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Home/ Questions/Q 4613532
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Editorial Team
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Editorial Team
Asked: May 22, 20262026-05-22T01:33:55+00:00 2026-05-22T01:33:55+00:00

In trying to refactor some I code I attempted to throw the exception in

  • 0

In trying to refactor some I code I attempted to throw the exception in the catch clause like so –

try {
....
}
catch(Exception exception){
.....
throw exception
}

However when I attempted to throw the exception on line “throw exception” the compiler complained with a message that I needed to surround my throw clause in a new try/catch like so –

try
{
  ....
}
catch (Exception exception)
{
  .....
  try
  {
    throw exception
  }
  catch (Exception e2)
  {
     ...
  }
}

Why does the compiler require this and what use does it provide ?

Thanks

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-22T01:33:55+00:00Added an answer on May 22, 2026 at 1:33 am

    My guess is that your trying to throw an exception sub class that isn’t declared by the method as an exception type it can throw.

    The following example works

    package test.example;
    
    public class ExceptionTest {
    
        public static void main(String[] args) throws Exception{
            try {
                int value = 1/0;
            } catch (Exception e) {
                System.out.println("woops the world is going to end");
                throw e;
            }
    
        }
    
    }
    

    However this example will give an error.

    package test.example;
    
    public class ExceptionTest {
    
        public static void main(String[] args) throws RuntimeException{
            try {
                int value = 1/0;
            } catch (Exception e) {
                System.out.println("woops the world is going to end");
                throw e;
            }
    
        }
    
    }
    

    Note in the second example I’m simply catching Exception not RuntimeException, it won’t compile as I throw Exception which is an undeclared throws, even though I do declare RuntimeException.

    Yes the exception is a RuntimeException but the compiler doesn’t know that.

    Just thought of a third working example to show you. This one also works because your throwing the same type as you declare. (note the only change is the catch block)

    package test.example;
    
    public class ExceptionTest {
    
        public static void main(String[] args) throws RuntimeException{
            try {
                int value = 1/0;
            } catch (RuntimeException e) {
                System.out.println("woops the world is going to end");
                throw e;
            }
    
        }
    
    }
    

    You need to understand the differences between all three of these answers

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