一个简单且类似Unix的教学操作系统，作为指导您实现间接块以支持大文件管理。在现有实现时，单个间接块可以处理对大文件无效的有限块 经营在这个分配中，您将把xv6文件的最大大小增加实现用于进一步扩展的双重间接块。我们建议您在编写代码之前先阅读第8章：文件系统。 准备工作mkfs程序创建xv6文件系统磁盘映像，并确定总块数文件系统具有；此大小由kernel/param.h中的FSSIZE控制该实验室的存储库设置为200000个块。您应该看到以下来自生成输出中的mkfs/mkfs：1 nmeta 70（引导、超级、日志块30索引节点块13、位图块25）块1
CSC3150-Instruction-A4
Introduction
This assignment uses xv6, a simple and Unix-like teaching operating system, as the platform to
guide you in implementing the indirect block to support big file management. In existing
implementation, singly-indirect blocks can handle limited blocks that are invalid for large file
management. In this assignment, you'll increase the maximum size of an xv6 file by
implementing doubly-indirect blocks for further extension.
We suggest you read Chapter 8: File system before writing code.
Preliminaries
The mkfs program creates the xv6 file system disk image and determines how many total blocks
the file system has; this size is controlled by FSSIZE in kernel/param.h. You'll see that FSSIZE in
the repository for this lab is set to 200,000 blocks. You should see the following output from
mkfs/mkfs in the make output:
1 nmeta 70 (boot, super, log blocks 30 inode blocks 13, bitmap blocks 25) blocks 1
This line describes the file system that mkfs/mkfs built: it has 70 meta-data blocks (blocks used
to describe the file system) and 199,930 data blocks, totaling 200,000 blocks.
If at any point during the lab you find yourself having to rebuild the file system from scratch, you
can run 'make clean', which forces make to rebuild fs.img.
Submission
• Due on: 23:59, 06 Dec 2023
• Plagiarism is strictly forbidden. Please note that TAs may ask you to explain the meaning of
your program to ensure that the codes are indeed written by yourself. Please also note that
we would check whether your program is too similar to your fellow students' code and
solutions available on the internet using plagiarism detectors.
• Late submission: A late submission within 15 minutes will not induce any penalty on your
grades. But 00:16 am-1:00 am: Reduced by 10%; 1:01 am-2:00 am: Reduced by 20%; 2:01
am-3:00 am: Reduced by 30% and so on. (e.g., Li Hua submitted a perfect attempt at 2:10
a.m. He will get (100+10 (bonus))*0.7=77p
Format guide
The project structure is illustrated below. You can also use ls command to check if your
structure is fine. Structure mismatch would cause grade deduction.
For this assignment, you don't need a specific folder for the bonus part. Tahe source folder
should contain four files: fs.c, file.h, fs.h, sysfile.c
1 main@ubuntu:~/Desktop/Assignment_4_120010001$ ls
2 Report.pdf source/
3
4 (One directory and one pdf.)
1 main@ubuntu:~/Desktop/Assignment_4_120010001/source$ ls
2 file.h fs.c fs.h sysfile.c
3
4 (two .c files and two .h file)
Please compress the folder containing all required files into a single zip file and name it using
your student ID as the code shown below and above, for example,
Assignment_4_120010001.zip. The report should be submitted in pdf format, together with
your source code. Format mismatch would cause grade deduction. Here is the sample step for
compressing your code.
1 main@ubuntu:~/Desktop$
2 zip -q -r Assignment_4_120010001.zip Assignment_4_120010001
3
4 main@ubuntu:~/Desktop$ ls
5 Assignment_4_120010001 Assignment_4_120010001.zip
Instruction Guideline
We limit your implementation within fs.c, file.h, fs.h, sysfile.c four files, starting with "TODO:"
comments. The entry (where you may start learning) of the test program is the main function in
bigfile.c, symlinktest.c (Bonus) under the 'xv6-labs-2022/user' directory.
Sections with (*) are introduction sections. These sections introduce tools and functions that
will help you understand what this system is about and how the system works with these
components. You might need to use it for this assignment. Do NOT CHANGE them except the
TODO parts.
1. For the introduction sections, please figure out how functions work and how to use them.
2. Be sure you have a basic idea of the content before starting your assignment. We believe that
those would be enough for handling this assignment.
3. (option) For students who are interested in the xv6 system and want to learn more about it,
you are welcome to read "xv6-book" to get more details.
a. https://pdos.csail.mit.edu/6.828/2022/xv6/book-riscv-rev3.pdf
Sections without (*) are TODO sections. In these sections, the logic of how this component/
function should work is a detailed list. You should implement functions in given places.
1. However, no sample code will be shown here. You need to figure out the implementation
based on the logic and APIs provided in the introduction sections.
Task1: Large Files
1. In this assignment, you'll increase the maximum size of an xv6 file. Currently, xv6 files are
limited to 268 blocks or 268*BSIZE bytes (BSIZE is 1024 in xv6).
a. This limit comes from the fact that an xv6 inode contains 12 "direct" block numbers and
one "singly-indirect" block number, which refers to a block that holds up to 256 more
block numbers for a total of 12+256=268 blocks.
2. The bigfile command creates the longest file it can and reports the size
a. The template we provide will fail to write 256 blocks. The test fails because bigfile expects
to be able to create a file with 65803 blocks, but unmodified xv6 limits files to 268 blocks.
3. You'll change the xv6 file system code to support a "doubly-indirect" block in each inode,
containing 256 addresses of singly-indirect blocks, each of which can contain up to 256
addresses of data blocks.
a. The result will be that a file can consist of up to 65803 blocks, or 256*256+256+11 blocks
(11 instead of 12 because we will sacrifice one of the direct block numbers for the doubleindirect block).
Definitions*
For more details, read <xv6-book> chapter 8.10
Following the hints and definitions above, we have provided you with the modified
structure. Please read the comments on the codes.
1 // Defined in kernel/fs.h
2 #define NDIRECT 11 // 12->11 By 3.a, we sacrifice 1 block for "doubly-indirec
3 #define NINDIRECT (BSIZE / sizeof(uint)) // = 1024/4 = 256
4 #define DNINDIRECT (NINDIRECT * NINDIRECT) // = 256*256
5 #define MAXFILE (NDIRECT + NINDIRECT + DNINDIRECT) // = 256*256 + 256 + 11
6
7 ///NOTE: Do not modify the structure
8 // On-disk inode structure
9 struct dinode { short type; // File type
10 short major; // Major device number (T_DEVICE only)
11 short minor; // Minor device number (T_DEVICE only)
12 short nlink; // Number of links to inode in file system
13 uint size; // Size of file (bytes)
14 ///NOTE: +2 instead of +1 because we NDIRECT is change from 12 to 11
15 uint addrs[NDIRECT+2]; // Data block addresses
16 };
17
18 //Defined in kernel/file.h
19 ///NOTE: Do not modify the structure
20 // in-memory copy of an inode
21 struct inode { uint dev; // Device number
22 uint inum; // Inode number
23 int ref; // Reference count
24 struct sleeplock lock; // protects everything below here
25 int valid; // inode has been read from disk?
26
27 short type; // copy of disk inode
28 short major;
29 short minor;
30 short nlink;
31 uint size;
32 uint addrs[NDIRECT+2]; ///NOTE: +2 instead of +1 because we NDIRECT is chan
33 };
We provide a simple figure for you to have a basic idea of what we are going to achieve
APIs*
Block Management
1 //Defined in fs.c
2
3 // Zero a block.
4 static void bzero(int dev, int bno);
5
6 // Blocks.
7 // Allocate a zeroed disk block.
8 // returns 0 if out of disk space.
9 static uint balloc(uint dev);
10
11 // Free a disk block.
12 static void bfree(int dev, uint b);
1 //Defined in bio.c
2
3 // Return a locked buf with the contents of the indicated block.
4 struct buf* bread(uint dev, uint blockno);
5
6 // Release a locked buffer.
7 // Move to the head of the most-recently-used list.
8 void brelse(struct buf *b);
1 //Defined in log.c
2
3 // Caller has modified b->data and is done with the buffer.
4 // Record the block number and pin in the cache by increasing refcnt.
5 // commit()/write_log() will do the disk write.
6 //
7 // log_write() replaces bwrite(); a typical use is:
8 // bp = bread(...)
9 // modify bp->data[]
10 // log_write(bp)
11 // brelse(bp)
12 void log_write(struct buf *b);
Learn from examples
For APIs provided above, they have been used to implement functions. You can learn how to use
those functions to develop our system.
You may take a look at how it is used in bmap(), itrunc(), bzero(), balloc(), bfree()
Especially read existing code in bmap() and itrunc() as these two functions are where we need
to modify, and they have already been implemented singly-indirect .
bmap()
See <xv6-book> chapter 8.10
1 // Inode content
2 //
3 // The content (data) associated with each inode is stored
4 // in blocks on the disk. The first NDIRECT block numbers
5 // are listed in ip->addrs[]. The next NINDIRECT blocks are
6 // listed in block ip->addrs[NDIRECT].
7
8 // Return the disk block address of the nth block in inode ip.
9 // If there is no such block, bmap allocates one.
10 // returns 0 if out of disk space.
11
12 // TODO: implement doubly-indirect
13 static uint bmap(struct inode *ip, uint bn);
bmap() is called both when reading and writing a file. When writing, bmap() allocates new
blocks as needed to hold file content, as well as allocating an indirect block if needed to hold
block addresses.
bmap() deals with two kinds of block numbers. The bn argument is a "logical block number" -- a
block number within the file, relative to the start of the file. The block numbers in ip->addrs[],
and the argument to bread(), are disk block numbers. You can view bmap() as mapping a file's
logical block numbers into disk block numbers.
itrunc()
See <xv6-book> chapter 8.10
itrunc frees a file’s blocks, resetting the inode’s size to zero. itrunc (kernel/fs.c:430) starts by
freeing the direct blocks(kernel/fs.c:436-441), then the ones listed in the indirect block
(kernel/fs.c:446- 449), and finally the indirect block itself (kernel/fs.c:451-452).
(TODO) Modify to support doubly-indirect block
1 ///TODO: modify it to support doubly-link
2 // Inode content
3 //
4 // The content (data) associated with each inode is stored
5 // in blocks on the disk. The first NDIRECT block numbers
6 // are listed in ip->addrs[]. The next NINDIRECT blocks are
7 // listed in block ip->addrs[NDIRECT].
8
9 // Return the disk block address of the nth block in inode ip.
10 // If there is no such block, bmap allocates one.
11 // returns 0 if out of disk space.
12 static uint bmap(struct inode *ip, uint bn);
Modify bmap() so that it implements a doubly-indirect block in addition to direct blocks and a
singly-indirect block.
You'll have to have only 11 direct blocks, rather than 12, to make room for your new doublyindirect block; you're not allowed to change the size of an on-disk inode.
i.e., Do NOT modify the structure or size of addrs in dinode or inode. We have already set it
up for you.
1 ///TODO: add discard of doubly-link correspondingly
2 // Truncate inode (discard contents).
3 // Caller must hold ip->lock.
4 void itrunc(struct inode *ip);
Hint
• The first 11 elements of ip->addrs[] should be direct blocks
• The 12th should be a singly-indirect block (just like the current one)
• The 13th should be your new doubly-indirect block. You are done with this exercise when
bigfile writes 65803 blocks
• Remember that it needs modification to release Double-Indirect blocks (modify itrunc())
Task2(Bonus): Symbolic links
In this exercise, you will add symbolic links to xv6.
• Symbolic links (or soft links) refer to a linked file by pathname; when a symbolic link is
opened, the kernel follows the link to the referred file.
• Symbolic links resembles hard links, but hard links are restricted to pointing to file on the
same disk, while symbolic links can cross disk devices.
• Although xv6 doesn't support multiple devices, implementing this system call is a good
exercise to understand how pathname lookup works.
(TODO) Implementation of symlink
You will implement the symlink(char *target, char *path) system call, which creates a new
symbolic link at path that refers to the file named by target. For further information, see the man
page symlink.
Your solution is complete when you pass all cases in symlinktest.
Hints
• Add a new file type (T_SYMLINK) to kernel/stat.h to represent a symbolic link. (We already
add it for you)
• Add a new flag to kernel/fcntl.h, (O_NOFOLLOW), that can be used with the open system call.
Note that flags passed to open are combined using a bitwise OR operator, so your new flag
should not overlap with any existing flags. This will let you compile user/symlinktest.c once
you add it to the Makefile. (We already define it for you)
• Implement the symlink(target, path) system call to create a new symbolic link at the path
that refers to target. Note that 'target' does not need to exist for the system call to succeed.
You will need to choose somewhere to store the target path of a symbolic link, for example,
in the inode's data blocks. symlink should return an integer representing success (0) or
failure (-1), similar to link and unlink.
• Modify the open system call to handle the case where the path refers to a symbolic link. If the
file does not exist, open must fail. When a process specifies O_NOFOLLOW in the flags to
open, open should open the symlink (and not follow the symbolic link).
• If the linked file is also a symbolic link, you must recursively follow it until a non-link file is
reached. If the links form a cycle, you must return an error code. You may approximate this
by returning an error code if the depth of links reaches some threshold (e.g., 10).
• Other system calls (e.g., link and unlink) must not follow symbolic links; these system calls
operate on the symbolic link itself.
• You do not have to handle symbolic links to directories for this lab.
Grading Rules
You can test the correctness of your code using the following commands under '~/xv6-labs-2022'
directory.
Test Task1
To run Task1, use the following command
1 make clean
2 make qemu
3 bigfile
By running the template we provide, you will receive the following information that tells you to
implement functions for big file.
1 $ bigfile
2 ..
3 wrote 268 blocks
4 bigfile: file is too small
5 $
When you finish Task1 correctly, you should see the following output