Java Class and Object
Question 1 |
Predict the output of following Java program?
class Test {
int i;
}
class Main {
public static void main(String args[]) {
Test t;
System.out.println(t.i);
}
0 | |
garbage value | |
compiler error | |
runtime error |
Discuss it
Question 1 Explanation:
t is just a reference, the object referred by t is not allocated any memory. Unlike C++, in Java all non-primitive objects must be explicitly allocated and these objects are allocated on heap. The following is corrected program.
Question 2 |
Predict the output of following Java program
class Test {
int i;
}
class Main {
public static void main(String args[]) {
Test t = new Test();
System.out.println(t.i);
}
}
garbage value | |
0 | |
compiler error | |
runtime error |
Discuss it
Question 2 Explanation:
In Java, fields of classes and objects that do not have an explicit initializer and elements of arrays are automatically initialized with the default value for their type (false for boolean, 0 for all numerical types, null for all reference types). Local variables in Java must be definitely assigned to before they are accessed, or it is a compile error.
Question 3 |
class demo
{
int a, b;
demo()
{
a = 10;
b = 20;
}
public void print()
{
System.out.println ("a = " + a + " b = " + b + "n");
}
}
class Test
{
public static void main(String[] args)
{
demo obj1 = new demo();
demo obj2 = obj1;
obj1.a += 1;
obj1.b += 1;
System.out.println ("values of obj1 : ");
obj1.print();
System.out.println ("values of obj2 : ");
obj2.print();
}
}
Compile error | |
values of obj1: a = 11 b = 21 values of obj2: a = 11 b = 21 | |
values of obj1: a = 11 b = 21 values of obj2: a = 10 b = 20 | |
values of obj1: a = 11 b = 20 values of obj2: a = 10 b = 21 | |
Run time error |
Discuss it
Question 3 Explanation:
Assignment of obj2 to obj1 makes obj2 a reference to obj1. Therefore, any change in obj1 will be reflected in obj2 also.
Question 4 |
Predict the output of following Java program.
class demoClass
{
int a = 1;
void func()
{
demo obj = new demo();
obj.display();
}
class demo
{
int b = 2;
void display()
{
System.out.println("na = " + a);
}
}
void get()
{
System.out.println("nb = " + b);
}
}
class Test
{
public static void main(String[] args)
{
demoClass obj = new demoClass();
obj.func();
obj.get();
}
}
a = 1 b = 2 | |
Compilation error
| |
b = 2 a = 1 |
Discuss it
Question 4 Explanation:
Members of inner class ‘demo’ can not be used in the outer class ‘Test’. Thus, get() of outer class can not access variable ‘b’ of inner class.
Question 5 |
Predict the output of the following program.
class First
{
void display()
{
System.out.println("Inside First");
}
}
class Second extends First
{
void display()
{
System.out.println("Inside Second");
}
}
class Test
{
public static void main(String[] args)
{
First obj1 = new First();
Second obj2 = new Second();
First ref;
ref = obj1;
ref.display();
ref = obj2;
ref.display();
}
}
Compilation error | |
Inside First Inside Second | |
Inside First Inside First | |
Runtime error |
Discuss it
Question 5 Explanation:
‘ref’ is a reference variable which obtains the reference of object of class First and calls its function display().
Then ‘ref’ refers to object of class Second and calls its function display().
Question 6 |
Predict the output of the following program.
class Test
{
int a = 1;
int b = 2;
Test func(Test obj)
{
Test obj3 = new Test();
obj3 = obj;
obj3.a = obj.a++ + ++obj.b;
obj.b = obj.b;
return obj3;
}
public static void main(String[] args)
{
Test obj1 = new Test();
Test obj2 = obj1.func(obj1);
System.out.println("obj1.a = " + obj1.a + " obj1.b = " + obj1.b);
System.out.println("obj2.a = " + obj2.a + " obj1.b = " + obj2.b);
}
}
obj1.a = 1 obj1.b = 2 obj2.a = 4 obj2.b = 3 | |
obj1.a = 4 obj1.b = 3 obj2.a = 4 obj2.b = 3 | |
Compilation error |
Discuss it
Question 6 Explanation:
obj1 and obj2 refer to same memory address.
Question 7 |
Predict the output of the following program.
class Test
{
int a = 1;
int b = 2;
Test func(Test obj)
{
Test obj3 = new Test();
obj3 = obj;
obj3.a = obj.a++ + ++obj.b;
obj.b = obj.b;
return obj3;
}
public static void main(String[] args)
{
Test obj1 = new Test();
Test obj2 = obj1.func(obj1);
System.out.println("obj1.a = " + obj1.a + " obj1.b = " + obj1.b);
System.out.println("obj2.a = " + obj2.a + " obj1.b = " + obj2.b);
}
}
obj1.a = 1 obj1.b = 2 obj2.a = 4 obj2.b = 3 | |
obj1.a = 4 obj1.b = 3 obj2.a = 4 obj2.b = 3 | |
Compilation error |
Discuss it
Question 7 Explanation:
obj1 and obj2 refer to same memory address.
Question 8 |
What is the output of the following JAVA program ?
Class Test {
public static void main(String[] args) {
Test obj = new Test();
obj.start();
}
void start() {
String stra = ”do”;
String strb = method(stra);
System.out.print(“: ”+stra + strb);
}
String method(String stra) {
stra = stra + ”good”;
System.out.print(stra);
return“ good”;
}
}
dogood : dogoodgood | |
dogood : gooddogood | |
dogood : dodogood | |
dogood : dogood |
Discuss it
Question 9 |
Java uses threads to enable the entire environment to be ______.
Symmetric | |
Asymmetric | |
Synchronous | |
Asynchronous |
Discuss it
Question 9 Explanation:
Java uses threads to enable the entire environment to be asynchronous. Asynchronous threading is pre-emptive i.e. a thread once start executing a task it can hold it in mid, save the current state and start executing another task (context switching) according to priority and other specified criteria and threading.
So, option (D) is correct.
Question 10 |
In Java, when we implement an interface method, it must be declared as:
Private | |
Protected | |
Public | |
Friend |
Discuss it
Question 10 Explanation:
In Java, when we implement an interface method, it must be declared as Public.
For more information on Java interface Refer:Interfaces in Java
option (C) is correct.
There are 11 questions to complete.