From: Martin Ambuhl <mambuhl AT earthlink DOT net>
Newsgroups: comp.os.msdos.djgpp
Subject: Re: bitfields
Date: Fri, 30 Apr 1999 01:49:44 -0400
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To: djgpp AT delorie DOT com
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Ralph Allan Rice wrote:
> 
> Hi...
> 
> In DJGPP for x86, I wanted to program a bitfield structure like this:
> 
> struct  codes
> {
>    int field1 : 4;
>    int field2: 4;
> };
> 
> when I declare sucha structure, lets say like this:
> 
> struct codes x;
> 
> x.field1 = 0xa;
> x.field2 = 0xb;
> 
> And I want to cast it to a short int, it does not convert as expected.
> 
> ex:
>   short int y = (short int) x;
> 
> when I print out y, it results in only half of the desired result.  I think
> it has to do with bitfield alignment.  Is there a compiler option to set the
> alignment so that y would give me a desired result (such as 0xab or 0xba,
> but not something wierd);

Try this:
#include <stdio.h>

typedef struct {
    unsigned field1:4;
    unsigned field2:4;
} fields;

int main(void)
{
    fields x = { field1:0xa, field2:0xb}; /* GNUism */
    unsigned short y;
    y = 0xff & *(unsigned short *)&x; /* To circumvent error of 
                                       * assigning aggregate to 
                                       * scalar; mask out high
                                       * order bytes */
    printf("x contains field1=%#x, field2=%#x\n"
            "y contains %#x\n", 
            x.field1, x.field2, y);
    return 0;
}

which yields:

x contains field1=0xa, field2=0xb
y contains 0xba

-- 
Martin Ambuhl (mambuhl AT earthlink DOT net)
Note: mambuhl AT tiac DOT net will soon be inactive