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Lab 8 File System: Large Files
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). 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.
Increase the maximum size of an xv6 file. 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. The result will be that a file will be able to consist of up to 256*256+256+11 blocks (11 instead of 12, because we will sacrifice one of the direct block numbers for the double-indirect block).
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 doubly-indirect block; you’re 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.
mkfs initializes the file system to have fewer than 2000 free data blocks, too few to show off the changes you’ll make. Modify kernel/param.h to change FSSIZE from 2000 to 200,000:#define FSSIZE 200000 // size of file system in blocks
Rebuild
mkfs so that is produces a bigger disk: $ rm mkfs/mkfs fs.img; make mkfs/mkfsChange NDIRECT from 12 to 11. In
file.h, modify in-memory copy of an inode struct: Change size of addrs from NDIRECT+1 to NDIRECT+2.bmap return the disk block address of the nth block in inode ip. If there is no such block, bmap allocates one.in
fs.h#define NDIRECT 11
#define NINDIRECT (BSIZE / sizeof(uint))
#define NDOUBLY_INDIRECT NINDIRECT*NINDIRECT
#define MAXFILE (NDIRECT + NINDIRECT + NDOUBLY_INDIRECT)
Improve
bmap to support this double indirect: 1. Load double indirect block, allocating if necessary. 2. Load 2nd layer block. 3. Find disk block. bn -= NINDIRECT;
if(bn < NDOUBLY_INDIRECT){
// Load double indirect block, allocating if necessary.
if((addr = ip->addrs[NDIRECT + 1]) == 0)
ip->addrs[NDIRECT + 1] = addr = balloc(ip->dev);
bp = bread(ip->dev, addr);
a = (uint*)bp->data;
// load 2nd layer block.
uint double_index = bn / NINDIRECT;
if((addr = a[double_index]) == 0){
a[double_index] = addr = balloc(ip->dev);
log_write(bp);
}
brelse(bp);
// now find disk block.
bp = bread(ip->dev, addr);
a = (uint*)bp->data;
uint pos = bn % NINDIRECT;
if ((addr = a[pos]) == 0) {
a[pos] = addr = balloc(ip->dev);
log_write(bp);
}
画图是最好的解答之一，再把画出的设计图实现，就是编程。
Last modified 3yr ago