Re: Clarifying bit-fields

From:
"James Kanze" <james.kanze@gmail.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Sat, 10 Mar 2007 10:55:19 CST
Message-ID:
<1173519708.598340.82290@v33g2000cwv.googlegroups.com>
On Mar 10, 9:56 am, "Gorc" <nebe...@yahoo.com> wrote:

On Mar 10, 2:55 am, "James Kanze" <james.ka...@gmail.com> wrote:

thank you a lot. A few more if I may:

Or may sizeof(bitf_1) be even
less than a byte?

How? The result is a size_t, which is an unsigned integral
type. So the only thing less than 1 is 0.


Bad question. Rephrased: having a

-------------------------------------------------------------------------
struct bitf_1
        {
        int m : 1;
        };

struct bitf_2
        {
        int m : 1;
        };

struct aggregate
        {
        bitf_1 b1;
        bitf_2 b2;
        };
-------------------------------------------------------------------------

is the compiler still conforming if *all* of the expressions bellow
are true:

-------------------------------------------------------------------------
sizeof( bitf_1 ) == 1;
sizeof( bitf_2 ) == 1;
sizeof( aggregate ) == 1; //!!
-------------------------------------------------------------------------


No. In aggregate, b1 and b2 are not bitfields, and can have
their address taken. And the addresses must be different.

I don't know of any reason why it should not be (consider the platform
with native bit arithmetic support for example).


Because the standard guarantees that the following works:

     aggregate a ;
     assert( &a.b1 != &a.b2 ) ;

<snip>

2) If constant-expression by a colon is larger than number of bits in

the objetc representation

<snip>

I'm not sure what you're trying to ask. The size of the struct
must be at least large enought to hold all of the bits asked
for.


<snip>

That was the question - must the size of the struct be "at least large
enought to hold all of the bits" or
"at least large enought to hold all of the bits *needed for the
types's value*".


My interpretation is the former. There's no may or anything in
the statement.

Personally, I'd prefer it be a diagnosable error. If I write
something like: "unsigned bits : 30", I expect to be able to use
values up to 2^30-1.

I understand your poistion but I am still in doubt as
padding is implementations specific. In theory padding that does not
break the standard is conforming, so for example - assuming 32 bit
integer -"downpadding" 128 bit bitfiled int m1 : 128; so that size of
the bitfield is 32 bits is IMHO legal.


I don't think that the C++ standard is as clear as it could be;
I can't even find any text which specifies the semantics of the
integral constant. Still, the C++ standard is based on C (90),
which says: "A bit-field is interpreted as an integral type
consisting of the specified number of bits." No more, no less.
I'm pretty sure that this is the intent. (Although I don't know
whether the intent in C was to ignore the extra bits, or simply
that the authors' of the C standard didn't think of the case.
Ignoring them is more or less implicite, however, since the
result of reading a bit field is an int; there is no such thing
as a type int:128, for example.)

BTW, not that it is a proof of
anything but many compilers do so.


G++ doesn't. (I think I tried it with Sun CC as well, but I'm
no longer sure, and I don't have it available here to double
check.)

Of course, it's the sort of thing that's likely to slip through
final testing, and not get a lot of complaints from customers,
either.

Padding which the standard is talking about here is
within a single element. And the wording in the standard
doesn't seem to make it optional.


I don't follow you here. What is "padding within a single element"?


Bits which don't participate in the value representation. See
?3.9.1/1: "For character types, all bits of the object
representation participate in the value representation. For
unsigned character types, all possible bit patterns of the value
representation represent numbers. These requirements do not hold
for other types." Off hand, the only example I can think of is
the now defunct Unysys (originally Burroughs) Series A: int's
were signed magnitude, in a 48 bit word, but only 40 bits
participated in the value representation. (The other 8 were
required to be 0. Otherwise, the hardware treated the number as
floating point.)

So on a machine with 32 bit int's, if I write something like:

     struct Toto
     {
         int titi : 1000 ;
     } ;

I have a 1000 bit int, but only 32 bits participate in the value
representation; the others are irrelevant padding. G++ says
that the size of the above is 128, on a 64 bit AMD machine with
sizeof int == 4. It also warns:

     bitfields.cc:13: warning: width of 'Toto::b' exceeds its type

which sounds right to me. (For once an error message that is
clear and understandable:-).)

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