I know that file system use clusters (n x sectors (512 B) usualy 4KB in size) for storing files. If I have file of size 5 KB then it use two cluster to store and remaining space is called slack space. My question is related to situation where user read file from disk, modify (add few characters) and save this file again. What will happened, will OS (overwrite) write file from location from it started to read file, or file will be writen in new cluster completely, and address of file starting cluster will be erased and replaced with new cluster address.
new part:
I just read in a book "Information technologie:An Introduction for Today’s Digital World" that if file use 2 bloks (clusters) and second file use 4 consecutive blocks after first file. First file is edited and modified, his file size increased to total of 3 blocks. This file will be writen after second file and previously occupied 2 blocks are free. But still don t know what will happend if I for example increase file with one character and file is still smaller then total of 2 blocks. Will this data be added on existing file, to existing first two blocks, or it will be stored on new disk physical location (new 2 blocks)?
When user store file it will occupy some space on disk (cluster = combine several sectors = 4 KB since sector is usually 512 bytes). If file take 3KB then 1KB stay unused in this cluster. Now what will happened if I increase little file adding some data to this file. Answer now depend of procedure that user use to modify file.
1. If I manualy add data to file (using echo "some text" >> filename) this data will add this data in existing cluster since there is 1KB of space availabile. If file site increase it will take another free sectors (file use "extents" to address all this sectors)
2. If i use text editor it will copy file on other location on disc (because of multiuser and situation when two users access same file in a same time). Previous location will be "free" (content in sector stay but File system don t have reference to that) and replace with new location on disk.
Since majority of users use some editor for editing file then scenario 2 is most common.
Related
I have to read a large. Sql file (50 GB) in Java and write the content in a database. I tried it with small files (200 MB), it works but only for the first file, by the second file it becomes too slowly and doesn't terminate correctly (OOM Java heap space) I changed the xmx to 6144m but it still slowly by the first file. Can I refresh the memory after every iteration.
I am working with the EXT2 File System and spent the last 2 days trying to figure out how to create a symbolic link. From http://www.nongnu.org/ext2-doc/ext2.html#DEF-SYMBOLIC-LINKS, "For all symlink shorter than 60 bytes long, the data is stored within the inode itself; it uses the fields which would normally be used to store the pointers to data blocks. This is a worthwhile optimization as it we avoid allocating a full block for the symlink, and most symlinks are less than 60 characters long"
To create a sym link at /link1 to /source I create a new inode and say it gets index 24. Since it's <60 characters, I placed the string "/source" starting at the i_block[0] field (so printing new_inode->i_block[0] in gdb shows "/dir2/source") and set i_links_count to 1, i_size and i_blocks to 0. I then created a directory entry at the inode 2 (root inode) with the properties 24, "link1", and file type EXT2_FT_SYMLINK.
A link called "link1" gets created but its a directory and when I click it it goes to "/". I'm wondering what I'm doing wrong...
A (very) late response, but just because the symlink's data is in the block pointers that doesn't mean the file size is 0! You need to set the i_size field in the symlink's inode equal to the length of the path
Let me explain clearly.
The following is my requirement:
Let's say there is a command which has an option specified as '-f' that takes a filename as argument.
Now I have 5 files and I want to create a new file merging those 5 files and give the new filename as argument for the above command.
But there is a difference between
reading a single file and
merging all files & reading the merged file.
There is more IO (read from 5 files + write to the merged file + any IO our command does with the given file) generated in the second case than IO (any IO our command does with the given file) generated in the first case.
Can we reduce this unwanted IO?
In the end, I really don't want the merged file at all. I only create this merged file just to let the command read the merged files content.
And to say, I also don't want this implementation. The file sizes are not so big and it is okay to have that extra negligible IO. But, I am just curious to know if this can be done.
So in order to implement this, I have following understanding/questions:
Generally what all the commands (that takes the filename argument) does is it reads the file.
In our case, the filename(filepath) is not ready, it's just an virtual/imaginary filename that exists (as the mergation of all files).
So, can we create such virtual filename?
What is a filename? It's an indirect inode entry for a storage location.
In our case, the individual files have different inode entries and all inode entries have different storage locations. And our virtual/imaginary file has in fact no inode and even if we could create an imaginary inode, that can only point to a storage in memory (as there is no reference to the storage location of another file from a storage location of one file in disk)
But, let's say using advanced programming, we are able to create an imaginary filepath with imaginary inode, that points to a storage in memory.
Now, when we give that imaginary filename as argument and when the command tries to open that imaginary file, it finds that it's inode entry is referring to a storage in memory. But the actual content is there in disk and not in the memory. So, the data is not loaded into memory yet, unless we read it explicitly. Hence, again we would need to read the data first.
Simply saying, as there is no continuity or references at storage in disk to the next file data, the merged data needs to be loaded to memory first.
So, with my deduction, it seems we would at least need to put the data in memory. However, as the command itself would need the file to be read (if not the whole file, at least a part of it until the commands's operation is done - let it be parsing or whatever). So, using this method, we could save some significant IO, if it's really a big file.
So, how can we create that virtual file?
My first answer is to write the merged file to tmpfs and refer to that file. But is it the only option or can we actually point to a storage location in memory, other than tmpfs? tmpfs is not option because, my script can be run from any server and we need to have a solution that work from all servers. If I mention to create merged file at /dev/shm in my script, it may fail in the server where it doesn't have /dev/shm. So I should be able to load to memory directly. But I think normal user will not have access to memory and so, it seems can not be done without shm.
Please let me know your comments and also kindly correct me if my understanding anywhere is wrong. Even if it is complicated for my level, kindly post your answer. At least, I might understand it after few months.
Create a fifo (named pipe) and provide its name as an argument to your program. The process that combines the five input files writes to this fifo
mkfifo wtf
cat file1 file2 file3 file4 file5 > wtf # this will block...
[from another terminal] cp wtf omg
Here I used cp as your program, and cat as the program combining the five files. You will see that omg will contain the output of your program (here: cp) and that the first terminal will unblock after the program is done.
Your program (here:cp) is not even aware that its 1st argument wtf refers to a fifo; it just opens it and reads from it like it would do with an ordinary file. (this will fail if the program attempts to seek in the file; seek() is not implemented for pipes and fifos)
I'm currently trying to code a FAT system in C on a Xillinx Kintex 7 card. It is equipped with a MicroBlaze and I've already managed to create most of the required functions.
The problem I'm facing is about the total capacity of a folder, I've read on the web that in FAT32 a folder should be able to contain more than 65 000 files but with the system I've put in place I'm limited to 509 files per folder. I think it's because of my comprehension of the way FAT32 works but here's what I've made so far:
I've created a format function that writes the correct data in the MBR (sector 0) and the Volume ID (sector 2048 on my disk).
I've created a function that writes the content of the root directory (first cluster that starts on sector 124 148)
I've created a function that writes a new folder that contains N files of size X. The name of the folder is written in the root directory (sector 124148) and the filenames are written on the next cluster (sector 124212 since I've set cluster size to 64 sectors). Finally, the content of the files (a simple counter) is written on the next cluster that starts on sector 124276.
Here, the thing is that a folder has a size of 1 cluster which means that it has a capacity of 64 sectors = 32KB and I can create only 512 (minus 2) files in a directory! Then, my question is: is it possible to change the size of a folder in number of cluster? Currently I use only 1 cluster and I don't understand how to change it. Is it related to the FAT of the drive?
Thanks in advance for your help!
NOTE: My drive is recognized by Windows when I plug it in, I can access and read every file (except those that exceed the 510 limit) and I can create new files through the Windows Explorer. It obviously comes from the way I understand file creation and folder creation!
A directory in the FAT filesystem is only a special type of file. So use more clusters for this "file" just as you would with any other file.
The cluster number of the root directory is stored at offset 0x2c of the FAT32 header and is usually cluster 2. The entry in the cluster map for cluster 2 contains the value 0x0FFFFFFF (end-of-clusters) if this is the only cluster for the root directory. You can use two clusters (for example cluster 2 and 3) for the root directory if you set cluster 3 in the cluster map as the next cluster for cluster 2 (set 0x00000003 as value for the entry of cluster 2 in the cluster map). Now, cluster 3 can either be the last cluster (by setting its entry to 0x0FFFFFFF) or can point in turn to another cluster, to make the space for the root directory even bigger.
The clusters do not need to be subsequent, but it usually has a performance gain on sequential reading (that's why defragmenting a volume can largly increase performance).
The maximum number of files within a directory of a FAT file system is 65,536 if all files have short filenames (8.3 format). Short filenames are stored in a single 32-byte entry.
That means the maximum size of a direcotry (file) is 65,536 * 32 bytes, i.e. 2,097,152 bytes.
Short filenames in 8.3 format consists of 8 characters plus optional a "." followed by maximum 3 characters. The character set is limited. Short filenames that contain lower case letters are additionally stored in a Long File Name entry.
If the filename is longer (Long File Name), it is spread over multiple 32-byte long entries. Each entry contains 13 characters of the filename. If the length of the filename is not a multiple of 13, the last entry is padded.
Additionally there is one short file name entry for each Long File Name entry.
2 32-byte entries are already taken by the "." and ".." entries in each directory (except root).
1 32-byte entry is taken as end marker?
So the actual maximum number of files in a directory depends on the length of the filenames.
I am working on user level filesystem using FUSE and my requirement is that :
When I issue read for File A, I want to superimpose the contents of another file (say File B) and present the contents of File B as File A's contents.
I have already achieved it by buffer modifications by capturing it in my fuse read and internally reading File B and copying the buffer contents to the passed in buffer for File A and not doing any actual read call for File A. So, File A call returns with File B's contents copied in its buffer.
Also, File A is of smaller size compared to File B.
When checked using debugger, File A buffer contents look fine (contains whole of File B contents), but when it gets displayed (say with Vi) for File A, I am able to see only those many characters as the File A's size, but as File B size is more, the whole data never gets shown even if the returned buffer to File A (the File B data copied) has more to display. This is because File A size is smaller and the display terminates when character count is reached for File A's filesize.
I tried looking into struct stat, but it is a read-only thing which shows me the size of File A which is smaller compared to File B.
struct stat stat1;
stat(fileA, &stat1);
So, my question is that how do I fake/change the size of File A on-the-fly, so that it is able to display whole the data (which got superimposed because File B was bigger).
You won't be able to do this because many applications request file size before reading the file and then read only the reported amount of data.