unsigned three = 1 in Linux kernel

Question

See this link (relevant function pasted below for the lazy).

/*
 * 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;
}

Can someone explain to me why this variable is initialised like this, and what this function is doing?

Answer 1

See line 00296 here. Comment says:

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  */

In short it seems to me three should be initialized to 1 to enable the function ext4_list_backups return 1.