Re: Bit field vs bit manipulation?

From:
=?Utf-8?B?RGF2aWQgQmVuZGVy?= <DavidBender@discussions.microsoft.com>
Newsgroups:
microsoft.public.vc.language
Date:
Mon, 13 Aug 2007 09:20:02 -0700
Message-ID:
<051B1374-EC3E-4E94-917D-9FD992A65C8B@microsoft.com>
"Ondrej Spanel" wrote:

Example:

struct twelve {
WORD high : 12;
WORD low : 12;
} parts;

low = (( byte.b & 0x0f) * 256) + byte.c;
high= ((byte.b & 0xf0) >> 4) + (byte.a * 16);


First is definitely more readable. Second may be preferred when you need to
maintain crossplatform compatibility of the data layout, as ordering of
items in bitfieds may vary between compilers / CPU architectures. Even if
you use second, I would probably avoid manipulating bytes and write the code
as:

    low = ( input& 0xfff );
    high = ( input & 0xfff000 )>>12;

This is much better readable and it will actually even perform a lot better
than your example.

If your intention was to have the data stored or accessed as three bytes in
the memory, then you will need to use a code like your second version, as
bitfield storage is always represented as 32b int.

Regards
Ondrej


 Thanks! That is a great help. I did figure it out later and simple
used the following to achieve my goal;

 union groups {
 struct bits {
 unsigned int high : 12;
 unsigned in low : 12;
const unisgned : 8;
} bit;
struct bytes {
 unsigned char msb;
 unsigned char nsb;
 unsigned char lsb;
} byte;
} group;

 The most important goal was not to corrupt data everytime I read a new
byte. Which is fixed now. Using the above "union" to build a table of
12 bit values, only once because I dont care about repeats at this time.

 Thanks.

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