For Windows OS users
Before we begin with a java program , you need to have following softwares :
- The Java SE Development Kit 6 (JDK 6) : If you do not have it ,then don’t worry .Download it from here .download
- A text editor : Notepad is sufficient .
Follow it upon installing jdk :
set PATH variable
- Right-click on My Computer icon.
- Select properties.
- Choose Advanced option.
- Select Environment variables option .
- Click on New .
- type PATH in variable name field.
- type path to the (bin folder of jdk) in the variable value field.
for example
Suppose jdk1.6.0_<version> is installed in C:\program files\Java\
Then you would type the following in the variable value field:
C:\program files\Java\ jdk1.6.0_<version>\bin;
- Finally click Ok.
Now, you are ready to do some programming in Java.
Please read it carefully .
All the Best ….
Follow the procedure :
1. Create a source file (i.e <filename>.java)
- Open Notepad.
- Type the following code :
class Sample
{
public static void main(String[] args)
{
System.out.println(”This is my first program”);
}
}
- Save the file as Sample.java in any directory.Let the directory be C:\Java\
Note:
* File name must be same as the name of the class which contains the main method .
* Since Java is case sensitive ,please be careful about the Lowercase and Uppercase letters.
2.Compiling the Source file (i.e <filename>.java)
- Click on start
- Press run
- Type cmd
- press enter key
- Now,go to the directory where you have saved java file.
In above example ,we saved Sample.java in C:\Java\
So,we have
C:\Java>
- Now type the following command to compile the program:
javac <filename>.java
Here, javac is java compiler that translates source file into a class file which contains the bytecode that the Java Virtual Machine (JVM) can understand.
i.e javac Sample.java for above example.
So,overall it looks like :
C:\Java>javac Sample.java
- press enter key
- Hence compilation is complete.
Now , what happens upon compilation ??
- After compilation ,open the directory where you have saved java file.
- Now,you will see a file Sample.class along with the Sample.java file.
- This class file contains the bytecodes corresponding to source file.
3.Run the program
- To run Sample.java , we have the following command :
java Sample
Here, java is java interpreter.
- On command prompt , we have :
C:\Java>java Sample
- It produces the following output :
This is my first program
Program Explanation
class Sample
{
public static void main(String[] args)
{
System.out.println(”This is my first program”);
}
}
Class Definition
- The first line
class Sample
declares a class which is an object oriented construct.
- Everything must be within a class.
- class is a keyword.
- Sample is a Java identifier that specifies the name of the class to be defined.
Braces { }
- Every class definition in Java begins with an opening brace “{” and ends with a matching closing brace “}” .
The main Line
- The third line
public static void main (String [] args)
defines a method named main.
- This is starting point for java interpreter to begin the execution of the program .
- A java program can have any number of classes ,but only one class must include a main method to initiate the execution.
- Above line contains a number of keywords :
public : It is one of the access modifier. Here ,in the main Line ,it allows JVM to access the main method. It should always be present in the main Line.
static : The main method must be always be declared as static since the interpreter uses this method before any objects are created to invoke the main method.
void : The type modifier void states that the main method does not return any value.
The Output Line
- The only executable statement in the program is
System.out.println(”This is my first program”);
This is similar to printf() statement of C .
or cout<< construct of C++.
- The method println always appends a newline character to the end of the string.
- This means that any subsequent output will start on a new line.
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