[Solved] COMP8051 Assignment 2 XV6 Processes & Files

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Question 1 Inode and Block layer in xv6

xv6 divides the disk into several sections, as shown in the Figure below.

The file system does not use block 0 (it holds the boot sector). Block 1 is called the superblock, it contains metadata about the file system (the file system size in blocks, the number of data blocks, the number of inodes, and the number of blocks in the log). Blocks starting at 2 hold inodes. After those come bitmap blocks tracking which data blocks are in use. Most of the remaining blocks are data blocks. The blocks at the end of the disk hold the logging layers log.

The term inode can have one of two related meanings. It might refer to the ondisk data structure ( struct dinode, fs.h ) containing a files size and list of data block numbers. Or inode might refer to an in-memory inode ( struct inode, file.h ), which contains a copy of the on-disk inode as well as extra information needed within the kernel.

The on-disk inode is defined by a struct dinode. The type field distinguishes between files, directories, and special files (devices). A type of zero indicates that an on-disk inode is free. The nlink field counts the number of directory entries that refer to this inode, in order to recognize when the on-disk inode and its data blocks should be freed. The size field records the number of bytes of content in the file. The addrs array records the block numbers of the disk blocks holding the files content.

The function readi is essentially called by the user level read function to read an amount of data from a file. The readi parameters are:

readi ( struct inode *ip, char *dst, uint off, uint n)

The inode pointed to by ip abstracts the file layout/structure in the filesystem. Readi uses the addrs array to find the block numbers that are associated with the file. It then reads in all the blocks from the disk to satisfy the read request, in xv6 each block is the same size as a disk sector, which is 512 bytes.

The function Bmap makes it easy for readi and writei to get at an inodes data. Readi (5503) starts by making sure that the offset and count are not beyond the end of the file. Reads that start beyond the end of the file return an error (55145515) while reads that start at or cross the end of the file return fewer bytes than requested (5516-5517) . The main loop processes each block of the file, copying data from the buffer into dst (5519-5524) . writei (5553) is identical to readi, with three exceptions: writes that start at or cross the end of the file grow the file, up to the maximum file size (5566-5567) ; the loop copies data into the buffers instead of out (5572) ; and if the write has extended the file, writei must update its size (5577-5580) .

Read chapter 6 of the xv6 book and briefly explain how the read calls in the following code taken from cat.c

void cat(int fd)

{ int n;

while((n = read(fd, buf, sizeof(buf))) > 0) { if (write(1, buf, n) != n) { printf(1, cat: write error
); exit();

} } if(n < 0){

printf(1, cat: read error
); exit();

}

}

are associated with sectors on the disk by the xv6 operating system.

here

You may use a diagram such as the one to illustrate your answer.

(10 marks)

Question 2 IDE Disk Driver

Read the sections on Drivers, Code: Drivers in chapter 3 and section 36.8 of filedevices.pdf, which gives a summary of the IDE disk controller protocol. See in particular the code in ide.c. See image below

The xv6 source code includes a working IDE driver in ide.c. For example the piece of code in idestart outb(0x1f6, 0xe0 | ((b->dev&1)<<4) | ((bsector>>24)&0x0f));

is associated with:

I/O Address 0x1F6 = 1B1D TOP4LBA: B=LBA, D=drive

The 0xe is to set 1B1 in bits 5,6,7 B=1 indicates that we are going to have the low 4 bits at address 0x1f6 set the top 4 bits of the Logical Block Address (LBA).

((bdev&1)<<4) sets the D bit.

((bsector>>24)&0x0f)) sets the top 4 bits of LBA to low four bits at address 0x1f6.

An IDE disk presents a simple interface to the Disk system, consisting of four types of register: control, command block, status, and error. These registers are available by reading or writing to specific I/O addresses (such as 0x3F6 ) using (on x86) the in and out I/O instructions.

On page 48 of the xv6 book we read The xv6 bootloader issues disk read commands and reads the disk controller status bits repeatedly until the data is ready (see Appendix B). This polling or busy waiting is fine in a boot loader, which has nothing better to do. In an operating system, however, it is more efficient to let another process run on the CPU and arrange to receive an interrupt when the disk operation has completed.

  1. Explain how the xv6 operating system uses interrupts to schedule I/Orequests to the disk?
  2. Explain how the bootloader interfaces with the IDE controller to load the xv6 operating system see Appendix B of the xv6 book (section Code: C bootstrap in particular) and bootmain.c in the xv6 source code?

(12 marks)

Question 3 Big Files in xv6

Currently xv6 files are limited to 140 sectors, or 71,680 bytes. 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 128 more block numbers, for a total of 12+128=140. Youll change the xv6 file system code to support a doubly-indirect block in each inode, containing 128 addresses of singly-indirect blocks, each of which can contain up to 128 addresses of data blocks. The result will be that a file will be able to consist of up to 16523 sectors (or about 8.5 megabytes). Preliminaries

Modify your Makefiles CPUS definition so that it reads:

CPUS := 1

Add

QEMUEXTRA = -snapshot right before QEMUOPTS

The above two steps speed up qemu tremendously when xv6 creates large files.

mkfs initializes the file system to have fewer than 1000 free data blocks, too few to show off the changes youll make. Modify param.h to set FSSIZE to:

#define FSSIZE 20000 // size of file system in blocks

Download big.c into your xv6 directory, add it to the UPROGS list, start up xv6, and run big. It creates as big a file as xv6 will let it, and reports the resulting size. It should say 140 sectors.

What to Look At

here

The format of an on-disk inode is defined by struct dinode in fs.h. Youre particularly interested in NDIRECT, NINDIRECT, MAXFILE, and the addrs[] element of struct dinode. Lookfor a diagram of the standard xv6 inode.

The code that finds a files data on disk is in bmap() in fs.c. Have a look at it and make sure you understand what its doing. 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 a block number 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 files logical block numbers into disk block numbers.

Your Job

Modify bmap() so that it implements a doubly-indirect block, in addition to direct blocks and a singly-indirect block. Youll have to have only 11 direct blocks, rather than 12, to make room for your new doubly-indirect block; youre not allowed to change the size of an on-disk inode. 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 dont have to modify xv6 to handle deletion of files with doubly-indirect blocks.

If all goes well, big will now report that it can write 16523 sectors. It will take big a few dozen seconds to finish.

Hints

Make sure you understand bmap(). Write out a diagram of the relationships between ip->addrs[], the indirect block, the doubly-indirect block and the singly-indirect blocks it points to, and data blocks. Make sure you understand why adding a doubly-indirect block increases the maximum file size by 16,384 blocks (really 16383, since you have to decrease the number of direct blocks by one).

Think about how youll index the doubly-indirect block, and the indirect blocks it points to, with the logical block number.

If you change the definition of NDIRECT, youll probably have to change the size of addrs[] in struct inode in file.h. Make sure that struct inode and struct dinode have the same number of elements in their addrs[] arrays.

If you change the definition of NDIRECT, make sure to create a new fs.img, since mkfs uses NDIRECT too to build the initial file systems. If you delete fs.img, make on Unix (not xv6) will build a new one for you.

If your file system gets into a bad state, perhaps by crashing, delete fs.img (do this from Unix, not xv6). make will build a new clean file system image for you.

Dont forget to brelse() each block that you bread().

You should allocate indirect blocks and doubly-indirect blocks only as needed, like the original bmap().

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[Solved] COMP8051 Assignment 2 XV6 Processes & Files
$25