. if I have a class stored in ".../project/this/is/package/name", it would only give me "/project/" and ignores the package name where the actual .class files lives.
Any suggestions?
EDIT:
Here's the explanation, taken from the exercise description
sourcedir can be either absolute (starting with “/”) or relative to where we run the program from
sourcedir is a given argument for the main method. how can I find that path?

Use this.getClass().getCanonicalName() to get the full class name.
Note that a package / class name ("a.b.C") is different from the path of the .class files (a/b/C.class), and that using the package name / class name to derive a path is typically bad practice. Sets of class files / packages can be in multiple different class paths, which can be directories or jar files.

There is a class, Class, that can do this:
Class c = Class.forName("MyClass"); // if you want to specify a class
Class c = this.getClass(); // if you want to use the current class
System.out.println("Package: "+c.getPackage()+"\nClass: "+c.getSimpleName()+"\nFull Identifier: "+c.getName());
If c represented the class MyClass in the package mypackage, the above code would print:
Package: mypackage
Class: MyClass
Full Identifier: mypackage.MyClass
You can take this information and modify it for whatever you need, or go check the API for more information.

The fully-qualified name is opbtained as follows:
String fqn = YourClass.class.getName();
But you need to read a classpath resource. So use
InputStream in = YourClass.getResourceAsStream("resource.txt");

use this.getClass().getName() to get packageName.className and use this.getClass().getSimpleName() to get only class name