COP 3530 - Fall 2001 - Program 2

In this program, we will develop a simulated directory-file name storage system.

Here are the operators:

md(<name>)		creates a new subdirectory with given
                    name in the current directory

cd(<name>)		change current directory to <name>

rd(<name>)		remove directory called <name> and
				delete all of its contents
rn(<old-name, new-name>)  rename command

save(<fname>)		save file named fname in current dir

attrib(<name>, <cmd-str>);  set <name>'s attribute string
      to  cmd-str  (check for valid chars).

del(<name>)         remove file named <name> from cur dir

copy(<new-dir>, <old-name>, <opt-new-name>)  make a copy of
    the file info for <old-name> in new dir (default uses
    same <old-name>, option uses new name. 

move(<new-dir>, <name>, <opt_new-name>)  make a copy of
file called <name> in <new-dir> and remove copy from
orig. dir

find(<name>,<where>)  report on existence of file or dir 
     with given name by displaying the complete path, 
     otherwise display "file not found".  Whenever you
     encounter this command, pause execution and ask me to 
     hit any key to continue.

dir(<opt-dir>,<opt-cmds>)  list contents of directory.  
     opt-cmds are A (for all) S (for recurse through 
     subdirectories).  Without A, do not list hidden files.
     Each file/directory should be listed on one line along 
     with its attribute string.  Whenever you encounter 
     this command, pause execution and ask me to hit any 
     key to continue.

Your directory will be a tree rooted in the "MYCOMP" node.

Each entry will contain the 8 char name, the 3 char extension, 
and an attribute string (up to 5 characters long).

Attribute string may contain the following characters:

F	for file
S	for system
H	for hidden
R	for read only
D	for directory

Duplicate names will not be allowed in any single directory 
by your system but may occur in different directories.

Your system must be able to read a command file and construct 
a corresponding file system tree. 

Tree structure details:

Each tree node will contain data plus four references.
The references will be:

left-most child, 
right-sibling
left-sibling
parent

Sample command file:

md C
cd C
md TEMP
cd TEMP
save GEORGE
save HENRY.TXT
save HENRY.DOC
rn GEORGE GEORGE.JPG
attrib GEORGE.JPG H
cd C
save BILL.JPG
move C C\TEMP\GEORGE.JPG
copy C C\TEMP\HENRY.TXT

Tree at this point will look like:


MYCOMP
  |
  C 
  |
TEMP --- BILL.JPG --- GEORGE.JPG(H) --- HENRY.TXT
  |
HENRY.TXT --- HENRY.DOC


dir C *.* S      (this is next line in input file)

	Output will be
C	(DIR)
	TEMP	(DIR)
		HENRY.TXT	(FILE)
		HENRY.DOC	(FILE)
	BILL.JPG	(FILE)
	HENRY.TXT	(FILE)


dir C *.* AS	   (this is next line in input file)

	Output will be
C	(DIR)
	TEMP	(DIR)
		HENRY.TXT	(FILE)
		HENRY.DOC	(FILE)
	BILL.JPG	(FILE)
	GEORGE.JPG	(HIDDEN FILE)
	HENRY.TXT	(FILE)

find HENRY.TXT		(this is next line in input file)

	Output will be
C:\HENRY.TXT
C:\TEMP\HENRY.TXT

quit		(this is the last line in the input file - just
               tell me to hit any key to quit and when I do you 
               exit your program.)




TREE NODE STRUCTURE outline:

ref Parent (Parent ref)
TreeNode (Left-most child ref)
Data: Name (string8) 
      Ext  (string3) 
	 Atb  (string5) 
ref RSib (Right sib ref)
ref LSib (Left sib ref)

For the final node representing GEORGE.JPG
   Parent is ref to C
   TreeNode is null
   Name is GEORGE
   Ext is JPG
   Atb is FH
   RSib is ref to HENRY.TXT node
   LSib is ref to BILL.JPG node

For the final node representing TEMP
   Parent is ref to C node
   TreeNode is ref to HENRY.TXT
   Name is TEMP
   Ext is null string
   Atb is D
   RSib is ref to BILL.JPG
   LSib is null
   


Save your main class under the name MY_FS.java

Ask for the name of the input command file.

Due date:  October 14, 2001