Re: bit fields in a structure

From: Peter (phaywood_at_alphalink.com.au.NO.SPAM)
Date: 05/22/04 

Date: Sat, 22 May 2004 01:32:14 GMT

Groovy hepcat Hetal was jivin' on 17 May 2004 14:45:01 -0700 in
comp.lang.c.
bit fields in a structure's a cool scene! Dig it!

>I have a structure as below:
>
>typedef struct
>{
> UInt8 data1;
> UInt8 data2_4b : 4,
> data2_nxt4b : 4;
> UInt16 data3;
>} data_T;
>
>assuming ..
>UInt8 = unsigned char
>UInt16 = unsigned short int
>
>My code is running on a Windows system (Little Endian)
>When I write code to store data in this structure and view the memory
>windows in Visual Studio ... this is wht I get.
>
>when I set data2_4b = 7; and data2_nxt4b = 3; the memory window shows
>...
>
>0011 0111 (0x37)
>Shouldn't I be seeing 0111 0011 (0x78). Thus looks like when splitting
>the byte .. the processor is storing the first 4 bits in the least
>significat half byte and the second four bits in the most significant
>half byte.

  Of course. That's because it's not only the bytes that have little
endian order, but the bits within each byte too. Or, to put it another
way, bytes are "backwards" on Intel hardware. So, the "first" few bits
occupy the "back" of the byte, while the "last" bits occupy the
"front".
  Of course, all this is off topic.

>Is this a processor specific thing ? That is will this code execute

  Yes. Some other processors have the same bit order architecture.
Others don't.

>differently on diffeent systems ?

  On some, yes. 

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