Need some help on basics

jagan7 · #1 (**permalink**) July 27th, 2007, 05:49 AM

Hi,
In an java application if we specify the class that is having a main() method as public must be saved with the name same as class name. If we won't specify the class as public and saved with another name it is compiled. what's the reason behind it.
Eg:
//Test.java
public class Test.java{ //here the program compiled
public static void main(String[] args){

//Test.java
public class NewName{ //here program won't compile
public static void main(String[] args){

//Test.java
class NewName{ //suprisingly program compiled!
public static void main(String[] args){

bobit79 · #2 (**permalink**) August 22nd, 2007, 10:33 AM

no problem for compilation,
it will compile.. and will create NewName.class
BUT if u trying run ( Test ) it will produce run time exception...

http://studyjava.org/forums/

ciderpunx · #3 (**permalink**) September 6th, 2007, 11:51 AM

You are allowed to define more than one class in a file, but only one public class,which must be named as the file. So you could if you wanted do:

Code:

//Test.java
public class Test {  // Named as the file (nb .java is not required)
    public static void main(String[] args){ System.out.println("Test"); }
}

class NewName{     //because we're allowed to define other classes in this file
    public static void main(String[] args){ System.out.println("NewName"); /*Won't run*/}
}

Cheers,
Charlie

--
Charlie Harvey's website - linux, perl, java, anarchism and punk rock: http://charlieharvey.org.uk

jdiff · #4 (**permalink**) September 9th, 2007, 02:44 AM

the generated class will be NewName.class and
$>java NewName

will also produce the desired output.

However

$>java Test

will fail at runtime.

Rgds
Jo

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