尝试在 xv6 操作系统中实现三重间接寻址时 XV6 崩溃
XV6 crashes when trying to implement triple indirection in xv6 operating system
原始xv6-rev7操作系统包含：
12 个定向块
1 个间接块（指向 128 个块）
这意味着我们有 140 个块。
每个块的大小为 512KB ==> 512 * 140 = 71,680 ~= 70KB 是 xv6 中文件大小的限制。
我想在 xv6 中实现三重间接访问以支持大小为 40MB 的文件。
为了做到这一点，我需要在三重间接之前实施双重间接。
所以我从我拥有的 12 个定向块中取出了 2 个。
1 个用于双重间接，另一个用于三次间接。
这就是我现在拥有的：
直接：10 块
单间接：128
双间接：128*128
三重间接：4*128*128（我使用 4 而不是 128，因为这足以容纳 40MB）
这就是为什么 #define NDIRECT 10 和 uint addrs[NDIRECT+3];
文件大小限制 = (10 + 128 + 128*128 + 4*128*128)*512kb = 42,013,696 ~= 42MB
所以我明白了这个概念。
三重间接寻址的实现在文件 fs.c.
中的函数 bmap 中
这是它的样子：
出于某种原因，当我尝试创建大小为 8.5MB 的文件时失败了：
我用的是bochs模拟器
我也不确定我需要在 mkfs.c 中更改哪些值：
int nblocks = 20985;
int nlog = LOGSIZE;
int ninodes = 200;
int size = 21029;
fs.h:
// On-disk file system format.
// Both the kernel and user programs use this header file.
// Block 0 is unused.
// Block 1 is super block.
// Blocks 2 through sb.ninodes/IPB hold inodes.
// Then free bitmap blocks holding sb.size bits.
// Then sb.nblocks data blocks.
// Then sb.nlog log blocks.
#define ROOTINO 1 // root i-number
#define BSIZE 512 // block size
// File system super block
struct superblock {
uint size; // Size of file system image (blocks)
uint nblocks; // Number of data blocks
uint ninodes; // Number of inodes.
uint nlog; // Number of log blocks
};
#define NDIRECT 10
#define NINDIRECT (BSIZE / sizeof(uint))
#define MAXFILE (NDIRECT + NINDIRECT + NINDIRECT*NINDIRECT + 4*NINDIRECT*NINDIRECT)
// On-disk inode structure
struct dinode {
short type; // File type
short major; // Major device number (T_DEV only)
short minor; // Minor device number (T_DEV only)
short nlink; // Number of links to inode in file system
uint size; // Size of file (bytes)
uint addrs[NDIRECT+3]; // Data block addresses
};
// Inodes per block.
#define IPB (BSIZE / sizeof(struct dinode))
// Block containing inode i
#define IBLOCK(i) ((i) / IPB + 2)
// Bitmap bits per block
#define BPB (BSIZE*8)
// Block containing bit for block b
#define BBLOCK(b, ninodes) (b/BPB + (ninodes)/IPB + 3)
// Directory is a file containing a sequence of dirent structures.
#define DIRSIZ 14
struct dirent {
ushort inum;
char name[DIRSIZ];
};
fs.c:
// Return the disk block address of the nth block in inode ip.
// If there is no such block, bmap allocates one.
static uint
bmap(struct inode *ip, uint bn)
{
uint addr, *a;
struct buf *bp;
if(bn < NDIRECT){
if((addr = ip->addrs[bn]) == 0)
ip->addrs[bn] = addr = balloc(ip->dev);
return addr;
}
bn -= NDIRECT;
if(bn < NINDIRECT){
// Load indirect block, allocating if necessary.
if((addr = ip->addrs[NDIRECT]) == 0)
ip->addrs[NDIRECT] = addr = balloc(ip->dev);
bp = bread(ip->dev, addr);
a = (uint*)bp->data;
if((addr = a[bn]) == 0){
a[bn] = addr = balloc(ip->dev);
log_write(bp);
}
brelse(bp);
return addr;
}
/* Double indirect */
bn -= NINDIRECT;
if(bn < NINDIRECT*NINDIRECT){
// Load 2nd indirect block, allocating if necessary.
if((addr = ip->addrs[NDIRECT+1]) == 0) // 2d block. NDIRECT+1 is to get the index vector
ip->addrs[NDIRECT+1] = addr = balloc(ip->dev);
bp = bread(ip->dev, addr);
a = (uint*)bp->data;
if ((addr = a[bn/(NINDIRECT)]) == 0) { /* get index for 1st
indirection. (NINDIRECT is 128) */
a[bn/(NINDIRECT)] = addr = balloc(ip->dev);
log_write(bp);
}
brelse(bp); /* release the double indirect table
(main level) */
bp = bread(ip->dev, addr);
a = (uint*)bp->data;
if ((addr = a[bn%(NINDIRECT)]) == 0) { /* get the 2nd level table */
a[bn%(NINDIRECT)] = addr = balloc(ip->dev);
log_write(bp);
}
brelse(bp);
return addr;
}
/* Triple indirect */
bn -= NINDIRECT*NINDIRECT;
if(bn < 4*NINDIRECT*NINDIRECT){
// Load 3rd indirect block, allocating if necessary.
if((addr = ip->addrs[NDIRECT+2]) == 0) // 3d block. NDIRECT+2 is to get the index vector
ip->addrs[NDIRECT+2] = addr = balloc(ip->dev);
bp = bread(ip->dev, addr);
a = (uint*)bp->data;
if ((addr = a[bn/(NINDIRECT*4)]) == 0) { /* get index for 2st
indirection. (NINDIRECT is 128) */
a[bn/(NINDIRECT*4)] = addr = balloc(ip->dev);
log_write(bp);
}
brelse(bp);
bp = bread(ip->dev, addr);
a = (uint*)bp->data;
if ((addr = a[bn/(NINDIRECT*NINDIRECT*4)]) == 0) {
a[bn/(NINDIRECT*NINDIRECT*4)] = addr = balloc(ip->dev);
log_write(bp);
}
brelse(bp);
if ((addr = a[bn%(NINDIRECT*NINDIRECT*4)]) == 0) {
a[bn%(NINDIRECT*NINDIRECT*4)] = addr = balloc(ip->dev);
log_write(bp);
}
brelse(bp);
return addr;
}
panic("bmap: out of range");
}
mkfs.c:
#define stat xv6_stat // avoid clash with host struct stat
#include "types.h"
#include "fs.h"
#include "stat.h"
#include "param.h"
int nblocks = 20985;
int nlog = LOGSIZE;
int ninodes = 200;
int size = 21029;
bigfile.c:
#include "types.h"
#include "stat.h"
#include "user.h"
#include "fcntl.h"
void
help()
{
printf(1, "usage:\nfiles <name> <letter> <num>\n"
"e.g. nfiles foo a 40\n creates a file foo, with 40 times the letter a\n");
}
void
num2str(int i, char str[3])
{
str[2]=i%10+'0';
i=i/10;
str[1]=i%10+'0';
i=i/10;
str[0]=i%10+'0';
i=i/10;
}
#define BUF_SZ 512
int
main(int argc, char *argv[])
{
int i, count, fd, n;
// char *name;
// char c;
char buf[BUF_SZ];
if (argc !=4) {
help();
exit();
}
count = atoi(argv[3]);
if((fd=open(argv[1], O_CREATE|O_RDWR))<0) {
printf(2,"Failed to open file: %s\n", argv[1]);
exit();
}
for (i=0; i<BUF_SZ;i++)
buf[i]=argv[2][0];
for (i=0; i<count/BUF_SZ;i++)
if ((n=write(fd,buf,BUF_SZ)) != BUF_SZ)
{
printf(2,"Failed 1 to Write count=%d\n",i*BUF_SZ);
exit();
}
for (i=0; i<count%BUF_SZ;i++)
if ((n=write(fd,argv[2],1)) != 1)
{
printf(2,"Failed 2 to Write count=%d\n",count-i);
exit();
}
exit();
}
1.The mkfs.c 中定义的 nblock 数量不足。
int nblocks = 20985;
int nlog = LOGSIZE;
int ninodes = 200;
int size = 21029;
您已定义：
#define MAXFILE (NDIRECT + NINDIRECT + NINDIRECT*NINDIRECT + 4*NINDIRECT*NINDIRECT
等于：10+128+128^2+4*128^2 = 82058.
只需选择大于 82058 的 nblocks 个数字，并相应地更新 size。
2.In 你的 bmap() 函数，在三重间接代码中，你的第一级间接是一个四条目数组（正如你在图表中提到的）。
一旦您知道您应该访问这四个条目中的哪一个，您就回到了您已经解决的双重间接问题。
所以，为了知道您应该访问四个条目中的哪一个，您可以使用：
if((addr = a[bn/(NINDIRECT*NINDIRECT)]) == 0){
a[bn/(NINDIRECT*NINDIRECT)] = addr = balloc(ip->dev);
log_write(bp);
}
然后您可以像这样减少 bn：
bn -= ((NINDIRECT*NINDIRECT)*(bn/(NINDIRECT*NINDIRECT)));
并再次解决双重间接寻址问题
原始xv6-rev7操作系统包含：
12 个定向块
1 个间接块（指向 128 个块）
这意味着我们有 140 个块。
每个块的大小为 512KB ==> 512 * 140 = 71,680 ~= 70KB 是 xv6 中文件大小的限制。
我想在 xv6 中实现三重间接访问以支持大小为 40MB 的文件。
为了做到这一点，我需要在三重间接之前实施双重间接。
所以我从我拥有的 12 个定向块中取出了 2 个。
1 个用于双重间接，另一个用于三次间接。
这就是我现在拥有的：
直接：10 块
单间接：128
双间接：128*128
三重间接：4*128*128（我使用 4 而不是 128，因为这足以容纳 40MB）
这就是为什么 #define NDIRECT 10 和 uint addrs[NDIRECT+3];
文件大小限制 = (10 + 128 + 128*128 + 4*128*128)*512kb = 42,013,696 ~= 42MB
所以我明白了这个概念。
三重间接寻址的实现在文件 fs.c.
中的函数 bmap 中
这是它的样子：
出于某种原因，当我尝试创建大小为 8.5MB 的文件时失败了：
我用的是bochs模拟器
我也不确定我需要在 mkfs.c 中更改哪些值：
int nblocks = 20985;
int nlog = LOGSIZE;
int ninodes = 200;
int size = 21029;
fs.h:
// On-disk file system format.
// Both the kernel and user programs use this header file.
// Block 0 is unused.
// Block 1 is super block.
// Blocks 2 through sb.ninodes/IPB hold inodes.
// Then free bitmap blocks holding sb.size bits.
// Then sb.nblocks data blocks.
// Then sb.nlog log blocks.
#define ROOTINO 1 // root i-number
#define BSIZE 512 // block size
// File system super block
struct superblock {
uint size; // Size of file system image (blocks)
uint nblocks; // Number of data blocks
uint ninodes; // Number of inodes.
uint nlog; // Number of log blocks
};
#define NDIRECT 10
#define NINDIRECT (BSIZE / sizeof(uint))
#define MAXFILE (NDIRECT + NINDIRECT + NINDIRECT*NINDIRECT + 4*NINDIRECT*NINDIRECT)
// On-disk inode structure
struct dinode {
short type; // File type
short major; // Major device number (T_DEV only)
short minor; // Minor device number (T_DEV only)
short nlink; // Number of links to inode in file system
uint size; // Size of file (bytes)
uint addrs[NDIRECT+3]; // Data block addresses
};
// Inodes per block.
#define IPB (BSIZE / sizeof(struct dinode))
// Block containing inode i
#define IBLOCK(i) ((i) / IPB + 2)
// Bitmap bits per block
#define BPB (BSIZE*8)
// Block containing bit for block b
#define BBLOCK(b, ninodes) (b/BPB + (ninodes)/IPB + 3)
// Directory is a file containing a sequence of dirent structures.
#define DIRSIZ 14
struct dirent {
ushort inum;
char name[DIRSIZ];
};
fs.c:
// Return the disk block address of the nth block in inode ip.
// If there is no such block, bmap allocates one.
static uint
bmap(struct inode *ip, uint bn)
{
uint addr, *a;
struct buf *bp;
if(bn < NDIRECT){
if((addr = ip->addrs[bn]) == 0)
ip->addrs[bn] = addr = balloc(ip->dev);
return addr;
}
bn -= NDIRECT;
if(bn < NINDIRECT){
// Load indirect block, allocating if necessary.
if((addr = ip->addrs[NDIRECT]) == 0)
ip->addrs[NDIRECT] = addr = balloc(ip->dev);
bp = bread(ip->dev, addr);
a = (uint*)bp->data;
if((addr = a[bn]) == 0){
a[bn] = addr = balloc(ip->dev);
log_write(bp);
}
brelse(bp);
return addr;
}
/* Double indirect */
bn -= NINDIRECT;
if(bn < NINDIRECT*NINDIRECT){
// Load 2nd indirect block, allocating if necessary.
if((addr = ip->addrs[NDIRECT+1]) == 0) // 2d block. NDIRECT+1 is to get the index vector
ip->addrs[NDIRECT+1] = addr = balloc(ip->dev);
bp = bread(ip->dev, addr);
a = (uint*)bp->data;
if ((addr = a[bn/(NINDIRECT)]) == 0) { /* get index for 1st
indirection. (NINDIRECT is 128) */
a[bn/(NINDIRECT)] = addr = balloc(ip->dev);
log_write(bp);
}
brelse(bp); /* release the double indirect table
(main level) */
bp = bread(ip->dev, addr);
a = (uint*)bp->data;
if ((addr = a[bn%(NINDIRECT)]) == 0) { /* get the 2nd level table */
a[bn%(NINDIRECT)] = addr = balloc(ip->dev);
log_write(bp);
}
brelse(bp);
return addr;
}
/* Triple indirect */
bn -= NINDIRECT*NINDIRECT;
if(bn < 4*NINDIRECT*NINDIRECT){
// Load 3rd indirect block, allocating if necessary.
if((addr = ip->addrs[NDIRECT+2]) == 0) // 3d block. NDIRECT+2 is to get the index vector
ip->addrs[NDIRECT+2] = addr = balloc(ip->dev);
bp = bread(ip->dev, addr);
a = (uint*)bp->data;
if ((addr = a[bn/(NINDIRECT*4)]) == 0) { /* get index for 2st
indirection. (NINDIRECT is 128) */
a[bn/(NINDIRECT*4)] = addr = balloc(ip->dev);
log_write(bp);
}
brelse(bp);
bp = bread(ip->dev, addr);
a = (uint*)bp->data;
if ((addr = a[bn/(NINDIRECT*NINDIRECT*4)]) == 0) {
a[bn/(NINDIRECT*NINDIRECT*4)] = addr = balloc(ip->dev);
log_write(bp);
}
brelse(bp);
if ((addr = a[bn%(NINDIRECT*NINDIRECT*4)]) == 0) {
a[bn%(NINDIRECT*NINDIRECT*4)] = addr = balloc(ip->dev);
log_write(bp);
}
brelse(bp);
return addr;
}
panic("bmap: out of range");
}
mkfs.c:
#define stat xv6_stat // avoid clash with host struct stat
#include "types.h"
#include "fs.h"
#include "stat.h"
#include "param.h"
int nblocks = 20985;
int nlog = LOGSIZE;
int ninodes = 200;
int size = 21029;
bigfile.c:
#include "types.h"
#include "stat.h"
#include "user.h"
#include "fcntl.h"
void
help()
{
printf(1, "usage:\nfiles <name> <letter> <num>\n"
"e.g. nfiles foo a 40\n creates a file foo, with 40 times the letter a\n");
}
void
num2str(int i, char str[3])
{
str[2]=i%10+'0';
i=i/10;
str[1]=i%10+'0';
i=i/10;
str[0]=i%10+'0';
i=i/10;
}
#define BUF_SZ 512
int
main(int argc, char *argv[])
{
int i, count, fd, n;
// char *name;
// char c;
char buf[BUF_SZ];
if (argc !=4) {
help();
exit();
}
count = atoi(argv[3]);
if((fd=open(argv[1], O_CREATE|O_RDWR))<0) {
printf(2,"Failed to open file: %s\n", argv[1]);
exit();
}
for (i=0; i<BUF_SZ;i++)
buf[i]=argv[2][0];
for (i=0; i<count/BUF_SZ;i++)
if ((n=write(fd,buf,BUF_SZ)) != BUF_SZ)
{
printf(2,"Failed 1 to Write count=%d\n",i*BUF_SZ);
exit();
}
for (i=0; i<count%BUF_SZ;i++)
if ((n=write(fd,argv[2],1)) != 1)
{
printf(2,"Failed 2 to Write count=%d\n",count-i);
exit();
}
exit();
}
1.The mkfs.c 中定义的 nblock 数量不足。
int nblocks = 20985;
int nlog = LOGSIZE;
int ninodes = 200;
int size = 21029;
您已定义：
#define MAXFILE (NDIRECT + NINDIRECT + NINDIRECT*NINDIRECT + 4*NINDIRECT*NINDIRECT
等于：10+128+128^2+4*128^2 = 82058.
只需选择大于 82058 的 nblocks 个数字，并相应地更新 size。
2.In 你的 bmap() 函数，在三重间接代码中，你的第一级间接是一个四条目数组（正如你在图表中提到的）。 一旦您知道您应该访问这四个条目中的哪一个，您就回到了您已经解决的双重间接问题。
所以，为了知道您应该访问四个条目中的哪一个，您可以使用：
if((addr = a[bn/(NINDIRECT*NINDIRECT)]) == 0){
a[bn/(NINDIRECT*NINDIRECT)] = addr = balloc(ip->dev);
log_write(bp);
}
然后您可以像这样减少 bn：
bn -= ((NINDIRECT*NINDIRECT)*(bn/(NINDIRECT*NINDIRECT)));
并再次解决双重间接寻址问题