尝试在xv6操作系统中实现三重间接时，XV6崩溃

最初的xv6-rev7操作系统包含：
12个定向块
1间接blcok（指向128个区块）

这意味着我们有140个街区。
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现双重间接。
所以我从12支队伍中拿走了2个定向区块。
1表示双重间接，1表示三重间接。
这就是我现在所拥有的：
直接：10个街区
单个间接：128
双重间接：128 * 128
三重间接：4 * 128 * 128（我使用的是4而不是128，因为这足够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

所以我理解这个概念。 三重间接的实现在文件fs.c中的函数bmap中。
这就是它的样子：

出于某种原因，当我尝试创建大小为8.5MB的文件时，它失败了：
我正在使用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文件中：

 // 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 addrs[bn]) == 0) ip->addrs[bn] = addr = balloc(ip->dev); return addr; } bn -= NDIRECT; if(bn 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 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 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   \n" "eg 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(); } 

Answer 1

1.在mkfs.c中定义的mkfs.c数量不足。

 int nblocks = 20985; int nlog = LOGSIZE; int ninodes = 200; int size = 21029; 

你已经定义了：

 #define MAXFILE (NDIRECT + NINDIRECT + NINDIRECT*NINDIRECT + 4*NINDIRECT*NINDIRECT 

等于：10 + 128 + 128 ^ 2 + 4 * 128 ^ 2 = 82058。

只需选择一些大于82058的nblock，并相应地更新size 。

2.在你的bmap（）函数中，在三重间接代码中，你的第一级间接是一个四项数组（正如你在图中提到的那样）。 一旦你知道你应该访问这四个条目中的哪一个，你就会回到已经解决的双重间接问题。

因此，为了知道您应该访问的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); } 

然后你可以像这样减少bn ：

 bn -= ((NINDIRECT*NINDIRECT)*(bn/(NINDIRECT*NINDIRECT))); 

并再次解决双重间接问题。