-> What will be the output of the program?
class Exc0 extends Exception { }
class Exc1 extends Exc0 { } /* Line 2 */
public class Test
{
public static void main(String args[])
{
try
{
throw new Exc1(); /* Line 9 */
}
catch (Exc0 e0) /* Line 11 */
{
System.out.println("Ex0 caught");
}
catch (Exception e)
{
System.out.println("exception caught");
}
}
}
Explanation:
An exception Exc1 is thrown and is caught by the catch statement on line 11. The code is executed in this block. There is no finally block of code to execute.
-> 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();
}
}
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.
-> What will be the output of the program (when you run with the -ea option) ?
public class Test
{
public static void main(String[] args)
{
int x = 0;
assert (x > 0) : "assertion failed"; /* Line 6 */
System.out.println("finished");
}
}
Explanation:
An assertion Error is thrown as normal giving the output "assertion failed". The word "finished" is not printed (ensure you run with the -ea option)
Assertion failures are generally labeled in the stack trace with the file and line number from which they were thrown, and also in this case with the error's detail message "assertion failed". The detail message is supplied by the assert statement in line 6.
-> What will be the output of the program?
class BoolArray
{
boolean [] b = new boolean[3];
int count = 0;
void set(boolean [] x, int i)
{
x[i] = true;
++count;
}
public static void main(String [] args)
{
BoolArray ba = new BoolArray();
ba.set(ba.b, 0);
ba.set(ba.b, 2);
ba.test();
}
void test()
{
if ( b[0] && b[1] | b[2] )
count++;
if ( b[1] && b[(++count - 2)] )
count += 7;
System.out.println("count = " + count);
}
}
Explanation:
The reference variables b and x both refer to the same boolean array. count is incremented for each call to the set() method, and once again when the first if test is true. Because of the && short circuit operator, count is not incremented during the second if test.
-> What will be the output of the program?
class BitShift
{
public static void main(String [] args)
{
int x = 0x80000000;
System.out.print(x + " and ");
x = x >>> 31;
System.out.println(x);
}
}
Explanation:
Option A is correct. The >>> operator moves all bits to the right, zero filling the left bits. The bit transformation looks like this:
Before: 1000 0000 0000 0000 0000 0000 0000 0000
After: 0000 0000 0000 0000 0000 0000 0000 0001
Option C is incorrect because the >>> operator zero fills the left bits, which in this case changes the sign of x, as shown.
Option B is incorrect because the output method print() always displays integers in base 10.
Option D is incorrect because this is the reverse order of the two output numbers.
