why bit-fields for same data types have less in size compared to bit-fields for mix-data types

Question

I am curious to know why bit fields with same data type takes less size than with mixed data types.

struct xyz 
{ 
  int x : 1; 
  int y : 1; 
  int z : 1; 
}; 


struct abc 
{ 
  char x : 1; 
  int y : 1; 
  bool z : 1; 
};

sizeof(xyz) = 4 sizeof(abc) = 12.

I am using VS 2005, 64bit x86 machine.

A bit machine/compiler level answer would be great.

Answer 1

Alignment.

Your compiler is going to align variables in a way that makes sense for your architecture. In your case, char, int, and bool are different sizes, so it will go by that information rather than your bit field hints.

There was some discussion in this question on the matter.

The solution is to give #pragma directives or __attributes__ to your compiler to instruct it to ignore alignment optimizations.

Answer 2

The C standard (1999 version, §6.7.2.1, page 102, point 10) says this:

An implementation may allocate any addressable storage unit large enough to hold a bit-field. 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.

There does not seem to be any wording to allow the packing to be affected by the types of the fields. Thus I would conclude that this is a compiler bug.

gcc makes a 4 byte struct in either case, on both a 32-bit and a 64-bit machine, under Linux. I don't have VS and can't test that.