Re: [PATCH 04/10] sscanf: fix overflow

[Alexey Dobriyan]

On Tue, May 5, 2015 at 12:51 PM, Rasmus Villemoes
<linux@xxxxxxxxxxxxxxxxxx> wrote:
> On Sat, May 02 2015, Alexey Dobriyan <adobriyan@xxxxxxxxx> wrote:
>
>> Fun fact:
>>
>>       uint8_t val;
>>       sscanf("256", "%hhu", &val);
>>
>> will return 1 (as it should), and make val=0 (as it should not).
>>
>
> What do you base these "should" and "should not" on? Both C99 and POSIX
> say that the behaviour is undefined - the kernel can obviously define
> its own semantics for scanf, but what do you think they should be?

POSIX can say whatever it wants, it's about common sense.

sscanf(), both kernel and libc, in this situation returns 0
when "0" character is nowhere to be found in the string!
It should either return 25 or do not return anything
because situation is ambiguous (read: set ERANGE).

> If we want to correctly handle overflow, the only sane way is to make
> sscanf return 0 in the above case (no conversions done). This also
> seems to be what your patch does, but then I'm confused by your
> first "as it should".
>
>> Apart from correctness, patch allows to remove checks and switch
>> to proper types in several (most?) cases:
>>
>>       grep -e 'scanf.*%[0-9]\+[dioux]' -n -r .
>>
>> Such checks can be incorrect too, checking for 3 digits with %3u
>> for parsing uint8_t is not enough.
>
> Yeah, and it may be too much; sscanf("0042", "%hhu", &val") should give
> 42, not 4. I agree that one should be able to rely on scanf doing range
> checking as part of matching.
>
> Actually, I think one should go through all the callers of sscanf which
> use a field width with an integer conversion and see if we can get rid
> of it, and then rip it away from the sscanf implementation. Otherwise
> there's another bug which would need fixing, namely
>
> int x;
> char rest[50];
> sscanf("12345678901234567890", "%3d%s", &x, rest)
>
> should successfully return 2 (storing 123 in x), but it can't when the
> strategy is to convert as much as possible (which may then give an early
> failure due to overflow), then divide by 10 until we haven't consumed
> more than we're allowed.
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