Re: [current tinderbox] failure on ...all...

From: Giorgos Keramidas <keramida_at_freebsd.org>
Date: Fri, 10 Jun 2005 14:53:40 +0300
On 2005-06-10 14:02, Ruslan Ermilov <ru_at_freebsd.org> wrote:
>On Fri, Jun 10, 2005 at 01:28:13PM +0300, Giorgos Keramidas wrote:
>> That would be very hard to guarantee if two different modules that use
>> the types are compiled with different alignment options, right?
>>
>> 	/* header1.h */
>> 	struct t1 {short t1s; int t1a;};
>>
>> 	/* header2.h */
>> 	struct t2 {short t2s; int t2a;};
>>
>> 	/* module1.c */
>> 	#include "header1.h"
>> 	struct t1 x;
>>
>> 	/* module2.c */
>> 	#include "header1.h"
>> 	#include "header2.h"
>> 	extern struct t1 x;
>> 	struct t2 y;
>>
>> If the two modules are compiled with different options that may affect
>> struct member alignment, how would one ensure that it is correct to
>> use code like this in module.c?
>>
>> 	y.t2s = 10;
>> 	y.t2a = 100;
>> 	memcpy(&x, &y, sizeof(x));
>>
>> Even the use of sizeof(x) is tricky here, since there is no guarantee
>> that sizeof(x) < sizeof(y).
>
> And if both modules use the same t1 but compiled with different alignment
> options, the memcpy() in module.c won't work either, even a simple assignment
> could break.  :-)

Probably no.  Two struct t1 objects have the same type and assignment
between the two may be implemented as "element-at-a-time memcpy)", as
mentioned in $6.2.6.1(6) and footnote (42) of the C99 standard.  This
may or may not break though, but I couldn't find any good reference
to it.
Received on Fri Jun 10 2005 - 09:53:46 UTC

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