In following example size of struct B2 is 16. Why It is not 13? Is this a bug?

align(1) struct A1
{   byte a;
    int  b;
}
align(1) struct A2
{	byte a;
    int  b;
    A1 c;
}
align(4) struct B1
{   byte a;
    int  b;
}
align(1) struct B2
{   byte a;
    int  b;
    B1 c;
}

void main()
{
	printf("%d %d ",A1.sizeof,A2.sizeof); //prints 5 10
	printf("%d %d ",B1.sizeof,B2.sizeof); //prints 8 16
}

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"Artem Rebrov" <ar_other@mail.ru> wrote in message news:op.sy506hcuncj208@localhost...
> In following example size of struct B2 is 16. Why It is not 13? Is this a bug?
>
> align(1) struct A1
> {   byte a;
>     int  b;
> }
> align(1) struct A2
> { byte a;
>     int  b;
>     A1 c;
> }
> align(4) struct B1
> {   byte a;
>     int  b;
> }
> align(1) struct B2
> {   byte a;
>     int  b;
>     B1 c;
> }
>
> void main()
> {
> printf("%d %d ",A1.sizeof,A2.sizeof); //prints 5 10
> printf("%d %d ",B1.sizeof,B2.sizeof); //prints 8 16
> }
>

I might be wrong, but I think it's because B1 itself is aligned on 4-byte boundaries.  Thus, the memory layout of B2 is:

abbb bxxx cccc cccc

Where x is unused padding, and each letter represents one byte.

In article <op.sy506hcuncj208@localhost>, Artem Rebrov says...
>
>In following example size of struct B2 is 16. Why It is not 13? Is this a bug?

It's not 13 because of the order you have the variables declared.  If you change it to this:

align(1) struct B2
{
int  b;
B1 c;
byte a;
}

Then it will have a size of 13.  The problem was the padding between a and b.


Sean

Hi.
Check Sean's post: he's right, it's the padding between 'a' and 'b', not
between 'b' and 'c' like you've mentioned:

axxx bbbb cccc cccc

L.

"Jarrett Billingsley" <kb3ctd2@yahoo.com> wrote in message news:djjn9f$1mnf$1@digitaldaemon.com...
> "Artem Rebrov" <ar_other@mail.ru> wrote in message news:op.sy506hcuncj208@localhost...
>> In following example size of struct B2 is 16. Why It is not 13? Is this a bug?
>>
>> align(1) struct A1
>> {   byte a;
>>     int  b;
>> }
>> align(1) struct A2
>> { byte a;
>>     int  b;
>>     A1 c;
>> }
>> align(4) struct B1
>> {   byte a;
>>     int  b;
>> }
>> align(1) struct B2
>> {   byte a;
>>     int  b;
>>     B1 c;
>> }
>>
>> void main()
>> {
>> printf("%d %d ",A1.sizeof,A2.sizeof); //prints 5 10
>> printf("%d %d ",B1.sizeof,B2.sizeof); //prints 8 16
>> }
>>
>
> I might be wrong, but I think it's because B1 itself is aligned on 4-byte boundaries.  Thus, the memory layout of B2 is:
>
> abbb bxxx cccc cccc
>
> Where x is unused padding, and each letter represents one byte.
> 

####
align(4) struct B1
{   byte a;
    int  b;
}
align(1) struct B2
{   byte a;
    int  b;
    B1 c;
}

int main()
{
  printf("%d %d \n",B2.a.offsetof,B2.b.offsetof);
}
####

This prints "0 1". This suggests that Sean is wrong, and Artem is right.  :)

Ciao
uwe

In article <op.sy7vt4so6yjbe6@sandmann.maerchenwald.net>, Uwe Salomon says...
>
>####
>align(4) struct B1
>{   byte a;
>     int  b;
>}
>align(1) struct B2
>{   byte a;
>     int  b;
>     B1 c;
>}
>
>int main()
>{
>   printf("%d %d \n",B2.a.offsetof,B2.b.offsetof);
>}
>####
>
>This prints "0 1". This suggests that Sean is wrong, and Artem is right. :)

This may be.  Though for what it's worth, I ran the test program with the struct I posted and its sizeof was 13.


Sean

I'm an idiot! I forgot the align(1), which was the whole point :-S

L.

"Uwe Salomon" <post@uwesalomon.de> wrote in message news:op.sy7vt4so6yjbe6@sandmann.maerchenwald.net...
> ####
> align(4) struct B1
> {   byte a;
>     int  b;
> }
> align(1) struct B2
> {   byte a;
>     int  b;
>     B1 c;
> }
>
> int main()
> {
>   printf("%d %d \n",B2.a.offsetof,B2.b.offsetof);
> }
> ####
>
> This prints "0 1". This suggests that Sean is wrong, and Artem is right. :)
>
> Ciao
> uwe

> This may be.  Though for what it's worth, I ran the test program with the struct
> I posted and its sizeof was 13.

Yes, and that's perfectly right.

struct B2
{
  int a; // sizeof 4
  B1 b; // sizeof is 8
  byte c; // sizeof is 1
}

As b is already aligned to a 4-byte-boundary, there is no padding between a and b. Try this:

struct B2
{
  byte a;
  B1 b;
  int c;
}

Now there are 3 bytes padding between a and b, because a.sizeof is 1 and b is aligned on a 4-byte-boundary. Together that makes 12 bytes, and the 4 from int c result in 16 bytes B2.sizeof.

Ciao
uwe

Oops.  Now I understand what you intended to do.  Sorry about that.


Sean

In article <djn80a$8hs$1@digitaldaemon.com>, Lionello Lunesu says...
>
>I'm an idiot! I forgot the align(1), which was the whole point :-S
>
>L.
>
>"Uwe Salomon" <post@uwesalomon.de> wrote in message news:op.sy7vt4so6yjbe6@sandmann.maerchenwald.net...
>> ####
>> align(4) struct B1
>> {   byte a;
>>     int  b;
>> }
>> align(1) struct B2
>> {   byte a;
>>     int  b;
>>     B1 c;
>> }
>>
>> int main()
>> {
>>   printf("%d %d \n",B2.a.offsetof,B2.b.offsetof);
>> }
>> ####
>>
>> This prints "0 1". This suggests that Sean is wrong, and Artem is right. :)
>>
>> Ciao
>> uwe
>
>

On Tue, 25 Oct 2005 02:46:11 +0400, Sean Kelly <sean@f4.ca> wrote:

> In article <op.sy506hcuncj208@localhost>, Artem Rebrov says...
>> In following example size of struct B2 is 16. Why It is not 13? Is this a
>> bug?
> It's not 13 because of the order you have the variables declared.  If you change
> it to this:
>
> align(1) struct B2
> {
> int  b;
> B1 c;
> byte a;
> }
>
> Then it will have a size of 13.  The problem was the padding between a and b.

Unfortunately I can't change order of struct members. I use functions from
a library, in which almost all structs aligned on 4-byte boundaries.
Actually length of members is equal to 4.  Although there is small number
of structs, that have smaller members and struct-members.


When I write D module based on C header file I must keep in mind that each
struct can be a member of other struct. And I must add align(1) attribute
to it even if its size is multiple of 4. Or when I use a struct as a member
type I must check alignment and, if necessary, correct declaration.

I rewrite my example in C to determine the source of such effect.
=============
#include <stdio.h>
#pragma pack(4)
struct B1
{   char a;
    int  b;
};
#pragma pack(1)
struct B2
{   char a;
    int  b;
    struct B1 c;
};
void main()
{
    printf("%d %d ",sizeof(struct B1),sizeof(struct B2)); //prints 8 16
}
=============
With DMC I get the same result, but with Microsoft CL it prints "8 13" as I expect.
Therefore it matches the behavior of the companion C compiler.

Is this more natural to pack my struct B2 as "abbb bccc cccc c" rather than
"axxx bbbb cccc cccc"? The existing behaviour looks like "align(1) B2" takes
no effect. Are there any related standards for C with which DMC is complied?
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