Throwing an exception in catch block - control flow

Question

I am trying to understand the flow throwing an exception inside catch.

public static void main(String[] args) {
        try
        {
            throw new IOException();
        }
        catch(IOException e)
        {
            System.out.println("catch 1");
            throw new ArrayIndexOutOfBoundsException();
        }
        catch (ArrayIndexOutOfBoundsException e)
        {
            System.out.println("catch 2");
        }
        finally {
            System.out.println("finally 1");
        }
    }
}

I am using Java 8 and IntelliJ IDEA 2016. I am seeing a the below 3 lines printed in the IDE console in different order every time I execute. Is there something I am missing here.

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException
at com.boot.config.Test.main(Test.java:17)
    at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
    at java.lang.reflect.Method.invoke(Method.java:498)
    at com.intellij.rt.execution.application.AppMain.main(AppMain.java:144)
catch 1
finally 1

I don't understand why would the stacktrace is thrown before "catch 1"?

Does the exception class use either printStackTrace() or System.err to print out the message in console? — Pradeep Kumaresan, Feb 16 '18 at 17:26
If an unhandled exception escapes `main()` then the main thread's unhandled exception handler will be triggered, and the default handler dumps the stack trace to stderr and exits the thread. — Daniel Pryden, Feb 16 '18 at 17:43
Possibly useful reference: https://docs.oracle.com/javase/7/docs/api/java/lang/Thread.UncaughtExceptionHandler.html — Daniel Pryden, Feb 16 '18 at 17:50

Throwing an exception in catch block - control flow

0 Answers0