Re: [HACKERS] Open 6.4 items

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От Tom Lane
Тема Re: [HACKERS] Open 6.4 items
Дата
Msg-id 22954.907945289@sss.pgh.pa.us
обсуждение исходный текст
Ответ на Re: [HACKERS] Open 6.4 items  (Tom Ivar Helbekkmo <tih@nhh.no>)
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Tom Ivar Helbekkmo <tih@nhh.no> writes:
> Now I'm explicitly asking for at least one byte more than I need, and
> I'm pretty damn sure that I never touch that extra byte, but something
> seems to, since the problem goes away.  It's arrogant of me, I know,
> but barring a complete misunderstanding on my part of how variable
> size records work (or a stupid bug that I've been staring at for hours
> without seeing, of course), I'd say that something outside my code is
> at fault.  Any ideas as to how to try to find out?

Well, I hate to ruin your day, but coming pre-armed with the knowledge
that the code is writing one byte too many, it's pretty obvious that the
first loop in inet_net_pton_ipv4 does indeed do that.  Specifically at
            else if (size-- > 0)
                *++dst = 0, dirty = 0;
where, when size (the number of remaining destination bytes) is reduced
to 0, the code nonetheless advances dst and clears the next byte.
The loop logic is fundamentally faulty: you can't check for emsgsize
overflow until you get a digit that is supposed to go into another byte.
I'd try something like

    tmp = 0;
    ndigits = 0;           // ndigits is # of hex digits seen for cur byte
    while (ch = next hex digit)
    {
        n = numeric equivalent of ch;
        assert(n >= 0 && n <= 15);
        tmp = (tmp << 4) | n;
        if (++ndigits == 2)
        {
            if (size-- <= 0)
                goto emsgsize;
            *dst++ = (u_char) tmp;
            tmp = 0, ndigits = 0;
        }
    }
    if (ndigits)
        goto enoent;    // odd number of hex digits is bogus


BTW, shouldn't this routine clear out all "size" bytes of the
destination, even if the given data doesn't fill it all?  A memset
at the top might be worthwhile.

            regards, tom lane

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