----Original Message----
From: boost-bounces@lists.boost.org
[mailto:boost-bounces@lists.boost.org] On Behalf Of Andy Little Sent:
07 March 2006 17:43 To: boost@lists.boost.org
Subject: Re: [boost] [bitfield] Initial bitfield proposal
availableinthevault
"Martin Bonner" wrote
Andy Little wrote
"Martin Bonner" <martin.bonner@pitechnology.com> wrote in message
news:D997BF79D1E92C4793B7FCC04B4F90A51D79B6@pigeon.pi.local...
----Original Message----
From: Emile Cormier
The bitfield mechanism relies on this assumption: Unions of
non-polymorphic, non-derived objects, having the exact same
underlying data member type, will have the same size as this
underlying data member type. I'm no language lawyer, so please let
me know if this is a safe and portable assumption.
I'm not quite sure what you mean, but given:
struct a { unsigned char ch; };
struct b { unsigned char ch; };
union u { a theA; b theB };
then you are not guaranteed that sizeof(u) == sizeof(unsigned
char).
Though in practise you can use:
BOOST_STATIC_ASSERT(sizeof(u) == sizeof(unsigned char))
My point was exactly that you CANNOT use that. (On a certain class
of machine).
Why not?. Will it a) compile but be incorrect or b) not compile but
be incorrect
or c) not compile but be correct?
The assert will fire.
How do you store an unsigned char then? (And whatever way that is
just pretend to the hardware that the struct is an unsigned char)
Storing an unsigned char is expensive. It involves extra bit twiddling.
(See below)
On word addressed machines (which /are/ still being built), it is
almost certain that the minimum size for a struct is a complete
word. This is because the C and C++ standards effectively promise
that pointers to structs are all of the same size (the size of a
pointer-to-struct does not depend on the contents of the struct).
It is desirable that a pointer-to-struct be the smaller,
cheaper-to-dereference pointer to word (rather than the larger
more-expensive-to-dereference pointer to char), so the smallest
struct has to occupy a whole word.
I dont see why the size of a pointer to a struct affects the size of
a struct which in the case of an empty struct is often 1 byte?
I don't think you have understood what a word addressed machine is!
On most modern archictectures there are 8 bits stored at (for
example) 0x100 and another 8 bits at 0x101. The 32 bits at 0x100
cover 0x100, 0x101, 0x102, and 0x103.
On a word addressed machine, there may be 36 bits stored at 02000 and
another (different) 36 bits stored at 02001. A simple 36-bit pointer
can address individual words, but not sub-units within those words.
To address individual bytes, you need a double-word pointer. One
word identifies the word, and a few bits within the second word
identifies which byte you are addressing.
On such a machine, it makes sense for an empty struct to occupy a
whole word (which is four nine-bit bytes), so that a pointer to
struct can (always) be a single word pointer.
Sounds like there is a choice. Either make unsigned char 36 bits and
use a small pointer or make unsigned char 9 bits and use a large
pointer.
Yup. And the COMPILER writer gets to make that choice.
I dont know whether C++ will allow both?
It will allow the compiler writer to make either of those choices,
It reminds me of the old Microchip PIC architecture
though. Last I looked they were working to make their hardware
compatible with C FWIW and just increasing the number of address
lines, because they were previously so difficult to deal with, with
the separate extra bits in an address and so on (though that was a
kind of paged memory IIRC). IOW in their case they
realised the downside of the idea as I understand it and moved to one
drop linear addressing for later designs.
I believe it is a similar sort of idea. This is the same sort of thing
as Prime changing their instruction set so that memset( ptr, 0,
sizeof(ptr) ) set ptr to a null pointer (the natural representation of a
null pointer on a Prime was 07777/000000).
Maybe I have got the wrong end of the proverbial stick again though ?
My point is that assuming
BOOST_STATIC_ASSERT(sizeof(u) == sizeof(unsigned char))
Means that there is a class of C++ implementations where the library
will not work. It is then up to the library author to consider whether
that class is suffiently important to change his implementation for (it
may well not be).
--
Martin Bonner
Martin.Bonner@Pitechnology.com
Pi Technology, Milton Hall, Ely Road, Milton, Cambridge, CB4 6WZ,
ENGLAND Tel: +44 (0)1223 203894
Martin Bonner