[Solved] SOLVED:Assignment 3 Linux OS interface and OS responsibilities

Complete the associated exercises before attempting the assignmentAimThis assignment and its associated exercises introduce the services offered by an OS and, through a simple simulation, provide some familiarity with how an OS supports a filesystem on disk.ObjectivesOn completion of this assignment and its associated exercise, you will be able to: Explain how an OS provides functionality through supervisor calls; Correctly use some of the basic Linux supervisor calls for process management and bytelevelI/O; Explain how the Linux file hierarchy can be explored and data on files retrieved; Explain how a simple file system can be implemented; Explain how the Linux i-node structures allow for more sophisticated approaches to the mapping of a file to disk blocks; Link code that you have written with supplied libraries.TaskYou are to implement a simple file management system and make it work with a supplied driver function that will create and delete files using your simulated system.A simple file management systemA file management system is a major component in any operating system. Modern operating systems typically support several different file system structures; you may see references to FAT32, NTFS, HFS, ISO9660, Linux ext3, or Linux ext4. All file systems have to maintain meta-data about files and free-space, and must arrange for the mapping of files to disk blocks.Your simulation program will be modelling a simple contiguous file allocation scheme.This was the kind of file system that was available on the micro-computer operating systems of the late 1970s and early 1980s. These used small exchangeable disks the original IBMPC and Apple computers had 8-inch floppy disks with only ~80kbytes (representing maybe 160 disk blocks); by 1984, the Macintosh had 2.5-inch 400Kbyte disks (maybe 400 disk blocks).These were flat file systems. There was no file hierarchy; instead all files were identified in a single directory. This directory occupied specific blocks at the start of the disk.The directory structure would have had two parts.The first part would be something like an array of structs:struct direntry {char filename[NAMESIZE];int start;int size;};typedef struct direntry Direntry;static Direntry directory[MAXFILES];The essential fields in a directory entry structure are a name (with a fixed maximum length NAMESIZE, which would have been limited to 6 or 8 characters in those days), the start block number, and the number of blocks allocated to the file. Usually, there were additional data fields in entries such as a creation date-time field and often a file type (which would in some way identify the application used to create and manipulate the file). Along with this array of structs, there would be a count for the number of entries in use.Such a directory structure has many limitations. All files must have to have unique names.There is a maximum limit on the number of files that you can have on disk.Another part of the directory structure at the start of the disk would have been a block map.This block map would have been a bit-map one bit for each block on the disk. A zero value would indicate that the block was free; a one value would indicate that the block was allocated to a file.A new disk would be formatted with an empty directory structure on the first disk blocks, and the bit-map entries set to indicate that all the remaining disk blocks were free.The system would provide a createFile function that would take as arguments a file-name and requested size. This function would fail if there were already a file with that name. If the name was unique, the system would try to find a contiguous set of blocks to satisfy the request.Finding free blocks is a matter of searching through the bit-map. Several approaches were used in the different systems, such as: First FitStart at the beginning (starting after the fixed directory blocks at the first block thatcan be allocated to a file) and searching in the bit-map for a sequence of free blocks greater or equal in size to the requested size. Best FitSearch the entire bit-map for the sequence of free blocks whose length equals therequest or exceeds the request by the smallest amount (can stop searching if find a sequence of free blocks that exactly matches the request size). Cyclic(This needs an extra field in the fixed directory structure that stores the blocknumber where the last file was allocated.)The search continues from the last allocation spot searching towards the end ofdisk, then if necessary resumes at the start of the disk until a sufficient sizedsequence of free blocks (greater than or equal to request) is found. (The searchterminates unsuccessfully if it gets back to its starting point.)The createFile request would fail if there was no adequate set of contiguous free blocks.If the space was available, a new entry would be made in the directory recording the filename, start block, and size; the bit-map would be updated to indicate that all the blocks were in use.The operating system would maintain some hidden structure associated with the program that had opened the file. Subsequent readBlock and writeBlock requests would work via this OS structure. The OS would verify that they were valid (not attempting to use a non-existent file-block) and would work out the actual disk block that was to be used for the transfer.(The integer file-descriptor used on Linux is an index into a table of such file access structures.)The corresponding deleteFile request would update the bit-map marking the blocks as again free and remove the directory entry. Typically, the directory would be re-written with all entries that had been after the file that was deleted being moved up one slot.A problem with this simple allocation approach is that it leads to fragmentation of the free space. There may be many free blocks on disk, more than enough for the next file request,but these blocks are all separated by blocks that are allocated to other files.Such file systems had to provide compaction operations. Compaction would be invoked at user command. The compaction process would shift files on disk, copying the data blocks and updating the directory entries as it worked. When the compaction process was completed, all the free space would be in one contiguous region.Your simulation overviewYour simulation is to be built as a NetBeans C application project.The main() function will end up being quite simple:(You dont need to start with this main, you can start with code that directly invokes the functions that you are developing for your file management component Disk.c).It is intended that once your Disk.c code works that you use driver function that are in a supplied .a archive library. The functions in this library will make sequences of create, read, write, delete, compact requests to your file system implementation.The file libfilesystemdriver2.a can be downloaded from the web site; it is compiled for the Ubuntu workstations in the laboratory.Why sleep(1)?Sometimes, when a C program terminates NetBeans will close its connection to the process before all of the output text has been flushed from buffers and displayed in NetBeans output pane.There are three functions in the driver that you can invoke: Run Simple SimulationMakes a small number of create and delete requests; then invokes functions (thatyou supply) that show the contents of the directory and give some details of howfiles are mapped to disk blocks. Run Some Error TestsMakes many requests to your implementation including invalid requests likecreate a file with a duplicate name, delete a file that was never created, writedata to a non-existent block of file. It will also do things like force compaction offiles and then check that the data blocks have been correctly copied. Run SimulationA thorough test with a couple of thousand requests to your file system.The following outputs are from my version. Your outputs should be similar but you can invent your own alternative approach to summarising the contents of the file directory and showing the mapping of files to disk blocks.I chose to output the directory as a tabular listing with file-name, start block, and size. I chose to have an extra field in my directory entries an identifier. My display of block allocations has a . for a free block, and the identifier letter of a file if the block is part of a file.(An identifier is just one of the letters [A-Za-z] allocated sequentially. My identifier was not unique once I got to z, I started over at A again. But this is rarely going to cause any confusion.)Following the display of file mapping to disk blocks, there are some summary statistics. You simply need to maintain counts of create and delete requests, the number of compactions, the number of files currently in the directory, and the total number of free blocks on disk.Run Simple SimulationRun Some Error TestsRun SimulationDetails, details, Your file system must support the following functions (you must use this Disk.h header file exactly as is given, the driver file has been compiled using these function definitions):#ifndef DISK_H#define DISK_H#define MAXFILES 64#define DISKBLOCKS 512enum filesyserrors { NO_ERR, DIRECTORY_FULL, CREATE_FAIL,NON_EXISTENT_FILE, DUPLICATE_NAME, INVALID_BLOCK,ZERO_SIZE };typedef enum filesyserrors FileSysErrors;void initialiseFileSystem();// createFile returns DIRECTORY_FULL if maxfiles used// Returns DUPLICATE_NAME if there is already a file with the name// Returns ZERO_SIZE if size <= 0// Returns CREATE_FAIL if unable to find space for the fileFileSysErrors createFile(const char* filename, int size);// deleteFile returns NON_EXISTENT_FILE if filename is invalidFileSysErrors deleteFile(const char* filename);// readBlock and writeBlock will return NON_EXISTENT_FILE if// filename is invalid, and INVALID_BLOCK if requested block// is invalid (<0, or = size of file)// blocks would have 512bytes but for simulation content of block represented by// single integer valueFileSysErrors writeBlock(const char* filename, int block, int value);// Read block stores value read in integer whose address is passed as// the third argumentFileSysErrors readBlock(const char* filename, int block, int* vp);// compactFiles returns number of free blocks should be all in one large group// at end of disk after compactionint compactFiles();// Display disk show directory contents and provide some form of map// identifying the mapping of files to disk blocksvoid displayDisk();// show history// number of files created, deleted, number of entries in directory,// number of blocks still free, number of compactions performedvoid showHistory();#endif /* DISK_H */This file system uses a small disk! It has a maximum of 64 entries in the directory(MAXSIZE). There are 512 blocks available for allocation to user files (DISKBLOCKS).(The directory doesnt occupy any space in this simulated version.)The simulation driver does write to blocks, and then read data back to check that things work.The compaction process will move everything around but subsequent reads should still get back the same data as was written before the compaction.In reality, blocks would have 512bytes (or more). But for the purposes of the simulation, you need use only an array of integers When you write to a block of file, a particular element is set in diskblockcontent[]. When you read, you get back the value from diskblockcontent[].The simulation is handled by the functions (functions returning FileSysErrors will return NO_ERR if things worked):void initialiseFileSystem();You initialise all the data structures that you will be using. (They may as well all be static arrays).FileSysErrors createFile(const char* filename, int size);All the file operations will use the file name. (Use #define NAMESIZE 32; names should all be shorter than 32 characters).FileSysErrors deleteFile(const char* filename);Removes the named file and tidies up.FileSysErrors writeBlock(const char* filename, int block, int value);Using data from the directory for this file, work out the actual disk block corresponding to block of file zero based array from file start. Write value into diskblockcontent for that block.FileSysErrors readBlock(const char* filename, int block, int* vp);Again, work out which entry in diskblockcontent is to be used, and copy the value from there into the integer variable whose address is passed in vp.int compactFiles();Reorganise the contents of your disk and update your directory.The display functions that you must also implement are:void displayDisk();Display the contents of your directory and provide some visual display of how files are mapped to blocks on disk.void showHistory();Provide summary of calls to your simulated system.Your implementationYou must define appropriate data structures and implement the functions as shown above.You will need several private auxiliary functions to handle details.My implementation was:The entries are the functions listed above compactFiles, createFile, deleteFile,displayDisk, initialiseFileSystem, readBlock, showHistory, writeBlock.The entries correspond to auxiliary functions that I added when implementing the main functions. For example, my displayDisk invokes listDirectoryAlphabetically to list the directory entries in alphabetic order.You do not need to implement the bit-map as a bit map! For this simulation, it is acceptable to use an integer array with 0/1 values to indicate whether a block is in use. (Using an actual bit-map just entails some fussy code to convert a block number into a byte identifier and a bit identifier for an array of unsigned bytes.)The hardest functions are likely to be createFile and compactFiles.My createFile function utilized some helper functions:My findspace function returns ve if it cannot find a set of contiguous free blocks of sufficient size; or returns the start block number. It uses nextfreeblock to find the next individual block that is free, and freeblocksize to determine how many blocks were available at a given point.The compactFiles function just careful house-keeping in some while-loops. An example:Directory listing:Filename Start Size (symbol)1 data4.c 44 4 G2 data5.gif 80 7 I3 hello6.txt 63 6 u4 hello7.jpg 24 4 N5 log11.gif 108 22 M6 log9.exe 28 8 p7 note1.o 56 4 B8 prog1.exe 49 7 h9 prog2.pdf 91 4 k10 report6.jpg 0 4 K11 test2.jpg 11 7 c12 test3.txt 39 5 f13 tmp1.txt 5 5 x14 tmp3.gif 95 13 H15 var4.pdf 18 6 JKKKK.xxxxx.cccccccJJJJJJNNNNppppppppfffffGGGG.hhhhhhhBBBBuuuuuu..IIIIIII.kkkkHHHHHHHHHHHHHMMMMMMMMMMMMMMMMMMMMMM…1) Move tmp1.txt (x) one blockKKKKxxxxx..cccccccJJJJJJNNNNppppppppfffffGGGG.hhhhhhhBBBBuuuuuu..IIIIIII.kkkkHHHHHHHHHHHHHMMMMMMMMMMMMMMMMMMMMMM..2) Move test2.jpg (c) two blocksKKKKxxxxxccccccc..JJJJJJNNNNppppppppfffffGGGG.hhhhhhhBBBBuuuuuu..IIIIIII.kkkkHHHHHHHHHHHHHMMMMMMMMMMMMMMMMMMMMMM..3) Move var4.pdf (J) two blocks4) Move hello7.jpg (N) two blocks5) Move log9.exe (p) two blocks6) Move test3.txt (f) 5 blocksKKKKxxxxxcccccccJJJJJJNNNNppppppppfffff..GGGG.hhhhhhhBBBBuuuuuu..IIIIIII.kkkkHHHHHHHHHHHHHMMMMMMMMMMMMMMMMMMMMMM..The start block field in the directory entry for each file is updated as the process continues; so for this partially completed compaction process one would get:Directory listing:Filename Start Size (symbol)1 data4.c 44 4 G2 data5.gif 80 7 I3 hello6.txt 63 6 u4 hello7.jpg 22 4 N5 log11.gif 108 22 M6 log9.exe 26 8 p7 note1.o 56 4 B8 prog1.exe 49 7 h9 prog2.pdf 91 4 k10 report6.jpg 0 4 K11 test2.jpg 9 7 c12 test3.txt 34 5 f13 tmp1.txt 4 5 x14 tmp3.gif 95 13 H15 var4.pdf 16 6 JYou will probably find it useful to have your compactFiles function start by creating a list of current directory entries ordered by start block.Building your programYou must copy the supplied library archive file into the directory for your NetBeans project.You must edit the projects properties. You need to specify additional libraries for the linking step.Right-click on the project and select properties.In the dialog select linker.Select Add Library File. Find the library.You will also need the maths library. The dialog used to add libraries has a special option for standard libraries:The program should build. The linker report should show linkage first with the .a library and then with -lm the maths library.Your implementation strategy1. Define the structs needed to represent your directory2. Write some code that to place entries in the directory (test code, will get changed later) and to display directory contents. You will need to see what is happening in your directory.3. Define your bit-map representation and create some form of display function that can indicate whether blocks are free or allocated. Test your display.4. Define a simple createFile function. New files will simply be allocated the next set of blocks. Make this simplified version work with your directory and bit map displays.5. Define a simple deleteFile function. Write a driver to create several files, delete a subset, and check that the results all display correctly.6. Implement a more realistic createFile function this one must now search for the appropriate place to create a new file (as deleteFile will have opened up gaps where a file might fit). Test your functions and your display of the results.7. You could now probably use the runSimpleSimulation() function from the supplied library file.8. Implement compactFiles. Test your code and display the results.9. Implement the read and write operations. (Your compactFiles function had better remember to update diskFileContents[] when it moves those files.)10. Try running with the runSomeErrorTests() function. If it complains, check your code.11. Run with the runSimulation() function.Assignment reportYou do not submit your code files!You are marked on a report that you write. This report presents details of your Disk.cimplementation, and demonstrates that it works with the supplied driver.