Why such output is coming?

Question

It might be easy but I don't understand why the output is coming as 1 4. And what is the function of the return statement at Line 9?

public static void main(String[] args) {
    try{
        f();
    } catch(InterruptedException e){
        System.out.println("1");
        throw new RuntimeException();
    } catch(RuntimeException e){
        System.out.println("2");
        return;                    \\ Line 9
    } catch(Exception e){
        System.out.println("3");
    } finally{
        System.out.println("4");
    } 
        System.out.println("5");
}   
        static void f() throws InterruptedException{
            throw new InterruptedException("Interrupted");
    }

Thanks in advance.

The runtime exception isn't catched because there is no `try` and line 9 not executed. — Narmer, Oct 14 '14 at 07:50
Also 5 is not going to be printed because `RuntimeException` is an unchecked exception, thus is internally catched (without explicit `try-catch` block) and terminating the program before `System.out.println("5");` — Narmer, Oct 14 '14 at 07:56

score 2 · Accepted Answer · edited May 23 '17 at 12:11

Your function f() throws InterruptedException, which is caught by the first catch block (hence it prints 1), but this catch block cannot throw other exceptions (if it is not thrown by your method), Hence, no other catch block can catch your excception and therefore finally is executed (finally executes in every case except those silly infinite loop cases). you can refer to Exception thrown inside catch block - will it be caught again?.

I hope it helps.

Just to summarize, you can throw any exception from try block & it will be caught (if there is a good catch block). But from catch block only those exceptions can be thrown (and consequently caught by) which your method throws.

If you throw exception from catch block which are not thrown by your method, it is meaning less and won't be caught (like in your case).

score 1 · Answer 2 · answered Oct 14 '14 at 07:49

1

As you can see f() throws InterruptedException, so it will first print 1 which is inside first catch block and then finally would execute so it will print 4.

answered Oct 14 '14 at 07:49

DeepInJava

1,871
3
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score 1 · Answer 3 · answered Oct 14 '14 at 07:49

1

The 1 is printed because f() throws an InterruptedException. Because the Exception is handled in the first catch block it is not handled in the other exception blocks belower anymore. The finally statement is always run, so the 4 is printed too.

answered Oct 14 '14 at 07:49

Marco

508
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score 1 · Answer 4 · answered Oct 14 '14 at 07:50

try{
    f();
} catch(InterruptedException e){
    System.out.println("1");
    throw new RuntimeException(); // this RuntimeException will not catch by 
                                  //following catch block
} catch(RuntimeException e){

You needs to change your code as follows to catch it.

 try{
        f();
    } catch(InterruptedException e){
        System.out.println("1");
        try {
            throw new RuntimeException();// this RuntimeException will catch 
                                         // by the following catch 
        }catch (RuntimeException e1){
            System.out.println("hello");
        }
    } catch(RuntimeException e){
        System.out.println("2");
        return;
    } catch(Exception e){
        System.out.println("3");
    } finally{
        System.out.println("4");
    }
    System.out.println("5");

Then your out put"

  1
  hello // caught the RunTimeException
  4 // will give you 2,but also going to finally block then top element of stack is 4
  5 // last print element out side try-catch-finally

score 1 · Answer 5 · answered Oct 14 '14 at 07:50

1

f() throws an InterruptedException so you get the 1. finally is always called at the end so you get the 4.

answered Oct 14 '14 at 07:50

Jens

67,715
15
98
113

score 1 · Answer 6 · answered Oct 14 '14 at 07:56

As f() throw InterruptedException which executes,

catch(InterruptedException e){
        System.out.println("1");
        throw new RuntimeException();
}

prints --> 1

and finally gets executed before exiting the program,

finally{
        System.out.println("4");
}

prints --> 4

Code after return wont execute but finally will get executed.

Why such output is coming?

6 Answers6