On Tue, 16 Nov 2010 13:19:36 +0200 Boaz Harrosh <bharrosh@xxxxxxxxxxx> wrote: > On 11/15/2010 09:12 PM, Eric Dumazet wrote: > > Le dimanche 14 novembre 2010 __ 18:06 -0800, Andrew Morton a __crit : > >> On Sun, 14 Nov 2010 12:25:33 +0300 Vasiliy Kulikov <segoon@xxxxxxxxxxxx> wrote: > >>> > >>> if (timeval) { > >>> - rtv.tv_sec = rts.tv_sec; > >>> - rtv.tv_usec = rts.tv_nsec / NSEC_PER_USEC; > >>> + struct timeval rtv = { > >>> + .tv_sec = rts.tv_sec, > >>> + .tv_usec = rts.tv_nsec / NSEC_PER_USEC > >>> + }; > >>> > >>> if (!copy_to_user(p, &rtv, sizeof(rtv))) > >>> return ret; > >> > >> Please check the assembly code - this will still leave four bytes of > >> uninitalised stack data in 'rtv', surely. > > > > Thats a good question. > > > > In my understanding, gcc should initialize all holes (and other not > > mentioned fields) with 0, even for automatic storage [C99 only mandates > > this on static storage] > > > > I tested on x86_64 and this is the case, but could not find a definitive > > answer in gcc documentation. > > > > This kind of construct is widely used in networking tree. > > > > Maybe we should ask to gcc experts if this behavior is guaranteed by > > gcc, or if we must review our code ;( > > > > CC Jakub > > > > Thanks ! > > > > This is what I thought too. If it is not there are tones of bugs I wrote > of code that relays on this behaviour. > > It would be interesting to know for sure Well. We certainly assume in many places that struct foo { int a; int b; } f = { .a = 1, }; will initialise b to zero. But I doubt if much code at all assumes that this initialisation patterm will reliably zero out *holes* in the struct. -- To unsubscribe from this list: send the line "unsubscribe kernel-janitors" in the body of a message to majordomo@xxxxxxxxxxxxxxx More majordomo info at http://vger.kernel.org/majordomo-info.html