On 5/20/14, 11:33 AM, David Sterba wrote:
> On Tue, May 20, 2014 at 02:36:48PM +0800, Anand Jain wrote:
>> From: Anand Jain <anand.j...@oracle.com>
>>
>> generally if you use
>>   echo "test" > /sys/fs/btrfs/<fsid>/label
>> it would introduce return char at the end and it can not
>> be part of the label. The correct command is
>>   echo -n "test" > /sys/fs/btrfs/<fsid>/label
>>
>> This patch will check for this user error
>>
>> Signed-off-by: Anand Jain <anand.j...@oracle.com>
>> ---
>>  v2: accepts review comments. Thanks Eric and Roman
>>
>>  fs/btrfs/sysfs.c |   20 +++++++++++++++++---
>>  1 files changed, 17 insertions(+), 3 deletions(-)
>>
>> diff --git a/fs/btrfs/sysfs.c b/fs/btrfs/sysfs.c
>> index c5eb214..ca63fcd 100644
>> --- a/fs/btrfs/sysfs.c
>> +++ b/fs/btrfs/sysfs.c
>> @@ -373,22 +373,36 @@ static ssize_t btrfs_label_store(struct kobject *kobj,
>>      struct btrfs_trans_handle *trans;
>>      struct btrfs_root *root = fs_info->fs_root;
>>      int ret;
>> +    char *label;
>> +    char *pos;
>>  
>> -    if (len >= BTRFS_LABEL_SIZE) {
>> +    label = kzalloc(len, GFP_NOFS);
> 
> You can avoid allocating the buffer entirely:
> 
> - search for '\n', if found, use only that amount of bytes
> - check for maximum size, copy to label if ok

that's probably better than my strstrip idea, which requires
writable memory.  Odds of finding \t are slim.  ;)

-Eric

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