Question 1

    
import java.awt.*;
class Ticker extends Component 
{
    public static void main (String [] args) 
    {
        Ticker t = new Ticker();
        /* Missing Statements ? */
    }
}

which two of the following statements, inserted independently, could legally be inserted into missing section of this code?

boolean test = (Component instanceof t);
boolean test = (t instanceof Ticker);
boolean test = t.instanceof(Ticker);
boolean test = (t instanceof Component);

Answer

Answer: D

(2) is correct because class type Ticker is part of the class hierarchy of t; therefore it is a legal use of the instanceof operator. (4) is also correct because Component is part of the hierarchy of t, because Ticker extends Component.

(1) is incorrect because the syntax is wrong. A variable (or null) always appears before the instanceof operator, and a type appears after it. (3) is incorrect because the statement is used as a method (t.instanceof(Ticker);), which is illegal.

Question 2

What will be the output of the program?

public class Test 
{ 
    public static void leftshift(int i, int j) 
    {
        i <<= j; 
    } 
    public static void main(String args[]) 
    {
        int i = 4, j = 2; 
        leftshift(i, j); 
        System.out.println(i); 
    } 
}

Answer

Answer: B

Java only ever passes arguments to a method by value (i.e. a copy of the variable) and never by reference. Therefore the value of the variable i remains unchanged in the main method.

If you are clever you will spot that 16 is 4 multiplied by 2 twice, (4 * 2 * 2) = 16. If you had 16 left shifted by three bits then 16 * 2 * 2 * 2 = 128. If you had 128 right shifted by 2 bits then 128 / 2 / 2 = 32. Keeping these points in mind, you don't have to go converting to binary to do the left and right bit shifts.

Question 3

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);
    }
}

Answer

Answer: C

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.

Question 4

What will be the output of the program?

class Two 
{
    byte x;
}

class PassO 
{
    public static void main(String [] args) 
    {
        PassO p = new PassO();
        p.start();
    }

    void start() 
    {
        Two t = new Two();
        System.out.print(t.x + " ");
        Two t2 = fix(t);
        System.out.println(t.x + " " + t2.x);
    }

    Two fix(Two tt) 
    {
        tt.x = 42;
        return tt;
    }
}

Answer

Answer: C

In the fix() method, the reference variable tt refers to the same object (class Two) as the t reference variable. Updating tt.x in the fix() method updates t.x (they are one in the same object). Remember also that the instance variable x in the Two class is initialized to 0.

Question 5

What will be the output of the program?

class Test 
{
    static int s;
    public static void main(String [] args) 
    {
        Test p = new Test();
        p.start();
        System.out.println(s);
    }

    void start() 
    {
        int x = 7;
        twice(x);
        System.out.print(x + " ");
    }

    void twice(int x) 
    {
        x = x*2;
        s = x;
    }
}

Answer

Answer: B

The int x in the twice() method is not the same int x as in the start() method. Start()'s x is not affected by the twice() method. The instance variable s is updated by twice()'s x, which is 14.

Questions for: Operators And Assignments