Hi Mrinal
When you are in command line (in a DOS window for e.g.) the "javac" is
able to compile any file in any directory you wish, as long as you
properly specify the absolute or relative directory, according to the
DOS rules.
For e.g. the file D:\foo\bar\pack\Point.java compiles :
-- either with "javac D:\foo\bar\pack\Point.java compiles"
-- or with "javac Point.java" if the current directory is "D:\foo\bar\pack\"
-- or with "javac pack/Point.java" if the current directory is "D:\foo\bar\"
However, if your .java file contains references to other classes, then
javac tries to resolve dependencies according to the Java rules for
classes and packages.
By default, javac considers that the "CLASSPATH" is the current
directory (i.e. ".").
If "Line.java" is in "D:\foo\bar\shapes\" and extends "Point" (that is
not in a package), you might wish to make the current directory
"D:\foo\bar\pack\" (cd D:\foo\bar\pack) and compile Line.java using
"javac D:\foo\bar\shapes\Line.java". In this way, the javac compiler
finds the file to compile by the full name and the referenced class
"Point" as the file Point.java or Point.class can be found in the
current directory.
If the class "Point" is in a package and is referenced as "pack.Point".
Then you migh like to "cd D:\foo\bar", then "javac
D:\foo\bar\shapes\Line.java". In this case the javac compile searches
for a "pack.Point" starting from the current directory (i.e. looks for a
sub-directory "pack" and a file "Point.java" ou "Point.classé in this
sub-directory).
You can also play with the CLASSPATH system variable or with "-cp"
command line option, but you will get lost easily if you don't respect
some simple discipline:
1./ Make a root directory for your project, such as "D:\foo\bar\prj1".
2./ Put all your .java file in sub-directories corresponding to the
package name. If "Point" is in the package "pack1" and "Line" is the
package "pack2", then you must have the files
"D:\foo\bar\prj1\pack1\Point.java" and "D:\foo\bar\prj1\pack2\Line.java".
3./ For compiling purposes, make the current directory to be your
project root: "cd D:\foo\bar\prj1".
4./ Compile using the package name as relative directory name: "javac
pack1/Point.java" and "javac pack2/Line.java".
And everything becomes simple!
Unless you wish to do it more sophisticated!
The compiler put the .class files in the same directory as the original
.java files unless you use the -d option to tell him where to put the
files. It is a good practice to separate the sources from the compiled
classes (you might wish to deliver to the customer only the compiled
code, not the sources and it would be easier to separate the file if
they are in different directories from the beginning).
In this case:
1./ Your project root directory might have two sub-directories, one for
sources, one for classes: "D:\foo\bar\prj1\src" and
"D:\foo\bar\prj1\cls" (for e.g.).
2./ You would have your source files in sub-sub-directories
corresponding to the package names:
"D:\foo\bar\prj1\src\pack1\Point.java" and
"D:\foo\bar\prj1\src\pack2\Line.java".
3./ Make the current directory to be the project root "cd
D:\foo\bar\prj1" (just like above - no change at this step).
4./ Compile using the "-d" option (specifying the destination directory
for the ".class" files) and the "-cp" option (specifying where to start
looking for referenced files - pack1.Point can no longer by found
starting from the current directory) :
javac -cp ./src;./cls -d ./cls src/pack1/Point.java
javac -cp ./src;./cls -d ./cls src/pack2/Line.java
As you can notice:
-- Of course, you must insert the "src/" before the name of the package
(so that the relative file name to be properly constructed).
-- The compiler will automatically create the "./cls/pack1" and
"./cls/pack2" sub-sub-directories (from "./cls", as specified by the
"-d" option).
-- When compiling the "Line.java", the compiler have to look for
"pack1.Point", i.e. for a sub-directory "pack2" (starting from a point
in the classpath) containing a file Point.class our Point.java. As the
parameter "-cp" gives a list of places to start looking from, javac will
first look for some file "./cls/pack1/Point.class" or
"./cls/pack1/Point.java", then for "./src/pack1/Point.class" or
"./src/pack1/Point.java", whichever can be found first.
-- Thus way of doing allows to compile file in any order (exactly as in
the "simple" case).
It seems a little complicated, but:
-- It is like bycicle: once you have understood how it works you never
forget.
-- Advanced tools like NetBeans and Eclipse can do this automatically -
you don't need to bother (although you can be sure to be able to do what
a machine can do - programmers are smarter then their tools).
-- Such a simple discipline (combined with the use of .jar files) allow
to manage sources and binaries of huge applications (using several
thusend of classes).
I have a train to catch, so I am in a sort of hurry. So I didn't test
all my sayings. It is possible to have some errors, but if you
understood the idea, debugging is a good exercice :-)
Hope it helps.
Mihai
On 06/09/2012 12:14, Mrinal Roy wrote:
Hi,
I have a Point.java file and Line.java file in package named
"roymrinal_2dShape" package. Line class in line.java expends Point
class; Point.java compiles with no error.
But Line.java produces error as bellow:
bad class file: D:\Working Data\Quellcode\roymrinal_2dShape\Point.class
class file contains wrong class: roymrinal_2dShape.Point
Please remove or make sure it appears in the correct subdirectory of
the classpath.
public class Line extends Point {
Please guide me : what is the correct directory to place the 2 class
code files?
Regards
Mrinal
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