Hi Steve,
Thanks for your prompt response and the valuable information. But I
tried following on Linux 32-bit gcc 2.96
struct temp {
long long i;
char c;
} and sizeof(struct temp) gave 12 not 16.
Regards,
Amit Dang
----- Original Message -----
From: "Steve Graegert" <[EMAIL PROTECTED]>
To: "Amit Dang" <[EMAIL PROTECTED]>
Cc: "linux-c-programming" <[email protected]>
Sent: Friday, August 05, 2005 12:02 PM
Subject: Re: Any pointer to Byte Alignment & Structure Padding?
> On 8/5/05, Amit Dang <[EMAIL PROTECTED]> wrote:
> > Hi Vadiraj,
> > The statement " Take it for granted you get either 4 byte or 8 byte
boundary
> > but never 1 byte." you made is it generic or just valid for the
structure in
> > question? If its generic then I have a question.
> > Why the size of
> > struct {
> > char i;
> > char j;
> > char k;
> > } is 3 ? (gcc 2.96 on Linux 32-bit machine).
>
> Because the alignment requirement for this structure is 1, it is
> byte-aligned. A structure is padded and properly aligned only if one
> of its members requires more than a single byte of storage. Take a
> look at this:
>
> struct a {
> char a;
> int :0;
> char b;
> };
>
> its size is 3, since it's byte-aligned also. This rule does not hold
> for the structure
>
> struct b {
> char a;
> short s;
> }
>
> since one member, here s, requires more than one byte of storage and
> must be aligned to a 4 byte boundary (4 is the smallest possible
> multiple of 2 larger than 3) resulting in sizeof(b) == 4.
>
> > What I have understood atleast for gcc compiler Linux 32-bit machine is
> > that, Maximum byte boundary is 4.
>
> True for int, but double and long long will always be 8 byte aligned by
default.
>
> > equal to minimum of (4 or field with maximum size (within the
structure)).
> > i.e. for the above example maximum field size if 1 and min (4, 1) = 1,
so
> > structure is aligned to 1 byte.
> > If I have following structure
> > struct {
> > short i;
> > char j;
> > } its size will be 4.
>
> True.
>
> > if i modify the above struct to
> > struct {
> > int i;
> > char j;
> > } its size will be 8.
>
> Yes, exactly.
>
> > Now modifying int to long long in the above structure will have a size
of
> > 12 not 16 because byte alignment min (4, 8) = 4.
>
> No, it will have a size of 16 because the alignment requirement for
> this structure is a multiple of its largest member. long long and
> double is always double word aligned.
>
> Regards
>
> \Steve
-
To unsubscribe from this list: send the line "unsubscribe linux-c-programming"
in
the body of a message to [EMAIL PROTECTED]
More majordomo info at http://vger.kernel.org/majordomo-info.html
