A simple (?) question about the size of structs

[Bryan Feeney]

This structure is, according to sizeof, 3 bytes long, which makes sense

struct test
{ char text[3];
};

This structure is, according to sizeof, 4 bytes long, which also makes sense

struct test
{ int number;
};

This structure is, according to sizeof, 8 bytes long, which is confusing the
hell out of me!

struct test
{ char text[3];
int number;
};

Where is the extra byte coming from? It's not something I need to know, but
I'd like to know what's going on. I'm using GCC3 on Windows via MinGW as
supplied by the Bloodshed Dev-C++ IDE (verson 4.9.8.0).
Thanks in advance
--
Bryan

[Chris McDonald]

"Bryan Feeney" <bf*****@oceanf ree.net> writes:

This structure is, according to sizeof, 3 bytes long, which makes sense struct test
{ char text[3];
}; This structure is, according to sizeof, 4 bytes long, which also makes sense struct test
{ int number;
}; This structure is, according to sizeof, 8 bytes long, which is confusing the
hell out of me! struct test
{ char text[3];
int number;
}; Where is the extra byte coming from? It's not something I need to know, but
I'd like to know what's going on. I'm using GCC3 on Windows via MinGW as
supplied by the Bloodshed Dev-C++ IDE (verson 4.9.8.0).

Your compiler and/or host architecture prefer or require to align
the integer field on a 4-byte boundary. At least under Unix on some
architectures, this may be required to avoid bus-errors occuring at
run-time, or preferred (for speed) so that the integers do not need
aligning at run-time.

That said, there can be other "gotchas" with some compilers/architectures:

struct {
char a;
int i;
char b;
}

The sizeof() this may still be 8 bytes, if the compiler chooses to pack
the two characters into the 4-byte space before (or after) the integer.

It is something of which you should be aware, if writing sizeof(structs)
to files or "down" a network connection.

_______________ _______________ _______________ _______________ _______________ ____
Dr Chris McDonald EMAIL: ch***@csse.uwa. edu.au
School of Computer Science & Software Engineering
The University of Western Australia WWW: http://www.csse.uwa.edu.au/~chris
Crawley, Western Australia, 6009 PH: [phone], FAX: [phone]

[Kevin Bracey]

In message <c7**********@e nyo.uwa.edu.au>
Chris McDonald <ch***@csse.uwa .edu.au> wrote:

That said, there can be other "gotchas" with some compilers/architectures:

struct {
char a;
int i;
char b;
}

The sizeof() this may still be 8 bytes, if the compiler chooses to pack
the two characters into the 4-byte space before (or after) the integer.

The rule about structures with common initial sequences in unions restricts
that slightly.

The addition of a member to the end of a structure is not allowed to change
the positions of earlier members.

Thus 'b' could be tucked in between 'a' and 'i', but 'a' couldn't be
placed after 'i'.

Is anyone actually aware of any implementations that do this though?

--
Kevin Bracey, Principal Software Engineer
Tematic Ltd Tel: [phone]
182-190 Newmarket Road Fax: [phone]
Cambridge, CB5 8HE, United Kingdom WWW: http://www.tematic.com/

[Dan Pop]

In <4b************ ****@tematic.co m> Kevin Bracey <ke**********@t ematic.com> writes:

In message <c7**********@e nyo.uwa.edu.au>
Chris McDonald <ch***@csse.uwa .edu.au> wrote:

That said, there can be other "gotchas" with some compilers/architectures:

struct {
char a;
int i;
char b;
}

The sizeof() this may still be 8 bytes, if the compiler chooses to pack
the two characters into the 4-byte space before (or after) the integer.
The rule about structures with common initial sequences in unions restricts
that slightly.

The addition of a member to the end of a structure is not allowed to change
the positions of earlier members.

Thus 'b' could be tucked in between 'a' and 'i', but 'a' couldn't be

^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^ ^placed after 'i'.

Nonsense!

13 Within a structure object, the non-bit-field members and the
units in which bit-fields reside have addresses that increase in
the order in which they are declared.

There is no way to have 'b' allocated before 'i'. Note that a correct
C program can trivially check this (e.g. using the offsetof macro) so
the as-if rule doesn't apply.

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de

[Thomas Matthews]

Bryan Feeney wrote:

This structure is, according to sizeof, 3 bytes long, which makes sense

struct test
{ char text[3];
};

This structure is, according to sizeof, 4 bytes long, which also makes sense

struct test
{ int number;
};

This structure is, according to sizeof, 8 bytes long, which is confusing the
hell out of me!

struct test
{ char text[3];
int number;
};

Where is the extra byte coming from? It's not something I need to know, but
I'd like to know what's going on. I'm using GCC3 on Windows via MinGW as
supplied by the Bloodshed Dev-C++ IDE (verson 4.9.8.0).
Thanks in advance
--
Bryan

Here is the rule on the sizes of structures:
The size of a structure may not be the sum of the sizes of its fields.

Reasoning:
Compilers are allowed to insert "padding" bytes between fields.
Compilers can also add "bookkeepin g" information in the structure
as well.

Some compilers like to keep the fields "aligned" to certain address
boundaries, such as 16-bit and 32-bits. This helps processors
access the data in a more efficient manner.

Mapping Structures To Data Or Hardware
Because of the padding rule, structures cannot be portably used to
map data to external formats (such as message formats or hardware
registers). One may keep the structure, but there must be code
to convert the external layout to the internal one.

Given a message header of:
command [8-bits]
length [16-bits]
The structure of:
struct Message
{
unsigned char command; /* CHAR_BITS == 8 */
unsigned short length;
};
allows the compiler to insert padding between the command and
length fields, so that they won't match the incoming data. And
not forgetting that an "unsigned short" or even "unsigned int"
may be larger than the 16-bits in the message.

However, once the data is placed into the structure variables,
the program can reference the fields with no problem. It's
just that the data has to be reformatted when outputting.

--
Thomas Matthews

C++ newsgroup welcome message:
http://www.slack.net/~shiva/welcome.txt
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Other sites:
http://www.josuttis.com -- C++ STL Library book

[Peter Ammon]

Bryan Feeney wrote:

This structure is, according to sizeof, 3 bytes long, which makes sense

struct test
{ char text[3];
};

This structure is, according to sizeof, 4 bytes long, which also makes sense

struct test
{ int number;
};

This structure is, according to sizeof, 8 bytes long, which is confusing the
hell out of me!

struct test
{ char text[3];
int number;
};

Where is the extra byte coming from? It's not something I need to know, but
I'd like to know what's going on. I'm using GCC3 on Windows via MinGW as
supplied by the Bloodshed Dev-C++ IDE (verson 4.9.8.0).
Thanks in advance
--
Bryan

I'm really surprised that nobody has pointed out that this is a FAQ.

<http://www.eskimo.com/~scs/C-faq/q2.13.html>

-Peter

[Mark McIntyre]

On Wed, 12 May 2004 14:21:30 +0100, in comp.lang.c , "Bryan Feeney"
<bf*****@oceanf ree.net> wrote:

This structure is, according to sizeof, 8 bytes long, which is confusing the
hell out of me!

struct test
{ char text[3];
int number;
};

Where is the extra byte coming from?

FAQ 2.12 and 2.13

--
Mark McIntyre
CLC FAQ <http://www.eskimo.com/~scs/C-faq/top.html>
CLC readme: <http://www.angelfire.c om/ms3/bchambless0/welcome_to_clc. html>
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[glen herrmannsfeldt]

Dan Pop wrote:

(someone wrote)

struct {
char a;
int i;
char b;

(someone else wrote)
Thus 'b' could be tucked in between 'a' and 'i', but 'a' couldn't be

Nonsense! 13 Within a structure object, the non-bit-field members and the
units in which bit-fields reside have addresses that increase in
the order in which they are declared. There is no way to have 'b' allocated before 'i'. Note that a correct
C program can trivially check this (e.g. using the offsetof macro) so
the as-if rule doesn't apply.

Can the compiler detect use of the offsetof macro?

There are stories related to SPEC and tricks compilers
use when they discover that they are compiling a SPEC program.

Or maybe only when a certain compiler flag is set to allow it.

-- glen

[Dan Pop]

In <9RDoc.37479$xw 3.2495220@attbi _s04> glen herrmannsfeldt <ga*@ugcs.calte ch.edu> writes:

Dan Pop wrote:

(someone wrote)

struct {
char a;
int i;
char b;
(someone else wrote)
Thus 'b' could be tucked in between 'a' and 'i', but 'a' couldn't be

Nonsense!

13 Within a structure object, the non-bit-field members and the
units in which bit-fields reside have addresses that increase in
the order in which they are declared.

There is no way to have 'b' allocated before 'i'. Note that a correct
C program can trivially check this (e.g. using the offsetof macro) so
the as-if rule doesn't apply.

Can the compiler detect use of the offsetof macro?

What the compiler cannot detect is the usage of the offsetof macro
in another translation unit using the same structure ;-)
There are stories related to SPEC and tricks compilers
use when they discover that they are compiling a SPEC program.

What is not a story is that many commercial compilers have their
optimisers tuned for the SPEC programs. As a result, the compiler
generates optimal code for the SPEC programs naturally, without making
any attempt to recognise them.

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de