De facto standard size of bit-field objects

[marktxx]

Although the C90 standard only mentions the use of 'signed int' and
'unsigned int' for bit-fields (use 'int' at your own risk) and C99
adds _Bool.
It seems that most compilers create the size of the bit-field object
from the size used to specify the field.

Could this be considered a defacto standard now (at least for 8 bit
sized bit-fields)?

Any recent compilers not allowing this?

struct
{
unsigned char x:4;
unsigned char y:4;
} nibs;
struct
{
unsigned int x:4;
unsigned int y:4;
} nibs;

see

http://www.keil.com/support/docs/928.htm

It refers the OP to the book
"The C Programming Language" by Kernighan & Ritchie but this book
doesn't mention using unsigned char in bit-fields.
[Soapbox]
Wouldn't it had more sense for the compiler folks to just add up the
specified number bits and use the smallest integer type that would fit
so code could follow the standard?
Obviously programmers want to use bit-field objects that may be
smaller or larger than standard integer size shouldn't the standard
support that.
(I know that one can always use masking and shifting but that isn't my
question.)
[end soapbox]

[Keith Thompson]

ma*****@yahoo.com writes:

Although the C90 standard only mentions the use of 'signed int' and
'unsigned int' for bit-fields (use 'int' at your own risk) and C99
adds _Bool.
It seems that most compilers create the size of the bit-field object
from the size used to specify the field.

Of course.

Could this be considered a defacto standard now (at least for 8 bit
sized bit-fields)?

Any recent compilers not allowing this?

struct
{
unsigned char x:4;
unsigned char y:4;
} nibs;
struct
{
unsigned int x:4;
unsigned int y:4;
} nibs;

Oh, you meant that the *type* of the bit-field determines the size.

see

http://www.keil.com/support/docs/928.htm

Apparently for that compiler the type of a bit field affects the size
of the enclosing structure (though it doesn't affect the size of the
bit field itself). In the second declaration, the structure
apparently is at least as large as unsigned int.

C99 6.7.2.1p9:

A bit-field is interpreted as a signed or unsigned integer type
consisting of the specified number of bits.

Allowing unsigned char bit fields is obviously a compiler extension.
Making a struct bigger than it needs to be based on the declared type
of a bit field, rather than its declared width, is an odd choice and
is not required, or even suggested, by the standard as far as I can
tell. (I've seen this behavior in other compilers, including gcc.)

[...]

[Soapbox]
Wouldn't it had more sense for the compiler folks to just add up the
specified number bits and use the smallest integer type that would fit
so code could follow the standard?
Obviously programmers want to use bit-field objects that may be
smaller or larger than standard integer size shouldn't the standard
support that.
(I know that one can always use masking and shifting but that isn't my
question.)
[end soapbox]

The width (":4" in the examples above) determines the size of a bit
field, not the declared type. I suggest that the compiler should be
smart enough to treat int and unsigned int bit fields properly without
wasting space. I see no need to change the standard.

Perhaps there's some sensible rationale for this behavior, but I'm not
seeing it.

--
Keith Thompson (The_Other_Keith) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <* <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[Thad Smith]

Keith Thompson wrote:

ma*****@yahoo.com writes:

>>struct
{
unsigned char x:4;
unsigned char y:4;
} nibs;
...
http://www.keil.com/support/docs/928.htm

Apparently for that compiler the type of a bit field affects the size
of the enclosing structure (though it doesn't affect the size of the
bit field itself). In the second declaration, the structure
apparently is at least as large as unsigned int.
...
Allowing unsigned char bit fields is obviously a compiler extension.
Making a struct bigger than it needs to be based on the declared type
of a bit field, rather than its declared width, is an odd choice and
is not required, or even suggested, by the standard as far as I can
tell. (I've seen this behavior in other compilers, including gcc.)
...
The width (":4" in the examples above) determines the size of a bit
field, not the declared type. I suggest that the compiler should be
smart enough to treat int and unsigned int bit fields properly without
wasting space. I see no need to change the standard.

Perhaps there's some sensible rationale for this behavior, but I'm not
seeing it.

The rationale, I think, is to be compatible with other compilers that
set the minimum size for a struct containing bit fields as the size of
an int. The extension, then, provides that compatibility and allows
generating a minimum-size struct.

That said, I think that the int-size minimum and compatibility attempt
are wrong-headed. I agree that compilers should use the minimum size of
a struct unless that causes a penalty somewhere.

--
Thad

[Arthur J. O'Dwyer]

On Fri, 23 Feb 2007 ma*****@yahoo.com wrote:

>
Although the C90 standard only mentions the use of 'signed int' and
'unsigned int' for bit-fields (use 'int' at your own risk) and C99
adds _Bool.
It seems that most compilers create the size of the bit-field object
from the size used to specify the field.

Could this be considered a defacto standard now (at least for 8 bit
sized bit-fields)?

I don't exactly see what you mean. I think you are saying that on
"most" compilers, the structure definition

struct {
unsigned char x:4;
unsigned char y:4;
} nibs;

yields a struct with a size of 8 bits, arranged as xxxxyyyy, but

struct {
unsigned int x:4;
unsigned int y:4;
} nibs;

yields a struct with a size of 32 bits, arranged as
xxxxyyyy000000000000000000000000.

This is indeed true for Keil's compiler:

http://www.keil.com/support/docs/928.htm

and GCC and yes, probably most modern compilers. However, there's
an extra wrinkle that you didn't mention: On "most" compilers, a
bitfield of declared type T will never span memory chunks of
size T. (The real type of a bitfield is simply a "bit-field type"; but
like you, I'm talking about the "unsigned char" or whatever that you
use in the struct definition.)
For example, an "unsigned char" bitfield will never span two bytes;
padding bits will be inserted if necessary to justify it in its own
byte. Therefore, the struct definition

struct {
unsigned char x : 5;
unsigned char y : 5;
} nabs;

will correspond on "most" compilers to xxxxx000yyyyy000, while

struct {
unsigned short x : 5;
unsigned short y : 5;
} nabs;

will correspond to xxxxxyyyyy000000.

[Soapbox]
Wouldn't it had more sense for the compiler folks to just add up the
specified number bits and use the smallest integer type that would fit
so code could follow the standard?

It would make about as much sense, I guess. I don't see how it would
make /more/ sense. If you care about that kind of micro-optimization,
you probably welcome the extra tiny bit of control over alignment given
to you by the "de-facto" standard.
The issue may originally have been that unaligned memory accesses are
terribly slow on most platforms; therefore, it makes sense to allow the
programmer to force byte-alignment or word-alignment with a minimum of
fuss. (C99 introduced anonymous bitfields to deal with the same issue.)
The issue now is certainly compatibility with other compilers. Peer
pressure is a strong force in the compiler field.

N869 section 6.7.2.1#9 seems to encourage the "de-facto" behavior:

[#9] An implementation may allocate any addressable storage
unit large enough to hold a bit-field. If enough space
remains, a bit-field that immediately follows another bit-
field in a structure shall be packed into adjacent bits of
the same unit. If insufficient space remains, whether a
bit-field that does not fit is put into the next unit or
overlaps adjacent units is implementation-defined. The
order of allocation of bit-fields within a unit (high-order
to low-order or low-order to high-order) is implementation-
defined. The alignment of the addressable storage unit is
unspecified.

Obviously programmers want to use bit-field objects that may be
smaller or larger than standard integer size shouldn't the standard
support that.

No. If there's no Standard support for 128-bit integer math, it
seems pretty silly to require implementations to support integer
math on bitfields of type "signed int foo : 128". That would put a
huge burden on implementors to deal with arbitrarily-wide integer
math, while making users jump through silly hoops to get at it.

Some compilers support "long long" bitfields. Interestingly, GCC
will pack "long long" bitfields across 8-byte boundaries, and will pad
them only to 4-byte boundaries (e.g., a struct containing two fields
of type "long long : 4" will have size 32 bits, not 64 bits). That
seems needlessly inconsistent to me, but I don't know what other
compilers do. I'll find out what ours does on Monday. ;)

-Arthur,
one of those compiler folks

[Keith Thompson]

"Arthur J. O'Dwyer" <aj*******@andrew.cmu.eduwrites:
[...]

(C99 introduced anonymous bitfields to deal with the same issue.)

[...]

C90 has anonymous bitfields.

--
Keith Thompson (The_Other_Keith) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <* <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[marktxx]

On Feb 23, 11:43 pm, "Arthur J. O'Dwyer" <[e-mail]>
wrote:

On Fri, 23 Feb 2007 [e-mail] wrote:

Although the C90 standard only mentions the use of 'signed int' and
'unsigned int' for bit-fields (use 'int' at your own risk) and C99
adds _Bool.
It seems that most compilers create the size of the bit-field object
from the size used to specify the field.

Could this be considered a defacto standard now (at least for 8 bit
sized bit-fields)?

I don't exactly see what you mean. I think you are saying that on
"most" compilers, the structure definition

struct {
unsigned char x:4;
unsigned char y:4;
} nibs;

yields a struct with a size of 8 bits, arranged as xxxxyyyy, but

struct {
unsigned int x:4;
unsigned int y:4;
} nibs;

yields a struct with a size of 32 bits, arranged as
xxxxyyyy000000000000000000000000.

This is indeed true for Keil's compiler:

http://www.keil.com/support/docs/928.htm

and GCC and yes, probably most modern compilers. However, there's
an extra wrinkle that you didn't mention: On "most" compilers, a
bitfield of declared type T will never span memory chunks of
size T. (The real type of a bitfield is simply a "bit-field type"; but
like you, I'm talking about the "unsigned char" or whatever that you
use in the struct definition.)
For example, an "unsigned char" bitfield will never span two bytes;
padding bits will be inserted if necessary to justify it in its own
byte.

bitfield of declared type T - I originally thought that bitfields
were of bit-field-type, that K&R could have created a new reserved/key
word for, such as 'bitf', if desired.

not real code:
struct {
unsigned bitf x : 5;
unsigned bitf y : 5;
signed bitf z : 6;
} nabs;

But it seems now that type has more meaning.

Therefore, the struct definition

>
struct {
unsigned char x : 5;
unsigned char y : 5;
} nabs;

will correspond on "most" compilers to xxxxx000yyyyy000, while

struct {
unsigned short x : 5;
unsigned short y : 5;
} nabs;

will correspond to xxxxxyyyyy000000.

If you care about that kind of micro-optimization,

you probably welcome the extra tiny bit of control over alignment given
to you by the "de-facto" standard.
The issue may originally have been that unaligned memory accesses are
terribly slow on most platforms; therefore, it makes sense to allow the
programmer to force byte-alignment or word-alignment with a minimum of
fuss. (C99 introduced anonymous bitfields to deal with the same issue.)
The issue now is certainly compatibility with other compilers. Peer
pressure is a strong force in the compiler field.

When you interface with hardware or send packed data between CPUs
the bits have to be exact which is probably the primary motivation
here.

>
N869 section 6.7.2.1#9 seems to encourage the "de-facto" behavior:

[#9] An implementation may allocate any addressable storage
unit large enough to hold a bit-field. If enough space
remains, a bit-field that immediately follows another bit-
field in a structure shall be packed into adjacent bits of
the same unit. If insufficient space remains, whether a
bit-field that does not fit is put into the next unit or
overlaps adjacent units is implementation-defined. The
order of allocation of bit-fields within a unit (high-order
to low-order or low-order to high-order) is implementation-
defined. The alignment of the addressable storage unit is
unspecified.

Obviously programmers want to use bit-field objects that may be
smaller or larger than standard integer size shouldn't the standard
support that.

No. If there's no Standard support for 128-bit integer math, it
seems pretty silly to require implementations to support integer
math on bitfields of type "signed int foo : 128". That would put a
huge burden on implementors to deal with arbitrarily-wide integer
math, while making users jump through silly hoops to get at it.

I stand corrected the standard shouldn't force the use of these bit-
field types but maybe have an optional supplement/section that would
encourage compilers to extend in the same manner.

[Malcolm McLean]

"Arthur J. O'Dwyer" <aj*******@andrew.cmu.eduwrote

No. If there's no Standard support for 128-bit integer math, it
seems pretty silly to require implementations to support integer
math on bitfields of type "signed int foo : 128". That would put a
huge burden on implementors to deal with arbitrarily-wide integer
math, while making users jump through silly hoops to get at it.

Not a huge burden. It's another job which someone writing a quick and
cheerful compiler could probably do without, but no harder than implenting
floating -point arithmetic on machines without hardware float registers, for
instance.