请参见this link(懒惰者在下面粘贴了相关功能)。
/*
* Check that all of the backup GDT blocks are held in the primary GDT block.
* It is assumed that they are stored in group order. Returns the number of
* groups in current filesystem that have BACKUPS, or -ve error code.
*/
static int verify_reserved_gdb(struct super_block *sb,
ext4_group_t end,
struct buffer_head *primary)
{
const ext4_fsblk_t blk = primary->b_blocknr;
unsigned three = 1;
unsigned five = 5;
unsigned seven = 7;
unsigned grp;
__le32 *p = (__le32 *)primary->b_data;
int gdbackups = 0;
while ((grp = ext4_list_backups(sb, &three, &five, &seven)) < end) {
if (le32_to_cpu(*p++) !=
grp * EXT4_BLOCKS_PER_GROUP(sb) + blk){
ext4_warning(sb, "reserved GDT %llu"
" missing grp %d (%llu)",
blk, grp,
grp *
(ext4_fsblk_t)EXT4_BLOCKS_PER_GROUP(sb) +
blk);
return -EINVAL;
}
if (++gdbackups > EXT4_ADDR_PER_BLOCK(sb))
return -EFBIG;
}
return gdbackups;
}
有人可以向我解释为什么要这样初始化此变量,以及该函数在做什么吗?
最佳答案
参见第00296行here。评论说:
00295 /*
00296 * Iterate through the groups which hold BACKUP superblock/GDT copies in an
00297 * ext4 filesystem. The counters should be initialized to 1, 5, and 7 before
00298 * calling this for the first time. In a sparse filesystem it will be the
00299 * sequence of powers of 3, 5, and 7: 1, 3, 5, 7, 9, 25, 27, 49, 81, ...
00300 * For a non-sparse filesystem it will be every group: 1, 2, 3, 4, ...
00301 */
简而言之,在我看来
three
应该初始化为1
,以使功能ext4_list_backups
返回1
。关于linux - 在Linux内核中无符号三= 1,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/22010537/