Linux内核中unsigned three = 1

概述见 this link(懒惰的相关功能粘贴在下面). /* * 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 cu 见 this link(懒惰的相关功能粘贴在下面).

/* * 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这个变量是这样初始化的,这个函数在做什么？

解决方法 见00296 here行.评论说：

00295 /*00296  * Iterate through the groups which hold BACKUP superblock/GDT copIEs in an00297  * ext4 filesystem.  The counters should be initialized to 1,5,and 7 before00298  * calling this for the first time.  In a sparse filesystem it will be the00299  * sequence of powers of 3,and 7: 1,3,7,9,25,27,49,81,...00300  * For a non-sparse filesystem it will be every group: 1,2,4,...00301  */

总之,在我看来应该将三个初始化为1来启用函数ext4_List_backups返回1.

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