Order of fields when using a bit field in C

Question

I have a struct of the following type

typedef struct
{
unsigned int a : 8;
unsigned int b : 6;
unsigned int c : 2;
}x, *ptr;

What i would like to do, is change the value of field c.

I do something like the following

x structure = { 0 };
x->c = 1;

When I look at the memory map, I expect to find 00 01, but instead I find 00 40. It looks like when arranging the second byte, it puts c field in the lowest bits and b field in the highest bits. I've seen this on both GCC and Windows compilers.

For now, what I do is the following, which is working OK.

unsigned char ptr2 = (unsigned char*) ptr
*(ptr2 + 1)  &= 0xFC
*(ptr2 + 1)  |= 0x01

Am I looking at the memory map wrong? Thank you for your help.

Answer 1

C standard allows compiler to put bit-fields in any order. There is no reliable and portable way to determine the order.

If you need to know the exact bit positions, it is better use plain unsigned variable and bit masking.

Here's one possible alternative to using bit-fields:

#include <stdio.h>

#define MASK_A    0x00FF
#define MASK_B    0x3F00
#define MASK_C    0xC000
#define SHIFT_A   0
#define SHIFT_B   8
#define SHIFT_C   14

unsigned GetField(unsigned all, unsigned mask, unsigned shift)
{
    return (all & mask) >> shift;
}

unsigned SetField(unsigned all, unsigned mask, unsigned shift, unsigned value)
{
    return (all & ~mask) | ((value << shift) & mask);
}

unsigned GetA(unsigned all)
{
    return GetField(all, MASK_A, SHIFT_A);
}

unsigned SetA(unsigned all, unsigned value)
{
    return SetField(all, MASK_A, SHIFT_A, value);
}

/* Similar functions for B and C here */

int main(void)
{
    unsigned myABC = 0;
    myABC = SetA(myABC, 3);
    printf("%u", GetA(myABC)); // Prints 3
}

Answer 2

I know this is an old one, but I would like to add my thoughts.

1. Headers in C are meant to be usable across objects, which means the compiler has to be somewhat consistent.

2. In my experience I have always seen bitfields in LSB order. This will put the bits in MSB->LSB order of c:b:a

The 16 bits, read in byte order is "00 40". Translated from Little-endian, this is a 16-bit value of 0x4000.

This is:

c == [15:14] == b'01
b == [13:8] == 0
a == [7:0] == 0

Answer 3

memory always depends on the underlying machine structure (endianness) and on the strategy for packing/arranging the structure the compiler is performing.

Answer 4

You set a C structure to raw bits at your peril.

You know that the bits are and what they mean, so you can fill out the fields of the structure. Yes it's more code than memcpy, but it won't break if someone adds a field, and if helps enforce bit-level specificity at the communcations level.