Re: [PATCH] kernfs: fix potential null-ptr-deref in kernfs_path_from_node_locked()
From: Leizhen (ThunderTown)
Date: Wed Nov 23 2022 - 21:52:55 EST
On 2022/11/24 10:28, Leizhen (ThunderTown) wrote:
>
>
> On 2022/11/24 10:24, Leizhen (ThunderTown) wrote:
>>
>>
>> On 2022/11/24 0:55, Tejun Heo wrote:
>>> On Wed, Nov 23, 2022 at 10:04:19AM +0800, Zhen Lei wrote:
>>>> Ensure that the 'buf' is not empty before strlcpy() uses it.
>>>>
>>>> Commit bbe70e4e4211 ("fs: kernfs: Fix possible null-pointer dereferences
>>>> in kernfs_path_from_node_locked()") first noticed this, but it didn't
>>>> fix it completely.
>>>>
>>>> Fixes: 9f6df573a404 ("kernfs: Add API to generate relative kernfs path")
>>>> Signed-off-by: Zhen Lei <thunder.leizhen@xxxxxxxxxx>
>>>
>>> I think the right thing to do is removing that if. It makes no sense to call
>>> that function with NULL buf and the fact that nobody reported crashes on
>>> NULL buf indicates that we in fact never do.
>>
>> OK.
>>
>> How about I remove "buf[0] = '\0';" too? It seems to be a useless operation.
>> When 'kn_from' and 'kn_to' have a common ancestor, there must be a path from
>> 'kn_from' to 'kn_to', and strlcpy() always fills in the terminator correctly,
>> even if the buf is too small to save the first path node.
>
> Sorry, I misanalyzed. The length used by "len < buflen ? buflen - len : 0" may
> be zero.
Ah, my brain is unstable today. The initial value of len is 0. So "buf[0] = '\0';"
can still be safely removed.
>
>>
>> static void test(void)
>> {
>> char buf[4];
>> int i, n, buflen;
>>
>> buflen = 1;
>> n = strlcpy(buf, "string", buflen);
>> for (i = 0; i < buflen; i++)
>> printk("%d: %02x\n", i, buf[i]);
>> printk("n=%d\n\n", n);
>>
>> buflen = sizeof(buf);
>> n = strlcpy(buf, "string", buflen);
>> for (i = 0; i < buflen; i++)
>> printk("%d: %02x\n", i, buf[i]);
>> printk("n=%d\n", n);
>> }
>>
>> Output:
>> [ 33.691497] 0: 00
>> [ 33.691569] n=6
>>
>> [ 33.691595] 0: 73
>> [ 33.691622] 1: 74
>> [ 33.691630] 2: 72
>> [ 33.691637] 3: 00
>> [ 33.691650] n=6
>>
>>>
>>> Thanks.
>>>
>>
>
--
Regards,
Zhen Lei