Computer Science 360 Assignment 3 solution


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Introduction In this assignment you will implement utilities that perform operations on a file system similar to Microsoft’s FAT file system. Sample File Systems You have been given four file system images: disk1.img, disk1_empty.img, disk2.img and disk2_empty.img. You should get comfortable examining the raw data in the file system images using the program xxd. Part I – 5 marks In part I, you will write a program that displays information about the file system. In order to complete part I, you will need to read the file system super block and use the information in the super block to read in the FAT. Your program for part I will be invoked as follows: ./diskinfo disk1_empty.img Sample output: Super block information: Block size: 512 Block count: 15360 FAT starts: 1 FAT blocks: 120 Root directory start: 121 Root directory blocks: 4 FAT information: Free Blocks: 15235 Reserved Blocks: 121 Allocated Blocks: 4 Please be sure to use the exact same output format as shown above.
Part II – 5 marks
In part II, you will write a program that displays the contents of the root directory in the file system. Your program for part II will be invoked as follows: ./disklist disk1.img The first column will contain an F for regular files and a D for directories, followed by a single space. Next you will use 10 characters to show the file size, followed by a single space and then 30 characters for the file name, followed by a single space, followed by the file modification date (we won’t display the file creation date). For example: F 91 readme.txt 2016/03/02 12:20:19 F 56 bar.txt 2016/03/02 12:20:43 F 25600 file3.txt 2016/03/02 12:20:51 F 51200 file4.txt 2016/03/02 12:20:55 F 38512 ls_mac 2016/03/02 12:21:42 Part III – 5 Marks
In part III you will write a program that copies a file from the file system to the current directory in Unix. If the specified file is not found in the root directory of the file system, you should output the message “File not found.” on a single line and exit. Your program for part III will be invoked as follows: ./diskget disk1.img foo.txt
Part IV – 5 marks
In part IV you will write a program that copies a file from the current Unix directory into the file system. If the specified file is not found, you should output the message “File not found.” on a single line and exit. Your program for part IV will be invoked as follows: ./diskput disk1.img foo.txt
Accessing the file system
Unix allows you to treat devices as if they were files, so we can use the stdio functions to manipulate our file system. You will want to become familiar with the library functions: fopen, fseek, fread and fwrite.
What to hand in You need to hand in a .tar file containing all your source code and a Makefile that produces the excutables for parts 1 – 4. Please include a readme.txt file that explains any bonus activities that you completed. Bonus There is lots of room for bonus marks in this assignment. As before, you must get permission from your instructor before you begin the bonus activities. Some things you may consider: – implementing directories other than the root directory – implementing fast searching for directories and/or free space – implementing the file system as a device driver for Linux (!hard!) – writing a shell to interact with the filesystem If you have any other ideas, please email your instructor.
The four programs created have equal weight. Each part is worth 10 points. We will test your implementation against the four disk images we have supplied as well as an additional disk image that meets the specifications given in this assignment.
File System Specification
The file system has 3 major components: the super block, the File Allocation Table (FAT) and the directory structure. Each of these 3 components is describe in the sections below.
File system Superblock
The first block (512 bytes) is reserved to contain information about the file system. The layout of the superblock is as follows:
Description Size Default Value File system identifier 8 bytes CSC360FS Block Size 2 bytes 0x200 File system size (in blocks) 4 bytes 0x00001400 Block where FAT starts 4 bytes 0x00000001 Number of blocks in FAT 4 bytes 0x00000028 Block where root directory starts 4 bytes 0x00000029 Number of blocks in root dir 4 bytes 0x00000008
Directory Entries
Each directory entry takes up 64 bytes, which implies there are 8 directory entries per 512 byte block. Each directory entry has the following structure:
Description Size Status 1 byte Starting Block 4 bytes Number of Blocks 4 bytes File Size (in bytes) 4 bytes Create Time 7 bytes Modify Time 7 bytes File Name 31 bytes unused (set to 0xFF) 6 bytes
This is bit mask that is used to describe the status of the file. Currently only 3 of the bits are used. Bit 0 – set to 0 if this directory entry is available, set to 1 if it is in use Bit 1 – set to 1 if this entry is a normal file Bit 2 – set to 1 if this entry is a directory It is implied that only one of bit 2 or bit 1 can be set to 1. That is, an entry is either a normal file or it is a directory, not both.
Starting Block
This is the location on disk of the first block in the file
Number of Blocks
The total number of blocks in this file
File Size
The size of the file, in bytes. The size of this field implies that the largest file we can support is 2^32 bytes long.
Create Time
The date and time when this file was created. The file system stores the times as integer values in the format: YYYYMMDDHHMMSS 2 bytes used to store YYYY 1 byte to store MM 1 byte to store DD 1 byte to store HH 1 byte to store MM 1 byte to store SS
Modify Time
The last time this file was modified. Stored in the same format as the Create Time shown above.
File Name
The file name, null terminated. Because of the null terminator, the maximum length of any filename is 30 bytes. Valid characters are upper and lower case letters (a-z, A-Z), digits (0-9) and the underscore character (_). File Allocation Table (FAT)
Each directory entry contains the starting block number for a file, lets say it is block X. To find the next block in the file, you should look at entry X in the FAT. If the value you find there does not indicate End-of-File (see below) then that value, call it Y, is the next block in the file. That is, the first block is block X, you look in the FAT table at position X and find the value Y. The second data block is block Y. Then you look in the FAT at position Y to find the next data block… continue this until you find the special value in the FAT indicating that you are at the end of the file. The FAT is really just a linked list, which the head of the list being stored in the directory entry, and the ‘next pointers’ being stored in the FAT. Fat entries are 4 bytes long (32 bits), which implies there are 128 FAT entries per block. Special values for FAT entries are:
Value Meaning 0x00000000 This block is available 0x00000001 This block is reserved 0x00000002 – 0xFFFFFF00 Allocated blocks as part of files 0xFFFFFFFF This is the last block in a file
Byte Ordering
Different hardware architectures store multi-byte data (like integers) in different orders. Consider the large integer: 0xDEADBEEF On the Intel architecture (Little Endian), it would be stored in memory as: EF BE AD DE On the PowerPC (Big Endian), it would be stored in memory as: DE AD BE EF Our file system will use Big Endian for storage. This will make debugging the file system by examining the raw data much easier. This will mean that you have to convert all your integer values to Big Endian before writing them to disk. There are utility functions in netinit/in.h that do exactly that. (When sending data over the network, it is expected the data is in Big Endian format.) See the functions htons, htonl, ntohs and ntohl. The side effect of using these functions will be that your code will work on multiple platforms. (On machines that natively store integers in Big Endian format the above functions don’t actually do anything – but you should still use them!)