Q: What will be the output of the program? public class ExamQuestion7 { static int j; static void methodA(int i) { boolean b; do { b = i<10 | methodB(4); /* Line 9 */ b = i<10 || methodB…

The lines to watch here are lines 9 & 10. Line 9 features the non-shortcut version of the OR operator so both of its operands will be evaluated and therefore methodB(4) is executed. However line 10 has the shortcut version of the OR operator and if the 1st of its operands evaluates to true (which in this case is true), then the 2nd operand isn't evaluated, so methodB(8) never gets called. The loop is only executed once, b is initialized to false and is assigned true on line 9. Thus j = 4.

Q: What will be the output of the program? public class Test { public static void main(String[] args) { final StringBuffer a = new StringBuffer(); final StringBuffer b = new StringBuffer(); new Thread() { …

It gives different output while executing the same compiled code at different times.

Q: What will be the output of the program? String a = "newspaper"; a = a.substring(5,7); char b = a.charAt(1); a = a + b; System.out.println(a);

Both substring() and charAt() methods are indexed with a zero-base, and substring() returns a String of length arg2 - arg1.

Q: 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…

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.

Q: What will be the output of the program? String a = "ABCD"; String b = a.toLowerCase(); b.replace('a','d'); b.replace('b','c'); System.out.println(b);

String objects are immutable, they cannot be changed, in this case we are talking about the replace method which returns a new String object resulting from replacing all occurrences of oldChar in this string with newChar. b.replace(char oldChar, char newChar); But since this is only a temporary String it must either be put to use straight away i.e. System.out.println(b.replace('a','d')); Or a new variable must be assigned its value i.e. String c = b.replace('a','d');

Q: What will be the output of the program? public class SqrtExample { public static void main(String [] args) { double value = -9.0; System.out.println( Math.sqrt(value)); } }

The sqrt() method returns NaN (not a number) when it's argument is less than zero.

Q: What will be the output of the program? interface Foo141 { int k = 0; /* Line 3 */ } public class Test141 implements Foo141 { public static void main(String args[]) { int i; Test141 test141 = new Test141(); …

The variable k on line 3 is an interface constant, it is implicitly public, static, and final. Static variables can be referenced in two ways: Via a reference to any instance of the class (line 11) Via the class name (line 12).

Question 1

What will be the output of the program? class A 
{ 
    public A(int x){} 
} 
class B extends A { } 
public class test 
{ 
    public static void main (String args []) 
    {
        A a = new B(); 
        System.out.println("complete"); 
    } 
}

Accepted Answer

No constructor has been defined for class B therefore it will make a call to the default constructor but since class B extends class A it will also call the Super() default constructor. Since a constructor has been defined in class A java will no longer supply a default constructor for class A therefore when class B calls class A's default constructor it will result in a compile error.

Question 2

What will be the output of the program? public class Test138 
{ 
    public static void stringReplace (String text) 
    {
        text = text.replace ('j' , 'c'); /* Line 5 */
    } 
    public static void bufferReplace (StringBuffer text) 
    { 
…

Accepted Answer

A string is immutable, it cannot be changed, that's the reason for the StringBuffer class. The stringReplace method does not change the string declared on line 14, so this remains set to "java". Method parameters are always passed by value - a copy is passed into the method - if the copy changes, the original remains intact, line 5 changes the reference i.e. text points to a new String object, however this is lost when the method completes. The textBuffer is a StringBuffer so it can be changed. This change is carried out on line 9, so "java" becomes "javac", the text reference on line 9 remains unchanged. This gives us the output of "javajavac"

Question 3

What will be the output of the program? public class Test178 
{ 
    public static void main(String[] args) 
    {
        String s = "foo"; 
        Object o = (Object)s; 
        if (s.equals(o)) 
        { 
            System.out.print("AAA"); 
 …

Accepted Answer

AAACCC

Question 4

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?

Accepted Answer

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 5

What will be the output of the program? public class ExamQuestion7 
{  
    static int j; 
    static void methodA(int i)
    {
        boolean b; 
        do
        { 
            b = i<10 | methodB(4); /* Line 9 */
            b = i<10 || methodB…

Accepted Answer

The lines to watch here are lines 9 & 10. Line 9 features the non-shortcut version of the OR operator so both of its operands will be evaluated and therefore methodB(4) is executed. However line 10 has the shortcut version of the OR operator and if the 1st of its operands evaluates to true (which in this case is true), then the 2nd operand isn't evaluated, so methodB(8) never gets called. The loop is only executed once, b is initialized to false and is assigned true on line 9. Thus j = 4.

Question 6

What will be the output of the program? public class Test 
{ 
    public static void main(String[] args) 
    {
        final StringBuffer a = new StringBuffer(); 
        final StringBuffer b = new StringBuffer();

new Thread() 
        { …

Accepted Answer

It gives different output while executing the same compiled code at different times.

Question 7

What will be the output of the program? String a = "newspaper";
a = a.substring(5,7);
char b = a.charAt(1);
a = a + b;
System.out.println(a);

Accepted Answer

Both substring() and charAt() methods are indexed with a zero-base, and substring() returns a String of length arg2 - arg1.

Question 8

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…

Accepted Answer

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 9

What will be the output of the program? String a = "ABCD"; 
String b = a.toLowerCase(); 
b.replace('a','d'); 
b.replace('b','c'); 
System.out.println(b);

Accepted Answer

String objects are immutable, they cannot be changed, in this case we are talking about the replace method which returns a new String object resulting from replacing all occurrences of oldChar in this string with newChar. b.replace(char oldChar, char newChar); But since this is only a temporary String it must either be put to use straight away i.e. System.out.println(b.replace('a','d')); Or a new variable must be assigned its value i.e. String c = b.replace('a','d');

Question 10

What will be the output of the program? public class SqrtExample 
{
    public static void main(String [] args) 
    {
        double value = -9.0;
        System.out.println( Math.sqrt(value));
    }
}

Accepted Answer

The sqrt() method returns NaN (not a number) when it's argument is less than zero.

Question 11

What will be the output of the program? interface Foo141 
{ 
    int k = 0; /* Line 3 */
} 
public class Test141 implements Foo141 
{
    public static void main(String args[]) 
    {
        int i; 
        Test141 test141 = new Test141(); 
       …

Accepted Answer

The variable k on line 3 is an interface constant, it is implicitly public, static, and final. Static variables can be referenced in two ways: Via a reference to any instance of the class (line 11) Via the class name (line 12).

Question 12

What will be the output of the program? class Q207 
{ 
    public static void main(String[] args) 
    {
        int i1 = 5; 
        int i2 = 6; 
        String s1 = "7"; 
        System.out.println(i1 + i2 + s1); /* Line 8 */
    } 
}

Accepted Answer

This question is about the + (plus) operator and the overriden + (string cocatanation) operator. The rules that apply when you have a mixed expression of numbers and strings are: If either operand is a String, the + operator concatenates the operands. If both operands are numeric, the + operator adds the operands. The expression on line 6 above can be read as "Add the values i1 and i2 together, then take the sum and convert it to a string and concatenate it with the String from the variable s1". In code, the compiler probably interprets the expression on line 8 above as: System.out.println( new StringBuffer() 
    .append(new Integer(i1 + i2).toString()) 
    .append(s1) 
    .toString() );

Question 13

What will be the output of the program? String s = "hello"; 
Object o = s; 
if( o.equals(s) )
{
    System.out.println("A"); 
} 
else
{
    System.out.println("B"); 
} 
if( s.equals(o) )
{
    System.out.println("C"); 
} 
else
{ 
    System.out.prin…

Accepted Answer

1 and 3

Question 14

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;
  …

Accepted Answer

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 15

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++) 
        {
       …

Accepted Answer

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.

Question 16

What will be the output of the program? class Tree { } 
class Pine extends Tree { } 
class Oak extends Tree { } 
public class Forest1 
{ 
    public static void main (String [] args)
    { 
        Tree tree = new Pine(); 
        if( tree instanceo…

Accepted Answer

The program prints "Pine".

Question 17

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 S…

Accepted Answer

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 18

What will be the output of the program? String x = "xyz";
x.toUpperCase(); /* Line 2 */
String y = x.replace('Y', 'y');
y = y + "abc";
System.out.println(y);

Accepted Answer

Line 2 creates a new String object with the value "XYZ", but this new object is immediately lost because there is no reference to it. Line 3 creates a new String object referenced by y. This new String object has the value "xyz" because there was no "Y" in the String object referred to by x. Line 4 creates a new String object, appends "abc" to the value "xyz", and refers y to the result.

Question 19

What will be the output of the program? public class StringRef 
{
    public static void main(String [] args) 
    {
        String s1 = "abc";
        String s2 = "def";
        String s3 = s2;   /* Line 7 */
        s2 = "ghi";
        System.out.…

Accepted Answer

After line 7 executes, both s2 and s3 refer to a String object that contains the value "def". When line 8 executes, a new String object is created with the value "ghi", to which s2 refers. The reference variable s3 still refers to the (immutable) String object with the value "def".

Question 20

What will be the output of the program? int i = (int) Math.random();

Accepted Answer

Math.random() returns a double value greater than or equal to 0 and less than 1. Its value is stored to an int but as this is a narrowing conversion, a cast is needed to tell the compiler that you are aware that there may be a loss of precision. The value after the decimal point is lost when you cast a double to int and you are left with 0.

Java.lang Class Questions