Q: What will be the output of the program? class Test { public static void main(String [] args) { int x=20; String sup = (x < 15)? "small" : (x < 22)? "tiny" : "huge"; System.out.println(sup); } }

This is an example of a nested ternary operator. The second evaluation (x < 22) is true, so the "tiny" value is assigned to sup.

Q: What will be the output of the program? class Test { public static void main(String [] args) { Test p = new Test(); p.start(); } void start() { boolean b1 = false; boolean b2 = fix(b1); …

The boolean b1 in the fix() method is a different boolean than the b1 in the start() method. The b1 in the start() method is not updated by the fix() method.

Question 1

What will be the output of the program? class SC2 
{
    public static void main(String [] args) 
    {
        SC2 s = new SC2();
        s.start();
    }

void start() 
    {
        int a = 3;
        int b = 4;
        System.out.print(" " +…

Accepted Answer

Because all of these expressions use the + operator, there is no precedence to worry about and all of the expressions will be evaluated from left to right. If either operand being evaluated is a String, the + operator will concatenate the two operands; if both operands are numeric, the + operator will add the two operands.

Question 2

What will be the output of the program? class Bitwise 
{
    public static void main(String [] args) 
    {
        int x = 11 & 9;
        int y = x ^ 3;
        System.out.println( y | 12 );
    }
}

Accepted Answer

The & operator produces a 1 bit when both bits are 1. The result of the & operation is 9. The ^ operator produces a 1 bit when exactly one bit is 1; the result of this operation is 10. The | operator produces a 1 bit when at least one bit is 1; the result of this operation is 14.

Question 3

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();
       …

Accepted Answer

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 4

What will be the output of the program? class Equals 
{
    public static void main(String [] args) 
    {
        int x = 100;
        double y = 100.1;
        boolean b = (x = y); /* Line 7 */
        System.out.println(b);
    }
}

Accepted Answer

The code will not compile because in line 7, the line will work only if we use (x==y) in the line. The == operator compares values to produce a boolean, whereas the = operator assigns a value to variables. Option A, B, and D are incorrect because the code does not get as far as compiling. If we corrected this code, the output would be false.

Question 5

What will be the output of the program? class Test 
{
    public static void main(String [] args) 
    {
        int x=20;
        String sup = (x < 15)? "small" : (x < 22)? "tiny" : "huge";
        System.out.println(sup);
    }
}

Accepted Answer

This is an example of a nested ternary operator. The second evaluation (x < 22) is true, so the "tiny" value is assigned to sup.

Question 6

What will be the output of the program? class Test 
{
    public static void main(String [] args) 
    {
        Test p = new Test();
        p.start();
    }

void start() 
    {
        boolean b1 = false;
        boolean b2 = fix(b1);
       …

Accepted Answer

The boolean b1 in the fix() method is a different boolean than the b1 in the start() method. The b1 in the start() method is not updated by the fix() method.

Question 7

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);
    }
}

Accepted Answer

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.

Question 8

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 = …

Accepted Answer

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.

Question 9

What will be the output of the program? class SSBool 
{
    public static void main(String [] args) 
    {
        boolean b1 = true;
        boolean b2 = false;
        boolean b3 = true;
        if ( b1 & b2 | b2 & b3 | b2 ) /* Line 8 */
         …

Accepted Answer

The & operator has a higher precedence than the | operator so that on line 8 b1 and b2 are evaluated together as are b2 & b3. The final b1 in line 10 is what causes that if test to be true. Hence it prints "dokey".

Question 10

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) 
    …

Accepted Answer

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 11

What will be the output of the program? class Test 
{
    public static void main(String [] args) 
    {
        int x= 0;
        int y= 0;
        for (int z = 0; z < 5; z++) 
        {
            if (( ++x > 2 ) && (++y > 2)) 
            {
    …

Accepted Answer

In the first two iterations x is incremented once and y is not because of the short circuit && operator. In the third and forth iterations x and y are each incremented, and in the fifth iteration x is doubly incremented and y is incremented.

Question 12

What will be the output of the program? class Test 
{
    public static void main(String [] args) 
    {
        int x= 0;
        int y= 0;
        for (int z = 0; z < 5; z++) 
        {
            if (( ++x > 2 ) || (++y > 2)) 
            {
    …

Accepted Answer

The first two iterations of the for loop both x and y are incremented. On the third iteration x is incremented, and for the first time becomes greater than 2. The short circuit or operator || keeps y from ever being incremented again and x is incremented twice on each of the last three iterations.

Question 13

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); 
     …

Accepted Answer

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 14

What will be the output of the program? class PassS 
{
    public static void main(String [] args) 
    {
        PassS p = new PassS();
        p.start();
    }

void start() 
    {
        String s1 = "slip";
        String s2 = fix(s1);
     …

Accepted Answer

When the fix() method is first entered, start()'s s1 and fix()'s s1 reference variables both refer to the same String object (with a value of "slip"). Fix()'s s1 is reassigned to a new object that is created when the concatenation occurs (this second String object has a value of "slipstream"). When the program returns to start(), another String object is created, referred to by s2 and with a value of "stream".

Question 15

What will be the output of the program? class PassA 
{
    public static void main(String [] args) 
    {
        PassA p = new PassA();
        p.start();
    }

void start() 
    {
        long [] a1 = {3,4,5};
        long [] a2 = fix(a1);
  …

Accepted Answer

Output: 15 15 The reference variables a1 and a3 refer to the same long array object. When the [1] element is updated in the fix() method, it is updating the array referred to by a1. The reference variable a2 refers to the same array object. So Output: 3+7+5+" "3+7+5 Output: 15 15 Because Numeric values will be added

Question 16

Which three statements are true? f1 == f2 f1 == f3 f2 == f1[1] x == f1[0] f == f1[0] import java.awt.Button;
class CompareReference 
{
    public static void main(String [] args) 
    {
        float f = 42.0f;
        float [] f1 = new float[…

Accepted Answer

(2) is correct because the reference variables f1 and f3 refer to the same array object. (4) is correct because it is legal to compare integer and floating-point types. (5) is correct because it is legal to compare a variable with an array element. (3) is incorrect because f2 is an array object and f1[1] is an array element.

Question 17

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 t…

Accepted Answer

(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 18

Which two statements are equivalent? 16*4 16>>2 16/2^2 16>>>2

Accepted Answer

(2) is correct. 16 >> 2 = 4 (4) is correct. 16 >>> 2 = 4 (1) is wrong. 16 * 4 = 64 (3) is wrong. 16/2 ^ 2 = 10

Question 19

Which two are equal? 32/4 (8 >> 2) << 4 2^5 128 >>> 2 2 >> 5

Accepted Answer

(2) and (4) are correct. (2) and (4) both evaluate to 32. (2) is shifting bits right then left using the signed bit shifters >> and <<. (4) is shifting bits using the unsigned operator >>>, but since the beginning number is positive the sign is maintained. (1) evaluates to 8, (3) looks like 2 to the 5th power, but ^ is the Exclusive OR operator so (3) evaluates to 7. (5) evaluates to 0 (2 >> 5 is not 2 to the 5th).

Question 20

Which of the following are legal lines of code? int w = (int)888.8; byte x = (byte)1000L; long y = (byte)100; byte z = (byte)100L;

Accepted Answer

Statements (1), (2), (3), and (4) are correct. (1) is correct because when a floating-point number (a double in this case) is cast to an int, it simply loses the digits after the decimal. (2) and (4) are correct because a long can be cast into a byte. If the long is over 127, it loses its most significant (leftmost) bits. (3) actually works, even though a cast is not necessary, because a long can store a byte.

Operators and Assignments Questions