Re: [uml-devel] [PATCH 07/15] hostfs: Remove open coded strcpy()

[Geert Uytterhoeven]

Hi Richard,

On Mon, Mar 16, 2015 at 1:18 PM, Richard Weinberger <richard@xxxxxx> wrote:
> Am 16.03.2015 um 13:03 schrieb Geert Uytterhoeven:
>> On Mon, Mar 16, 2015 at 12:41 PM, Richard Weinberger <richard@xxxxxx> wrote:
>>> --- a/fs/hostfs/hostfs_kern.c
>>> +++ b/fs/hostfs/hostfs_kern.c
>>> @@ -105,11 +105,10 @@ static char *__dentry_name(struct dentry *dentry, char *name)
>>
>> This code looks fishy to me...
>>
>> First we have:
>>
>>     len = strlen(root);
>>     strlcpy(name, root, PATH_MAX);
>>
>> (I notice the code used strncpy() before. One difference with strlcpy()
>>  is that strncpy() fills the remaining of the destination buffer with zeroes.)
>>
>> Then:
>>
>>>                 __putname(name);
>>>                 return NULL;
>>>         }
>>> -       if (p > name + len) {
>>> -               char *s = name + len;
>>
>> Unless strlcpy() truncated the string (which is unlikely, as root
>> cannot be longer
>> than PATH_MAX?), s = name + len now points to the zero terminator.
>> So the below would copy just one single byte:

Oops, that's of course not true, as s is the destination, not the source,
of the copy operation.

>>> -               while ((*s++ = *p++) != '\0')
>>> -                       ;
>>> -       }
>>> +
>>> +       if (p > name + len)
>>> +               strcpy(name + len, p);
>>> +
>>
>> What is this code really supposed to do?
>
> Hostfs' __dentry_name() builds the real path. i.e, the prefix on the host side
> plus the requested path in UML.
>
> "strlcpy(name, root, PATH_MAX);" copies the host prefix into name and then
> the "strcpy(name + len, p);" copies the requested path into it.
>
> The trick is that both share the same buffer, allocated by dentry_path_raw().

Ah, so the path is stored in the end of the buffer...

> Therefore this bounds check works:
>         if (len > p - name) {

... and if this is true, prefix and path would overlap, which means there's\
not enough space in the buffer.

>                 __putname(name);
>                 return NULL;
>         }
>
> Is it now clearer or did I miss something?
> I agree that this code is tricky. :)

Yes, thanks for your explanation!

Gr{oetje,eeting}s,

                        Geert

--
Geert Uytterhoeven -- There's lots of Linux beyond ia32 -- geert@xxxxxxxxxxxxxx

In personal conversations with technical people, I call myself a hacker. But
when I'm talking to journalists I just say "programmer" or something like that.
                                -- Linus Torvalds
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