digitalmars.D.bugs - Alignment
- Artem Rebrov (27/27) Oct 24 2005 In following example size of struct B2 is 16. Why It is not 13? Is this ...
- Jarrett Billingsley (6/31) Oct 24 2005 I might be wrong, but I think it's because B1 itself is aligned on 4-byt...
- Lionello Lunesu (7/42) Oct 24 2005 Hi.
- Uwe Salomon (19/19) Oct 25 2005 ####
- Sean Kelly (4/21) Oct 25 2005 This may be. Though for what it's worth, I ran the test program with th...
- Uwe Salomon (20/23) Oct 26 2005 Yes, and that's perfectly right.
- Lionello Lunesu (4/23) Oct 25 2005 I'm an idiot! I forgot the align(1), which was the whole point :-S
- Sean Kelly (3/29) Oct 26 2005 Oops. Now I understand what you intended to do. Sorry about that.
- Sean Kelly (11/13) Oct 24 2005 It's not 13 because of the order you have the variables declared. If yo...
- Artem Rebrov (38/53) Oct 26 2005 Unfortunately I can't change order of struct members. I use functions fr...
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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Oct 24 2005
"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.
Oct 24 2005
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.
Oct 24 2005
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
Oct 25 2005
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
Oct 25 2005
"Uwe Salomon" <post uwesalomon.de> 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
Oct 26 2005
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
Oct 25 2005
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
Oct 26 2005
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
Oct 24 2005
On Tue, 25 Oct 2005 02:46:11 +0400, Sean Kelly <sean f4.ca> wrote:In article <op.sy506hcuncj208 localhost>, Artem Rebrov says...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? -- Using Opera's revolutionary e-mail client: http://www.opera.com/mail/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.
Oct 26 2005