1. Abstract class acts as a base class for other classes of same type.
2. Abstract class can’t be instantiated.
3. There can be any abstract class without any abstract method but vice versa is not possible.
4. If we are extending a class from an abstract class then we have to provide definition of all the abstract method in the first concrete class (concrete class means not abstract class).
5. Abstract class can have constructors.
6. Abstract class can have main method also.

Let’s do some practical on Abstract keyword. It will clear our doubts. After all learning through practical implementations will make our concept stronger.

Program 1

class abs1
{
void disp();
}

Output

This program will give a compile time error because it is a simple class program where we need to give definition to function of the class. Here we didn’t define the function of the class. Hence we get compile time error.

Program 2

class abs2
{
abstract void disp();
}

Output

Compile time error.

Reason :- abs 2 should be declared abstract. As I said before, if we declare any method inside the class as abstract then we have to declare the same class as abstract otherwise it will give a compile time error.

Program 3

abstract class abs3
{
abstract void disp();
}

Output

It will compile successfully.

Reason: – Since we have declared both method and class as abstract so there won’t be a problem.

Program 4

abstract class abs4
{
static void show()
{
System.out.println(“Hello”);
}
public static void main(String[] args)
{
show();
}
}

Output

Hello

Reason: – It will print the Hello statement. Here we have used a static keyword with the function, so we didn’t create instance of class. Now check out the program 5, it will make you things more clear.

Program 5

abstract class abs5
{
void show()
{
System.out.println(“Hello”);
}
public static void main(String[] args)
{
new abs5.show();
}
}

Output

Compile time error.

Reason: – We tried to instantiate and its clearly written in the above statement of abstract that Abstract class can’t be instantiated. This program 5 is almost same to program 4. Here we just removed the static keyword. Since the method or function is not static now, so we will need to instantiate to call the method but we are not allowed to instantiate in abstract class. So we get compiler time error.

Program 6

abstract class abs7
{
abstract void disp();
}

class check extends abs7
{
public static void main(String[] ar)
{
new check().disp();
}
}

Output

Compile time error.

Reason: – Here we have created two class check and abs7. abs7 is the base class and abstract class. We are inheriting the abstract class here but we are not defining the abstract method in the derived class. Since we didn’t define the abstract class in the in the derive class it caused a compiler error. If you read the above statement of abstract class its clear written that “If we are extending a class from an abstract class then we have to provide definition of all the abstract method in the first concrete class (concrete class means not abstract class).”

Note: – I will update this post more later.

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1. Super keyword is used to call immediate parent.
2. Super keyword can be used with instance members i.e., instance variables and instance methods.
3. Super keyword can be used within constructor to call the constructor of parent class.

OK now let’s practically implement this points of super keyword.

Check out the difference between program 1 and 2. Here program 2 proofs our first  statement of Super keyword in Java.

Program 1

class base
{
int a = 100;
}

class sup1 extends base
{
int a = 200;
void show()
{
System.out.println(a);
System.out.println(a);
}
public static void main(String[] args)
{
new sup1().show();
}
}

Output : -

200

200

Now check out the program 2 and try to figure out the main difference.

Program 2

class base
{
int a = 100;
}

class sup2 extends base
{
int a = 200;
void show()
{
System.out.println(super.a);
System.out.println(a);
}
public static void main(String[] args)
{
new sup2().show();
}
}

Output

100

200

In the program 1, the output was only of the derived class. It couldn’t print the variable of base class or parent class. But in the program 2, we used a super keyword with the variable a while printing its output and instead of printing the value of variable a of derived, it printed the value of variable a of base class. So it proofs that Super keyword is used to call immediate parent.

OK now check out the difference between program 3 and program 4

Program 3

class base
{
int a = 100;
void show()
{
System.out.println(a);
}
}

class sup3 extends base
{
int a = 200;
void show()
{
System.out.println(a);
}
public static void main(String[] args)
{
new sup3().show();
}
}

Output

200

Here the output is 200. When we called the show function, the show function of derived class was called. But what should we do if we want to call the show function of the parent class. Check out the program 4 for solution.

Program 4

class base
{
int a = 100;
void show()
{
System.out.println(a);
}
}

class sup4 extends base
{
int a = 200;
void show()
{
super.show();
System.out.println(a);
}
public static void main(String[] args)
{
new sup4().show();
}
}

Output

100

200

Here we are getting two output 100 and 200. When the show function of derived class is invoke, it first calls the show function of parent class because inside the show function of derived class we called the show function of parent class by putting the super keyword before the function name.

Ok here’s a program for you. You need to predict the output of this program

class base
{
int a=100;
void show()
{
System.out.println(a);
}
}

class sup5 extends base
{
int a=200;
void show()
{
System.out.println(a);
super.show();
}
public static void main(String[] args)
{
new sup5().show();
}
}

So tell us, what will be the output? So now I think you too know what is super keyword in Java? How and where the super keyword is used in Java? So friends, if you find or face any difficulty in learning this super keyword of Java then let me know.

        

We are going to write total 3 programs to prove it.

Program 1: -

class fin4
{
final int a=10;
void show()
{
System.out.println(a);
}
static public void main(String[] args)
{
new fin4().show();
}
}

If you compile this program, you will get output of 10 because there is no any error in the program but check out the program no. 2 and try finding out the differences.

class fin4
{
final int a;
void show()
{
a=10;
System.out.println(a);
}
static public void main(String[] args)
{
new fin4().show();
}
}

Here, in this program we didn’t assigned any value to the data member a in the first place so we are assigning value of 10 to local variable a.

a=10;

Remember variable a is already declared as final in the first place so the value can’t be assigned to a in any other place except constructor. So we will get an error message that cannot assign value to final variable a. Also check out the screen shot of compilation of this final keyword program 2.

Now there is only one way to assign value to a i.e., within construction. Check out this program 3.

class fin5
{
final int a;
fin5()
{
a=10;
}
void show()
{
System.out.println(a);
}
static public void main(String[] args)
{
new fin5().show();
}
}

Here we have called a constructor

fin5()
{
a=10;
}

and is assigning the value to variable a. There is legal way in final keyword. If you compile this program, you will not get any error. The output will be zero. Also check out the screen shot of compilation and output of the java program.

So now we know that Delay initialization of Final Variable are Possible only within Constructor.

        

Check out this program code including final keyword: -

final class fin2
{

}
class derived extends fin2
{

}

Here we have two classes fin2 and derived. fin2 is a final class and derived is a simple class. They are trying to build inheritance relationship because

class derived extends fin2

Since fin2 is a final class, it cannot be inherited. If you compile this program, there will be an error message. Check out the compilation of this java program below

So now we know that final class can not be inherited in java language.

        

Check out this simple program code: -

class fin1
{
public static void main(String[] ar)
{
final int a=100;
}
}

If you compile this program, it will show no error.

Now check out this program and try to notice the main difference: -

class fin1
{
public static void main(String[] ar)
{
final int a=100;
a=200;
}
}

Now if you compile this program, it will show 1 error. What can be the error. We have already declared variable a as final and have assigned a value = 100. Since we have assigned final variable a = 100 and according to the statement of the final keyword

If we declare any variable as final, it will be constant.

So a = 100 is already constant, it can not be changed but we tried to change by reassigning a new value a = 200. So we got an error message.Check out the screen shot of the compilation below

This programming of the final keyword in Java proofs the statement : -

If we declare any variable as final, it will be constant.

So now you know how to practically proof it.

        

1. Final is a keyword which can be used with variables, methods and classes.
2. If we declare any variable as final, it will be constant.
3. If we declare any method as final, it cannot be overridden.
4. If we declare any class as final, it can’t be inheritance.
5. Final is the only modifier which can be used with local variables.
6. Delay initialization of final variable are possible only within constructor.
7. Final variables will never get default variables.

        

class th2
{
int a, b;
th2(int a, int b)
{
this.a=a;
this.b=b;
}
public void show()
{
System.out.println(a);
System.out.println(b);
}
public static void main(String[] args)
{
//new th2(20,30).show();
th2 t2=new th2(20,30);
t2.show();
}
}

The above program will display the output 20, 30. Check out the screen shot below

We have use this keyword in the constructor “th2″ because we want to refer to the parameter of the constructor not the data members. Due to the use of this keyword

this.a=a;

this.b=b;

The value passed to the parameter of the constructor is called rather than the data members value.

This program practically proved the statement 1 of the this keyword -

“this always refers to current object.”

Note: –

You can instantiate member referance in two ways

th2 t2 = new th2(20, 30);

t2.show();

or

new th2(20,30).show();

        

There will be a series of this program around 4 programs. You will need to read all the 4 this program to understand the concept.

Program 1: – (In this program, I didn’t use this keyword)

class th1
{
int a,b;
th1(int a, int b)
{
a=a;
b=b;
}
public void show()
{
System.out.println(a);
System.out.println(b);
}
public static void main(String[] ar)
{
new th1(10, 20).show();
}
}

Create a particular folder on the drive like myjava and save this file as “th1.java”

Now open command prompt to compile and run the program. You can also click run and type cmd. It will open a command prompt.

If you are going to compile the java program for the first time then you will need to set the path of the java for whenever you open the command prompt. Its very easy process. When you installed the jdk software of any version probably in c drive then it gets a path like it is installed under program files in the c drive. Click the program files and click the java and then click the jdk software and then click the bin. Now in the address bar, check out the path and copy it and paste in the command prompt like I did

c:\java>set path=C:\j2sdk1.4.0\bin

Also check out the screen shot

After that compile the program by writing this code

javac th1.java

If there is no error then run the program by writing the code

java th1

When you compile and run this program, you will get a output of

0

0

as shown in screen shot

Hows the result is 0,0

Here in this program, we have two variables as data members int a and int b. Another is the constructor paramertised variable int a and int b.

In the main method, we passed the value 10 and 20 (  new th1(10, 20).show();) to the respective varible of the constuctor paramter int a and int b.

Now in this statement: -

a=a;
b=b;

the value of constructor int a is replaced by the data member int a.

the value of constructor int b is replaced by the data member int b.

So when the program is run, the output is shown as the defaut value of data members.

        

Keyword is a reserved words which has a special meaning. There are 50 keywords and 3 literals in the java language.

Ok back to the point, the important points of the this keywords are: -

1. this always refers to current object.
2. this keyword can be used to resolve ambiguity between local variables and data members.
3. this keyword cannot be called within static context.
4. this keyword can be used to called the constructor of the same class.

        

When I was attending the class of Java, we are asked by the teacher. Why Java is called java? We don’t have answer to this question. Later the teacher explain us that:-

Once they were deciding on what to call it. They were at a coffee shop drinking coffee, and somehow they thought of naming it java, which is another word for coffee. As coffee from JAVA island, Indonesia is so famous that sometimes coffee is itself called Java in US slang, the name was a good option for a cool and attractive name.

Note: - Java was originally named Oak, but was changed to Java, in hopes of drawing more interest to the new language.

This article explains how Java got it’s name:
http://www.javaworld.com/jw-10-1996/jw-10-javaname.html