Question

Consider the following test case:

public class Main {

    static int a = 0;

    public static void main(String[] args) {
        try {
            test();
            System.out.println("---");
            test2();
        }
        catch(Exception e) {
            System.out.println(a + ": outer catch");
            a++;
        }
    }

    public static void test()
    {
        try {
            throw new Exception();
        }
        catch (Exception e) {
            System.out.println(a + ": inner catch");
            a++;
        }
        finally {
            System.out.println(a + ": finally");
            a++;
        }
    }

    public static void test2() throws Exception
    {
        try {
            throw new Exception();
        }
        finally {
            System.out.println(a + ": finally");
            a++;
        }
    }
}

With output:

0: inner catch
1: finally
---
2: finally
3: outer catch

What’s the explanation for why in test() catch happens before finally while in test2() it’s the other way around?

Answer 1

Because the try block in test2() doesn’t have a catch block, only a finally. The code won’t “jump back” to the caller to fall into catch and then “jump ahead” to the finally to continue there as you seem to think.