On Tue, 16 Nov 2010 13:19:36 +0200
Boaz Harrosh <bharrosh@panasas.com> wrote:

quoted text
> 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@openwall.com> 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.

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