Java Programming - Exceptions - Discussion

Discussion Forum : Exceptions - Finding the output (Q.No. 4)
4.
What will be the output of the program?
public class X 
{  
    public static void main(String [] args) 
    {
        try 
        {
            badMethod();  
            System.out.print("A");  
        } 
        catch (RuntimeException ex) /* Line 10 */
        { 
            System.out.print("B"); 
        } 
        catch (Exception ex1) 
        { 
            System.out.print("C"); 
        } 
        finally 
        {
            System.out.print("D"); 
        } 
        System.out.print("E"); 
    } 
    public static void badMethod() 
    { 
        throw new RuntimeException(); 
    } 
}
BD
BCD
BDE
BCDE
Answer: Option
Explanation:

A Run time exception is thrown and caught in the catch statement on line 10. All the code after the finally statement is run because the exception has been caught.

Discussion:
20 comments Page 2 of 2.

Rasika said:   1 decade ago
The options mentioned are wrong!

Runtime Exception is subclass of Exception so above piece of code will not compile.

Ayushisa36491 said:   9 years ago
Line 10 has been declared as a comment, but then also exception is caught, how that suppose to take place?

Laxmi said:   1 decade ago
Why not A got printed. Because after exception caught, program should follow natural flow?

Aamir said:   1 decade ago
Hi @Mukesh.

If the Exception is caught then the catch try and catch block no compile.

Priyanka said:   1 decade ago
If exception is not caught then all the code after finally statement is run or not?

Parthipan said:   9 years ago
Hi, I'm not clear with the given explanation, can any help to to understand this?

Mohan said:   1 decade ago
I have also same question why A was not printed?

Mounika said:   9 years ago
Can anyone clearly explain the entire program?

Anusha said:   9 years ago
Can any explain clearly.

Vikram said:   9 years ago
Define badMethod().


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