笔记019 - HW10: bigger files for xv6

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题意

xv6的inode包括12个直接的block number和1个间接的block，间接的block存储128个block number，最大支持的文件长度为12+128=140个sectors。
修改xv6，将其中一个block number修改为双层间接的block，使得最大支持的文件长度为11+128+128x128=16523个sectors。

解决思路

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Makefile

CPUS := 1
QEMUOPTS : 添加 -snapshot
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下载big.c，添加到Makefile的UPROGS。
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param.h 修改FSSIZE

//#define FSSIZE       1000  // size of file system in blocks
#define FSSIZE       21113  // size of file system in blocks
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fs.h 修改NDIRECT、MAXFILE、dinode.addrs，添加NDBINDIRECT

//#define NDIRECT 12
#define NDIRECT 11 
#define NDBINDIRECT (NINDIRECT * NINDIRECT)

#define NINDIRECT (BSIZE / sizeof(uint)) // 512/4=128
//#define MAXFILE (NDIRECT + NINDIRECT)
#define MAXFILE (NDIRECT + NINDIRECT +NDBINDIRECT)

// 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+1];   // Data block addresses
  uint addrs[NDIRECT+2];   // Data block addresses
};
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file.h 修改inode.addrs

struct inode {
  uint dev;           // Device number
  uint inum;          // Inode number
  int ref;            // Reference count
  struct sleeplock lock;
  int flags;          // I_VALID

  short type;         // copy of disk inode
  short major;
  short minor;
  short nlink;
  uint size;
  //uint addrs[NDIRECT+1];
  uint addrs[NDIRECT+2];
};

fs.c的bmap()函数负责根据逻辑block number返回对应的物理block（若对应的记录为空，则先分配）。原先的bmap()函数先检查逻辑block number是否在NDIRECT范围内（直接的block）；否则，逻辑block number减去NDIRECT，检查新的逻辑block number是否在NINDIRECT范围内（间接的block）。新添加的处理过程是：逻辑block number再减去NINDIRECT，检查新的逻辑block number是否在NDBINDIRECT范围内（双层间接的block）。如果符合要求，则根据需要获取间接block内容或分配最终的block。

xv6 inode的标准格式如下：

扩展后的xv6 inode的格式如下：

编译并运行：

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$ make qemu
$ big
``` 
{% asset_img 3.JPG 编译并运行 %}
可以看到，系统支持的size和nblocks都更新了，支持的最大文件长度为16523个sectors。

# 源代码

//fs.c

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;
}

//homework10
bn -= NINDIRECT;
if(bn < NDBINDIRECT){
if((addr = ip->addrs[NINDIRECT+1]) == 0)
ip->addrs[NINDIRECT+1] = addr = balloc(ip->dev);
bp = bread(ip->dev, addr);
a = (uint*)bp->data;
int first_block_index = bn/NINDIRECT;
if((addr = a[first_block_index]) == 0){
a[first_block_index] = addr = balloc(ip->dev);
log_write(bp);
}
brelse(bp);

bp = bread(ip->dev, addr);
a = (uint*)bp->data;
int second_block_index = bn%NINDIRECT;
if((addr = a[second_block_index]) == 0){
  a[second_block_index] = addr = balloc(ip->dev);
  log_write(bp);
} 
brelse(bp);
return addr;

}

panic(“bmap: out of range”);
}
`

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