JAVA AWT vs. JAVA Swing

JAVA AWT	JAVA Swing
Java AWT components are platform dependent.	Java Swing components are platform independent.
AWT components are heavyweight.	Swing components are lightweight.
AWT does not support pluggable look and feel.	Swing supports pluggable look and feel.
AWT does not follow MVC.	Swing follow MVC (Model View Controller) where model represents data, view represents presentation and controller acts as an interface between model and view.

How to avoid duplicate values from ArrayList in Java?

This is one of the good interview question. I have faced many times this question in HCL & TCS interview. So frnds, let's see the following answer:

ArrayListDemo1.java

import java.util.*;
public class ArrayListDemo1 {

public static void main(String[] args)
{
ArrayList al = new ArrayList();

     al.add("Prajukta");
     al.add("Bishnupriya");
     al.add("Nitasha");
     al.add("Prajukta");
     al.add("Prajukta");      //Duplicate
     al.add("Prajukta");     //Duplicate

     ArrayList al2 = new ArrayList(new HashSet(al));

    Iterator it= al2.iterator();
    while(it.hasNext())
    {
    System.out.println(it.next());
    }

}
}

Output 
Prajukta
Bishnupriya
Nitasha

Difference between Java and C

Difference Between Java and C

Java vs. C :

Java	C
Java is object-oriented	C is procedural-oriented
Java is both compiled and interpreted language	C is a compiled language.
Java is platform independent	C is platform dependent
Java is a high level language	C is a low-level language
Java uses bottom-up approach	C uses top-down approach
In Java, pointer is in back stage	C uses pointer explicitly
Java supports method overloading	C does not support method overloading
Java does not support pre-processor	C support pre-processor
Java handles exception	C does not support exception handling mechanism
Java collect garbage automatically	C collect garbage explicitly

Why Interface in JAVA ?

We need interface due to the followings:

1.    It supports the functionality of multiple inheritance

2.    It achieves the concept of loose coupling which is the main advantage of interface

3.    It achieves 100% concept of abstraction, because interface must contain abstract method, not normal method. 

What is Dynamic Method Dispatch?

Method overriding is a disadvantageous fact, because in this concept only sub class property is coming as output, super class property is hidden. So to eliminate this disadvantage we need run-time polymorphism which is also known as dynamic method dispatch.

• Dynamic method dispatch is a mechanism in which a call to an overriden method is reserved at runtime rather than compile time.
• This is a mechanism that represents how java implement runtime polymorphism.
• In this mechanism a super class reference variable can refer to a sub class object.
• When an overriden method is called through a super class reference, java determines which method is executed and this is happened at runtime.

Example:

class Test
{
void display()
{
system.out.println("JavaRace");
}
}
class Test1 extends Test
{
void display( )
{
system.out.println("Silan Software");
}
}
class Test2 extends Test
{
void display( )
{
system.out.println("Silan Technology");
}
}
class Dispatch {
public static void main(String[] args)
{
Test obj1=new Test( );
Test1 obj2=new Test1( );
Test2 obj3=new Test2( );
Test r;
r=obj1;
r.display();
r=obj2;
r.display();
r=obj3;
r.display();
}
}

Output

JavaRace

Silan Software

Silan Technology 

Can we write a Java program without main() method ?

Yes, We can write a java program without main method. it is possible by using static block only, which was supported in java 1.6 version. But not possible in Java 7 or Java 8.

class Test{  

  static{  

  System.out.println("Silan Software");  

  System.exit(0);  

  }  

}  

Output 

Silan Software (if not JDK7)

In JDK7 and above, the output will be:

Error: Main method not found in class Test, please define the main method as:public static void main(String[] args)

Why JAVA does not support global variables ?

The answer is to your question is, because Java doesn't support global variables, by design. Java was designed with object-oriented principles in mind and as such, every variable in Java is either local or a member of a class.

Static class members are globally-accessible, which is certainly one possible definition of a global variable, depending upon your interpretation of the term. To be pedantic, while Static class members are accessible via the class name and therefore across multiple scopes, they are still class members; and therefore not truly global variables as such.

Java's lack of support for global variables is a good thing, as using global variables is a design anti-pattern.

Java program to check whether a given mail id is valid or invalid.

import java.util.regex.*;

class RegexDemo7

{

   public static void main(String[] args) 

   {

      Pattern p=Pattern.compile("[a-zA-Z0-9][a-zA-Z0-9_.]*@[a-zA-Z0-9]+([.][a-zA-Z]+)+");

      Matcher m=p.matcher(args[0]);

      if(m.find() && m.group().equals(args[0]))

      {

        System.out.println("valid mail id");

       }

       else

       {

         System.out.println("invalid mail id");

        }

    }

}

Output

1st run : 

java RegexDemo7  trilochan4u@gmail.com

valid mail id

2nd run : 

java RegexDemo7  trilochan4u

nvalid mail id

   

Java program to check whether a given number is a valid mobile number or not.

import java.util.regex.*;
class RegexDemo6
{
   public static void main(String[] args)
    {
        Pattern p=Pattern.compile("(0/91)?[7-9][0-9]{9}");
        Matcher m=p.matcher(args[0]);
        if(m.find() && m.group().equals(args[0]))
        {
System.out.println("valid mobile number");
         }
        else
        {
System.out.println("invalid number");
         }
      }
   }

Output 

                         1st run : java RegexDemo6  9439202111
                         valid mobile number
                         2nd run : java RegexDemo6  929291929394
                         invalid number

Java program to check whether a given string is palindrome or not without taking reverse.

import java.util.*;
public class StringPalindrome {

public static void main(String[] args)
{
String str;
int l,c=0;
Scanner s= new Scanner(System.in);
System.out.println("enter a string");
str= s.nextLine();
l=str.length();
for(int i=0;i<(l-1)/2; i++)
{
if(str.charAt(i)==str.charAt(l-1-i))
{
continue;
}
else
{
c++;
}
}
if(c==0)
System.out.println("the given string is palindrome");
else
System.out.println("the given string is not palindrome");

}

}

Output
1st Run:
enter a string
katak
the given string is palindrome

2nd Run:
enter a string
Silan
the given string is not palindrome