Hello!
On Sun, May 21, 2023 at 04:35:00PM +0200, Alejandro Colomar wrote:
> On 5/21/23 15:09, Maxim Dounin wrote:
> > Hello!
> >
> > On Sun, May 21, 2023 at 11:31:32AM +0200, Alejandro Colomar wrote:
> >
> >> On 5/21/23 03:42, Maxim Dounin wrote:
> >>> # HG changeset patch
> >>> # User Maxim Dounin <mdounin@mdounin.ru>
> >>> # Date 1684633125 -10800
> >>> # Sun May 21 04:38:45 2023 +0300
> >>> # Node ID 68fa4b86ed46138dd1a8fcf2cfd80206de068bec
> >>> # Parent 235d482ef6bc8c40a956b2413865d42c94e0fc05
> >>> QUIC: better sockaddr initialization.
> >>>
> >>> The qsock->sockaddr field is a ngx_sockaddr_t union, and therefore can hold
> >>> any sockaddr (and union members, such qsock->sockaddr.sockaddr, can be used
> >>> to access appropriate variant of the sockaddr). It is better to set it via
> >>> qsock->sockaddr itself though, and not qsock->sockaddr.sockaddr, so static
> >>> analyzers won't complain about out-of-bounds access.
> >>
> >> Correct. The previous code was UB, due to memcpy(3) writing to the
> >> 'struct sockaddr' member. By writing to sockaddr, you were only
> >> allowed to alias via other members the sa_family_t field, but no
> >> others.
> >
> > Well, not really. There is no UB in the previous code, it simply
> > uses a valid (void *) address to fill the sockaddr (and does so
> > without breaking strict aliasing rules).
>
> While the data being written was correctly written via memcpy(3),
> you wouldn't be allowed to access it later as anything that is
> not 'struct sockaddr'. For example, the following is a
> strict-aliasing violation:
>
> struct s { int a; int b; };
> struct t { int a; };
> union u { struct s s; struct t t; };
>
> struct s x = {42, 42};
> union u y;
> int z;
>
> memcpy(&y.t, &x, sizeof(x)); // This is fine
>
> // We created an object of type 'struct t' in the union.
> // Unions allow aliasing, so we're allowed to reinterpret
> // that object as a 'struct s' via the other member.
>
> z = y.s.a; // This is fine.
>
> // But we're not allowed to reinterpret bytes that are
> // officially uninitialized (even though we know they are
> // initialized).
>
> z = y.s.b; // UB here.
>
> The reason for the UB is that the compiler is free to assume
> that since you wrote to the struct t member, the write can't
> possibly write to the second member of the struct (even if
> the size passed to memcpy(3) is larger than that). In other
> words, the compiler may assume that anything past
> sizeof(struct t) is uninitialized.
You haven't wrote to the struct t member, you wrote to the address
using memcpy(). There is a difference, see C99 (or C11, whichever
you prefer), 6.5 Expressions.
> Also, writing past an
> object is very dubious, even via memcpy(3), even if you know
> that the storage is there (thanks to the union). It's just
> safer writing to the union itself, or to the field that has
> the correct object type.
And that's why the patch. While it is correct to write to the
memory with any pointer, using the union itself is more obvious
and causes less confusion.
[...]
--
Maxim Dounin
http://mdounin.ru/
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