Finding The Output

Question 1 :

What will be the output of the program?

String x = new String("xyz");
String y = "abc";
x = x + y;
How many String objects have been created?


A). 2
B). 3
C). 4
D). 5
Answer : Option C

Explanation :

Line 1 creates two, one referred to by x and the lost String "xyz". Line 2 creates one (for a total of three). Line 3 creates one more (for a total of four), the concatenated String referred to by x with a value of "xyzabc".


Question 2 :

What will be the output of the program?

public class WrapTest 
{
    public static void main(String [] args) 
    {
        int result = 0;
        short s = 42;
        Long x = new Long("42");
        Long y = new Long(42);
        Short z = new Short("42");
        Short x2 = new Short(s);
        Integer y2 = new Integer("42");
        Integer z2 = new Integer(42);

        if (x == y) /* Line 13 */
            result = 1;
        if (x.equals(y) ) /* Line 15 */
            result = result + 10;
        if (x.equals(z) ) /* Line 17 */
            result = result + 100;
        if (x.equals(x2) ) /* Line 19 */
            result = result + 1000;
        if (x.equals(z2) ) /* Line 21 */
            result = result + 10000;

        System.out.println("result = " + result);
    }
}


A). result = 1
B). result = 10
C). result = 11
D). result = 11010
Answer : Option B

Explanation :

Line 13 fails because == compares reference values, not object values. Line 15 succeeds because both String and primitive wrapper constructors resolve to the same value (except for the Character wrapper). Lines 17, 19, and 21 fail because the equals() method fails if the object classes being compared are different and not in the same tree hierarchy.


Question 3 :

What will be the output of the program?

public class BoolTest 
{
    public static void main(String [] args) 
    {
        int result = 0;

        Boolean b1 = new Boolean("TRUE");
        Boolean b2 = new Boolean("true");
        Boolean b3 = new Boolean("tRuE");
        Boolean b4 = new Boolean("false");

        if (b1 == b2)  /* Line 10 */
            result = 1;
        if (b1.equals(b2) ) /* Line 12 */
            result = result + 10;
        if (b2 == b4)  /* Line 14 */
            result = result + 100;
        if (b2.equals(b4) ) /* Line 16 */
            result = result + 1000;
        if (b2.equals(b3) ) /* Line 18 */
            result = result + 10000;

        System.out.println("result = " + result);
    }
}


A). 0
B). 1
C). 10
D). 10010
Answer : Option D

Explanation :

Line 10 fails because b1 and b2 are two different objects. Lines 12 and 18 succeed because the Boolean String constructors are case insensitive. Lines 14 and 16 fail because true is not equal to false.


Question 4 :

What will be the output of the program?

public class ObjComp 
{
    public static void main(String [] args ) 
    {
        int result = 0;
        ObjComp oc = new ObjComp();
        Object o = oc;

        if (o == oc)  
            result = 1;
        if (o != oc)  
            result = result + 10;
        if (o.equals(oc) )  
            result = result + 100;
        if (oc.equals(o) )  
            result = result + 1000;

        System.out.println("result = " + result);
    }
}


A). 1
B). 10
C). 101
D). 1101
Answer : Option D

Explanation :

Even though o and oc are reference variables of different types, they are both referring to the same object. This means that == will resolve to true and that the default equals() method will also resolve to true.


Question 5 :

What will be the output of the program?

public class Example 
{
    public static void main(String [] args) 
    {
        double values[] = {-2.3, -1.0, 0.25, 4};
        int cnt = 0;
        for (int x=0; x < values.length; x++) 
        {
            if (Math.round(values[x] + .5) == Math.ceil(values[x])) 
            {
                ++cnt;
            }
        }
        System.out.println("same results " + cnt + " time(s)");
    }
}


A). same results 0 time(s)
B). same results 2 time(s)
C). same results 4 time(s)
D). Compilation fails.
Answer : Option B

Explanation :

Math.round() adds .5 to the argument then performs a floor(). Since the code adds an additional .5 before round() is called, it's as if we are adding 1 then doing a floor(). The values that start out as integer values will in effect be incremented by 1 on the round() side but not on the ceil() side, and the noninteger values will end up equal.