Why is the exception thrown in the finally block the one that is ultimately reported in Java?

Exception Handling in Java: Understanding the Impact of Exceptions Thrown in Catch and Finally Clauses

Consider the following code snippet:

class MyExc1 extends Exception {}
class MyExc2 extends Exception {}
class MyExc3 extends MyExc2 {}

public class C1 {
    public static void main(String[] args) throws Exception {
        try {
            System.out.print(1);
            q();
        }
        catch (Exception i) {
            throw new MyExc2();
        }
        finally {
            System.out.print(2);
            throw new MyExc1(); // throws MyExc1 here
        }
    }

    static void q() throws Exception {
        try {
            throw new MyExc1(); // throws MyExc1 here
        }
        catch (Exception y) {
        }
        finally {
            System.out.print(3);
            throw new Exception(); // throws Exception here
        }
    }
}

In this snippet, we have a sequence of exceptions being thrown in both the try-catch and finally clauses. The expected output is "132Exception in thread main MyExc1," not "13Exception in thread main MyExc2."

Understanding this result requires comprehending the fundamental principle of exception handling in Java:

When a new exception is thrown within a catch or finally clause that propagates out of that block, the current exception is aborted, and the new exception is propagated instead.

Essentially, the current exception is forgotten, and the new exception takes over the unwinding process up the stack.

In the given code, a MyExc1 exception is thrown in the innermost finally clause. This exception will eventually bubble up to the main finally clause. However, in that finally clause, another exception, MyExc1 is thrown, aborting the propagation of the original MyExc1. Therefore, the outermost finally clause's MyExc1 becomes the exception that gets propagated outward.

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