Question

I need to pack some bits in a byte in this fashion:

struct  
{  
  char bit0: 1;  
  char bit1: 1;  
} a;  

if( a.bit1 ) /* etc */

or:

if( a & 0x2 ) /* etc */

From the source code clarity it’s pretty obvious to me that bitfields are neater. But which option is faster? I know the speed difference won’t be too much if any, but as I can use any of them, if one’s faster, better.
On the other hand, I’ve read that bitfields are not guaranteed to arrange bits in the same order across platforms, and I want my code to be portable.

Notes: If you plan to answer ‘Profile’ ok, I will, but as I’m lazy, if someone already has the answer, much better.
The code may be wrong, you can correct me if you want, but remember what the point to this question is and please try and answer it too.

Answer 1

I would rather use the second example in preference for maximum portability. As Neil Butterworth pointed out, using bitfields is only for the native processor. Ok, think about this, what happens if Intel’s x86 went out of business tomorrow, the code will be stuck, which means having to re-implement the bitfields for another processor, say RISC chip.

You have to look at the bigger picture and ask how did OpenBSD manage to port their BSD systems to a lot of platforms using one codebase? Ok, I’ll admit that is a bit over the top, and debatable and subjective, but realistically speaking, if you want to port the code to another platform, its the way to do it by using the second example you used in your question.

Not alone that, compilers for different platforms would have their own way of padding, aligning bitfields for the processor where the compiler is on. And furthermore, what about the endianess of the processor?

Never rely on bitfields as a magic bullet. If you want speed for the processor and will be fixed on it, i.e. no intention of porting, then feel free to use bitfields. You cannot have both!