exception handling and finally block in java

Question

public class Confusion {
    Confusion(int i) {
        int j = 5;
        int[] a = new int[2];
        try {
            a[0] = 4;
            if (i <= 0) {
                int k = j / i;
            } else {
                System.out.println(j / i);
            }
        } catch (ArithmeticException sa) {
            System.out.println("Wrong value" + sa);
        } catch (ArrayIndexOutOfBoundsException e) {
            System.out.println("out of range massage in Class");
        } finally {
            System.out.println("Executing finally block in code");
        }
    }

    void k() {
        int[] a = new int[2];
        {
            try {
                a[4] = 4;
            } catch (ArrayIndexOutOfBoundsException e) {
                System.out.println("out of range");
            }
        }
    }
}

public class Nested {
    public static void main(String[] args) {
        Confusion c = new Confusion(2);
        Confusion c1 = new Confusion(0);
        c1.k();
        c.k();
    }
}

Output:

-2
Executing finally block in code
Wrong valuejava.lang.ArithmeticException: / by zero
Executing finally block in code
out of range
out of range

Whenever i am executing the finally{} block written in the code below it is getting executed twice. Don't know why this is happening. I want to run the finally block only once. Is there any way to throw multiple exception in single method?

Answer 1

It because you have two Confusion objects. Therefore, the constructor will be executed twice.

confusion c=new confusion(2);
confusion c1=new confusion(0);

Answer 2

confusion c=new confusion(2);
confusion c1=new confusion(0);

thats why u are getting 2 outputs from finally.

Answer 3

This is because you are creating two objects c1 and c

Answer 4

The finally block always executes when the try block exits.

http://docs.oracle.com/javase/tutorial/essential/exceptions/finally.html

So regardless of any exception it will execute finally block. As you create two objects, it will call the constructor twice and will also execute finally twice.

Answer 5

You are calling the code in the try twice.

confusion c=new confusion(2);
confusion c1=new confusion(0);

This means that as the finally happens every time the try is called, it'll print...

Executing finally block in code

If you called it again,

confusion c1=new confusion(3);

It'd print out the contents of the finally a third time.

Answer 6

You are creating two objects using following code. Thats why the finally block is executed two times.

confusion c=new confusion(2);
confusion c1=new confusion(0);

Try this one please

public static void main(String[] args){
   confusion c=new confusion(2);
   confusion c1=new confusion(0);
   confusion c1=new confusion(10);
   confusion c1=new confusion(5);
} 

Now finally block called 4 times.

Constructor in Java is block of code which is executed at the time of Object creation.

Please read constructor reference http://docs.oracle.com/javase/tutorial/java/javaOO/constructors.html

Answer 7

Because you are creating two Confusion objects. Therefore, the finally block will be executed twice.

confusion c=new confusion(2);
confusion c1=new confusion(0);

If you want to execute finally only once.Try this code

 public class Confusion {
    Confusion(int i) {
        int j = 5;
        int[] a = new int[2];
        a[0] = 4;
        if (i <= 0) {
           int k = j / i;
        } else {
           System.out.println(j / i);
        }
    }

    void k() {
       int[] a = new int[2];
            a[4] = 4;
   }
}

public class Nested {
   public static void main(String[] args) {
      try{
           Confusion c = new Confusion(2);
           Confusion c1 = new Confusion(0);
           c1.k();
           c.k();
         }catch (ArithmeticException sa) {
            System.out.println("Wrong value" + sa);
         } catch (ArrayIndexOutOfBoundsException e) {
            System.out.println("out of range massage in Class");
        } finally {
            System.out.println("Executing finally block in code");
        }
     }
  }