write a c program showblock which displays the disk blocks of a file in an ext2 file system
the website link
https://www.eecs.wsu.edu/~cs360/LAB6.html
Write a C program, showblock, which displays the disk blocks of a file
in an EXT2 file system. The program runs as follows
showblock DEVICE PATHNAME
--------- ------- ----------
e.g. showblock diskimage /a/b/c/d (diskimage contains an EXT2 FS)
It locates the file named PATHNAME and prints the disk blocks (direct,
indirect, double-indirect) of the file.
**************************************************************************/
HOWTO Traverse EXT2 File System Tree
1. How to traverse the EXT2 FS tree:
Given a device, mydisk, containing an ext2 FS, and a pathname, e.g.
/cs360/is/fun
of a file, find the file.
NOTE!!! To find a file amounts to finding its INODE.
From its inode, you have ALL the information of a file.
2. ALGORITHM:
(1). Open disk for READ ==> file descriptor fd as dev number
You already have a get_block(fd, blk, buf[ ]) function.
(2). Tokenize pathname into components, e.g. /cs360/is/fun ==>
char *name[0] name[1] name[2]; with n = 3
| | |
"cs360" "is" "fun"
(3). The following C code prints the entries of a DIRectory INODE:
INODE *ip -> INODE structure of a DIRectory
char dbuf[BLKSIZE], temp[256];
DIR *dp;
char *cp;
int i;
int dev = opened fd of disk
for (i=0; i < 12; i++){ // assume at most 12 direct blocks
if (ip->i_block[i] == 0)
break;
get_block(dev, ip->i_block[i], dbuf);
dp = (DIR *)dbuf;
cp = dbuf;
while (cp < dbuf + BLKSIZE){
strncpy(temp, dp->name, dp->name_len);
temp[dp->name_len] = 0;
printf("%4d %4d %4d %sn",
dp->inode, dp->rec_len, dp->name_len, temp);
cp += dp->rec_len;
dp = (DIR *)cp;
}
}
Modify it to wrtie a search() function
int search(INODE *ip, char *name)
which searches the DIrectory's data blocks for a name string;
return its inode number if found; 0 if not.
(4). Start from the root inode #2: (YOU should already know HOW to do this)
INODE *ip->root inode;
int ino, blk, offset;
int iblk = inodes_start_block number (YOU should also know HOW)
char ibuf[BLKSIZE];
for (i=0; i < n; i++){
ino = search(ip, name[i]);
if (ino==0){
printf("can't find %sn", name[i]); exit(1);
}
// Mailman's algorithm: Convert (dev, ino) to inode pointer
blk = (ino - 1) / 8 + iblk; // disk block contain this INODE
offset = (ino - 1) % 8; // offset of INODE in this block
get_block(dev, blk, ibuf[ ]);
ip = (INODE *)ibuf + offset; // ip -> new INODE
}
(5). When the above for loop ends, ip MUST point at the INODE of pathname.
(6). Extract information from ip-> as required:
Print direct block numbers;
Print indirect block numbers;
Print double indirect block numbers, if any
Pat yourself on the back and say: Good Job!
(7). SAMPLES SOLUTION in samples/LAB6/:
showblock.bin diskimage
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