Re: Any pointer to Byte Alignment & Structure Padding?

["Amit Dang" <amit_dang@intersolutions.stpn.soft.net>]

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).
What I have understood atleast for gcc compiler Linux 32-bit machine is
that, Maximum byte boundary is 4. Structure will be aligned to byte boundary
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.

if i modify the above struct to
struct {
    int i;
    char j;
} its size will be 8.
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.

Assumptions: sizeof(char) = 1 byte, sizeof(short) = 2 byte, sizeof(int) = 4
byte, sizeof(long long) = 8 bytes.
Valid for gcc 2.96 Linux 32-bit machine

Regards,
Amit Dang
Regards,
Amit Dang


----- Original Message ----- 
From: "Vadiraj" <vadiraj.cs@gmail.com>
To: "Prabhat Hegde" <pubs.hegde@gmail.com>
Cc: "linux-c-programming" <linux-c-programming@vger.kernel.org>
Sent: Thursday, August 04, 2005 8:53 PM
Subject: Re: Any pointer to Byte Alignment & Structure Padding?


> On 8/4/05, Prabhat Hegde <pubs.hegde@gmail.com> wrote:
> > According to what you have said, dont you think it depnds from which
> > address your affset begins.
> >   Considring your last example,
> >    struct temp
> > > {
> > >       char c;   /* 1 byte lenght */
> > >       int i;      /* 4 byte length */
> > >       long long d /* 8 bytes  lenght */
> > >       char c1;
> > > };
> >
> >   Since char (1 byte ) can start anywhere withour padding,
> >   Lets take x to be at address X wherein (X mod 4 = 3), then in that
> > case "i" will have no padding since char (1 byte ) will make the
> > starting address of the int to agree with a 4 byte boundary.
> >   Now again the next long having a 8 byte boundary willl be/ may not
> > be padded depanding on the starting address.
> >  So what i feel is that begin the variable which has the maximum byte
> > boundary requirement. Now since all nos divisible by 8 are divisible
> > by 4, we will be cutting out the padding for int. Char can come in the
> > last.
>
>   The starting address is assured to be 4 byte boundary. Some
> compilers provide 8 byte boundary too. They never start with 1 byte
> boundary.
>
> That is the reason why the structure gets padded even at the end. To make
it
> 4 bytes alligned.
>
>
> >   hence what if the structure is like the below:
> >    struct temp{
> >        long lond d;   // x + padding if required to suffice 8 byte
boundary
> >        int i;             // no padding
> >        char c;         // no padding
> >        char c1;       // no padding
> >        };
>
>   For this example the size should come to 8+4+1+1=14 bytes. But when
> you check it will be 16 bytes. Due to padding, to make it 4/8 byte
> boundry.
>
> Take it for granted you get either 4 byte or 8 byte boundry but never 1
byte.
>
> Arranging the structure as mentioned above will definitely remove
> portablility issue
> as for both 4 byte or 8 byte aligned compilers the size will be 16 bytes.
>
> -- 
> cheers,
> Vadi
> -
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