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Java | Exception Handling | Question 6
  • Difficulty Level : Easy
  • Last Updated : 24 Dec, 2015




class Test
{
    public static void main(String[] args)
    {
        try
        {
            int a[]= {1, 2, 3, 4};
            for (int i = 1; i <= 4; i++)
            {
                System.out.println ("a[" + i + "]=" + a[i] + "\n");
            }
        }
          
        catch (Exception e)
        {
            System.out.println ("error = " + e);
        }
          
        catch (ArrayIndexOutOfBoundsException e)
        {
            System.out.println ("ArrayIndexOutOfBoundsException");
        }
    }
}

(A) Compiler error
(B) Run time error
(C) ArrayIndexOutOfBoundsException
(D) Error Code is printed
(E) Array is printed


Answer: (A)

Explanation: ArrayIndexOutOfBoundsException has been already caught by base class Exception. When a subclass exception is mentioned after base class exception, then error occurs.

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