C/C++: Force Bit Field Order and Alignment

Bitfields should be avoided - they aren't very portable between compilers even for the same platform. from the C99 standard 6.7.2.1/10 - "Structure and union specifiers" (there's similar wording in the C90 standard):

An implementation may allocate any addressable storage unit large enough to hold a bitfield. If enough space remains, a bit-field that immediately follows another bit-field in a structure shall be packed into adjacent bits of the same unit. If insufficient space remains, whether a bit-field that does not fit is put into the next unit or overlaps adjacent units is implementation-defined. The order of allocation of bit-fields within a unit (high-order to low-order or low-order to high-order) is implementation-defined. The alignment of the addressable storage unit is unspecified.

You cannot guarantee whether a bit field will 'span' an int boundary or not and you can't specify whether a bitfield starts at the low-end of the int or the high end of the int (this is independant of whether the processor is big-endian or little-endian).

Prefer bitmasks. Use inlines (or even macros) to set, clear and test the bits.

No, it will not be fully-portable. Packing options for structs are extensions, and are themselves not fully portable. In addition to that, C99 §6.7.2.1, paragraph 10 says: "The order of allocation of bit-fields within a unit (high-order to low-order or low-order to high-order) is implementation-defined."

Even a single compiler might lay the bit field out differently depending on the endianness of the target platform, for example.

endianness are talking about byte orders not bit orders. Nowadays , it is 99% sure that bit orders are fixed. However, when using bitfields, endianness should be taken in count. See the example below.

#include <stdio.h>

typedef struct tagT{

    int a:4;
    int b:4;
    int c:8;
    int d:16;
}T;


int main()
{
    char data[]={0x12,0x34,0x56,0x78};
    T *t = (T*)data;
    printf("a =0x%x\n" ,t->a);
    printf("b =0x%x\n" ,t->b);
    printf("c =0x%x\n" ,t->c);
    printf("d =0x%x\n" ,t->d);

    return 0;
}

//- big endian :  mips24k-linux-gcc (GCC) 4.2.3 - big endian
a =0x1
b =0x2
c =0x34
d =0x5678
 1   2   3   4   5   6   7   8
\_/ \_/ \_____/ \_____________/
 a   b     c           d

// - little endian : gcc (Ubuntu 4.3.2-1ubuntu11) 4.3.2
a =0x2
b =0x1
c =0x34
d =0x7856
 7   8   5   6   3   4   1   2
\_____________/ \_____/ \_/ \_/
       d           c     b   a

Bit fields vary widely from compiler to compiler, sorry.

With GCC, big endian machines lay out the bits big end first and little endian machines lay out the bits little end first.

K&R says "Adjacent [bit-]field members of structures are packed into implementation-dependent storage units in an implementation-dependent direction. When a field following another field will not fit ... it may be split between units or the unit may be padded. An unnamed field of width 0 forces this padding..."

Therefore, if you need machine independent binary layout you must do it yourself.

This last statement also applies to non-bitfields due to padding -- however all compilers seem to have some way of forcing byte packing of a structure, as I see you already discovered for GCC.