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Thread: unexpected output

  1. #1
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    Default unexpected output

    Why doesn't the following code display "stop" ? If i specify the function sleep() without a try/catch, then an interrupted exception is thrown. So now when I've given the try/catch block, that exception should be caught in the catch block and the string "stop" should be printed. But the output isn't so. Can anyone please explain?
    public class A
    {
                     public static void main(String[] args)
                   {
                    Thread t=Thread.currentThread();
                    for(int i=1;i<6;i++)
                        {
     
                         System.out.println( i );
     
                           try
                           {
                            Thread.sleep(1000);
                           }
     
                             catch(Exception e)
                            {
                             System.out.println("stop");
                            }
     
                         }
     
                    }
    }
    Last edited by sumitroy; June 29th, 2014 at 08:47 AM.


  2. #2
    Super Moderator Norm's Avatar
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    Default Re: unexpected output

    What exception is being thrown that you expect to execute the catch block?

    Please edit your post and wrap your code with code tags:
    [code=java]
    YOUR CODE GOES HERE
    [/code]
    to get highlighting and preserve formatting.

    an interrupted exception is thrown.
    Please copy the full text of the error message and paste it here. It has important info about the error.

    The compiler will give an error message if the code doesn't wrap the sleep() in a try{}catch
    If you don't understand my answer, don't ignore it, ask a question.

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    Default Re: unexpected output

    the compiler gives the following error when i specify sleep() without try/catch{}:

    Exception in thread "main" java.lang.Error: Unresolved compilation problem:
    Unhandled exception type InterruptedException

    at com.home.A.main(A.java:13)

    so definitely an exception should be caught when i specify the try/catch block,but it's not happening.

  4. #4
    Super Moderator Norm's Avatar
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    Default Re: unexpected output

    Unhandled exception
    The compiler is telling you to wrap some part of the code in a try{}catch block because that code can throw an exception.

    without a try/catch, then an interrupted exception is thrown
    No exception is thrown during compilation. There is a statement that CAN throw an exception when it is executed.
    If you don't understand my answer, don't ignore it, ask a question.

  5. #5
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    Default Re: unexpected output

    thanks!

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