Solved

how to fix "The exception was: java.lang.ClassCastException" error?

Posted on 2003-11-18
17
494 Views
Last Modified: 2012-05-04
Hi experts,

I have some codes work on one latest verions of jdk and broken on older version.

The error is "The exception was: java.lang.ClassCastException", which I just can't see why it won't work. Is there a way to find out exactly what class the object is? So, I can compare with what it is cast to? thanks.
0
Comment
Question by:panpioneer
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 6
  • 4
  • 3
  • +1
17 Comments
 
LVL 35

Accepted Solution

by:
TimYates earned 25 total points
ID: 9771718
> Is there a way to find out exactly what class the object is?

if( o instanceof String )
{
    String s = (String)o ;
}

0
 
LVL 35

Expert Comment

by:TimYates
ID: 9771723
you can also do:

System.out.println( o.getClass.getName() ) ;

to see what it is...
0
 
LVL 35

Expert Comment

by:TimYates
ID: 9771725
Sorry...

System.out.println( o.getClass().getName() ) ;
0
Optimize your web performance

What's in the eBook?
- Full list of reasons for poor performance
- Ultimate measures to speed things up
- Primary web monitoring types
- KPIs you should be monitoring in order to increase your ROI

 

Author Comment

by:panpioneer
ID: 9771962

THanks. It is very strange that both object are of the same type after the testing, however, when I assign one object to another, like the following, then I got the "java.lang.ClassCastException" error. why?

*****************************************
Research ac = new Research();
ac = (Research)request.getAttribute("ab");      

System.out.println((request.getAttribute("ab")).getClass().getName() ) ;
System.out.println(ac.getClass().getName() ) ;       
0
 
LVL 86

Expert Comment

by:CEHJ
ID: 9772152
I'm not sure why if this:

>>System.out.println((request.getAttribute("ab")).getClass().getName() ) ;

is printing Research (in whatever package)

But there is a redundancy in your code:

>>Research ac = new Research();

creates a new instance. You should be doing:

Research ac = null;
try {
      Object o = request.getAttribute("ab");
      ac = (Research)o;
}
catch(ClassCastException e) {
      System.err.println("Object was of type " + o.getClass().getName());
}

0
 

Author Comment

by:panpioneer
ID: 9772303
thanks. But, if I copy your code, I will get another error message: o is not defined....
0
 
LVL 86

Expert Comment

by:CEHJ
ID: 9772315
Sorry!

Research ac = null;
Object o = null;
try {
     o = request.getAttribute("ab");
     ac = (Research)o;
}
catch(ClassCastException e) {
     System.err.println("Object was of type " + o.getClass().getName());
}
0
 

Author Comment

by:panpioneer
ID: 9772321
CHEJ,

one more thing, if i move

Object o = request.getAttribute("ab"); to out side of try block, I am getting java.lang.null.exception.
0
 

Author Comment

by:panpioneer
ID: 9772324

Oop! it is java.lang.NullPointerException
0
 
LVL 86

Expert Comment

by:CEHJ
ID: 9772361
Always test for null:

if (ac != null) {
   // you got it
}
0
 

Author Comment

by:panpioneer
ID: 9772505
after the test, I still get the same object type of "Research", but, it still gives me the same ClassCastException...
0
 

Author Comment

by:panpioneer
ID: 9772616

I know this sound unbelievable. But, I re-start the Tomcat server and clean up some of the files on our C drive to give C more space and it works fine now.
0
 
LVL 86

Expert Comment

by:CEHJ
ID: 9772724
Well Tom does cache files and classes...
0
 
LVL 92

Expert Comment

by:objects
ID: 9774578
> ac = (Research)request.getAttribute("ab");    

Probably better to test that the stored attribute is actually a Research instance (as mentioned by tim earlier), rather than handle the exception when it occurs.

Object o = request.getAttribute("ab")
if (o instanceof Research)
{
   Research ac = (Research) o;
   // do what you need with Research instance
}
else
{
    // Not a Research instance
}
0

Featured Post

Salesforce Made Easy to Use

On-screen guidance at the moment of need enables you & your employees to focus on the core, you can now boost your adoption rates swiftly and simply with one easy tool.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Java had always been an easily readable and understandable language.  Some relatively recent changes in the language seem to be changing this pretty fast, and anyone that had not seen any Java code for the last 5 years will possibly have issues unde…
Introduction This article is the second of three articles that explain why and how the Experts Exchange QA Team does test automation for our web site. This article covers the basic installation and configuration of the test automation tools used by…
This tutorial will introduce the viewer to VisualVM for the Java platform application. This video explains an example program and covers the Overview, Monitor, and Heap Dump tabs.
This tutorial explains how to use the VisualVM tool for the Java platform application. This video goes into detail on the Threads, Sampler, and Profiler tabs.
Suggested Courses

623 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question