What is the difference between abstraction and encapsulation?

• Abstraction focuses on the outside view of an object (i.e. the interface) Encapsulation (information hiding) prevents clients from seeing it’s inside view, where the behavior of the abstraction is implemented.
• Abstraction solves the problem in the design side while Encapsulation is the Implementation.
• Encapsulation is the deliverables of Abstraction. Encapsulation barely talks about grouping up your abstraction to suit the developer needs.

What is the difference between JDK and JRE?

What is the difference between JDK and JRE?

The JRE is the Java RunTime Environment that is a plug-in needed for running java programs. The JRE is an implementation of the Java Virtual Machine which actually executes Java programs.

The JDK is the Java Development Kit for Java application developers. The JDK is bundle of software which contains one (or more) JRE's along with the various development tools like the Java source compilers, bundling and deployment tools, debuggers, development libraries, etc.

Spring Tutorial

Spring interview question: http://www.developersbook.com/

http://static.springsource.org/docs/Spring-MVC-step-by-step/overview.html#overview-whats-covered

http://java9s.com/

http://www.theserverside.com/news/1364527/Introduction-to-the-Spring-Framework

String pool in java

public class DemoStringCreation {

  public static void main (String args[]) {
    String str1 = "Hello"; 
    String str2 = "Hello";
    System.out.println("str1 and str2 are created by using string literal.");   
    System.out.println("    str1 == str2 is " + (str1 == str2)); 
    System.out.println("    str1.equals(str2) is " + str1.equals(str2)); 

   
    String str3 = new String("Hello"); 
    String str4 = new String("Hello");
    System.out.println("str3 and str4 are created by using new operator.");   
    System.out.println("    str3 == str4 is " + (str3 == str4)); 
    System.out.println("    str3.equals(str4) is " + str3.equals(str4)); 
   
    String str5 = "Hel"+ "lo"; 
    String str6 = "He" + "llo";
    System.out.println("str5 and str6 are created by using string

constant expression.");   
    System.out.println("    str5 == str6 is " + (str5 == str6)); 
    System.out.println("    str5.equals(str6) is " + str5.equals(str6));

    String s = "lo";
    String str7 = "Hel"+ s; 
    String str8 = "He" + "llo";
    System.out.println("str7 is computed at runtime.");          
    System.out.println("str8 is created by using string constant

expression.");   
    System.out.println("    str7 == str8 is " + (str7 == str8)); 
    System.out.println("    str7.equals(str8) is " + str7.equals(str8));
   
  }
}

The output result is:

str1 and str2 are created by using string literal.
    str1 == str2 is true
    str1.equals(str2) is true
str3 and str4 are created by using new operator.
    str3 == str4 is false
    str3.equals(str4) is true
str5 and str6 are created by using string constant expression.
    str5 == str6 is true
    str5.equals(str6) is true
str7 is computed at runtime.
str8 is created by using string constant expression.
    str7 == str8 is false
    str7.equals(str8) is true

The creation of two strings with the same sequence of letters without the use of the new keyword will create pointers to the same String in the Java String literal pool. The String literal pool is a way Java conserves resources.

REF:

http://www.xyzws.com/Javafaq/what-is-string-literal-pool/3

http://www.dreamincode.net/forums/topic/187631-the-string-pool/

Interview Question -http://www.java-questions.com

Reference for   Interview Question
http://www.java-questions.com

Calculate factorial and power(x,n) function‎ Using Recursion

power(x,n) function:
public int raisePower( int base, int power )
{
if (power == 1)
{
return base;
}
else
{
base *= raisePower( base,power - 1 );
}

return base;
}

Calculate factorial :

public int factorial(int n)
    {
        if (n == 1) {
            return n;
        }
        else {
            return n * factorial(n - 1);
        }
    }

What is deadlock in Java ? How to fix it ?

What is deadlock in Java ? How to fix it ?

1digg

This is one of the question which is flavor of the season for multithreading , asked more at a senior level and with lots of follow up questions , though question looks very basic but most of developer get stuck once you start going deep.

questions starts with "What is deadlock ?"
answer is simple , when two or more threads waiting for each other to release lock and get stuck for infinite time , situation is called deadlock . it will only happen in case of multithreading.

How do you detect deadlock ?
though this could have many answers , my version is first I would look the code if I see nested synchronized block or calling one synchronized method from other or trying to get lock on different object then there is good chance of deadlock if developer is not very careful.

other way is to find it when you actually get locked while running the application , try to take thread dump , in linux you can do this by command "kill -3" , this will print status of all the thread in application log file and you can see which thread is locked on which object.

other way is to use jconsole , jconsole will show you exactly which threads are get locked and on which object.

once you answer this , they may ask you to write code which will result in deadlock ?
here is one of my version

public void method1(){
synchronized(String.class){
System.out.println("Aquired lock on String.class object");

synchronized (Integer.class) {
System.out.println("Aquired lock on Integer.class object");
}
}
}

public void method2(){
synchronized(Integer.class){
System.out.println("Aquired lock on Integer.class object");

synchronized (String.class) {
System.out.println("Aquired lock on String.class object");
}
}
}

If method1() and method2() both will be called by two or many threads , there is a good chance of deadlock becuase if thead 1 aquires lock on Sting object while executing method1() and thread 2 aquires lock on Integer object while executing method2() both will be waiting for each other to release lock on Integer and String to proceed further which will never happen.

now interviewer comes to final part , one of the most important in my view , How to fix deadlock ?

if you have looked above code carefully you may have figured out that real reason for deadlock is not multiple threads but the way they access lock , if you provide an ordered access then problem will be resolved , here is
the fixed version.

public void method1(){
synchronized(Integer.class){
System.out.println("Aquired lock on Integer.class object");

synchronized (String.class) {
System.out.println("Aquired lock on String.class object");
}
}
}

public void method2(){
synchronized(Integer.class){
System.out.println("Aquired lock on Integer.class object");

synchronized (String.class) {
System.out.println("Aquired lock on String.class object");
}
}
}

Now there would not be any deadlock because both method is accessing lock on Integer and String object in same order . so if thead A aquires lock on Integer object , thread B will not proceed until thread A releases Integer lock , same way thread A will not be blocked even if thread B holds String lock because now thread B will not expect thread A to release Integer lock to proceed further.

hope this would be useful.