Question

Sizeof() doesn’t work when applied to bitfields:

# cat p.c
  #include<stdio.h>
  int main( int argc, char **argv )
  {
    struct { unsigned int bitfield : 3; } s;
    fprintf( stdout, "size=%d\n", sizeof(s.bitfield) );
  }
# gcc p.c -o p
  p.c: In function ‘main’:
  p.c:5: error: ‘sizeof’ applied to a bit-field

…obviously, since it can’t return a floating point partial size or something. However, it brought up an interesting question. Is there an equivalent, in C, that will tell you the number of bits in a variable/type? Ideally, it would also work for regular types as well, like char and int, in addition to bitfields.

Update:

If there’s no language equivalent of sizeof() for bitfields, what is the most efficient way of calculating it – at runtime! Imagine you have loops that depend on this, and you don’t want them to break if you change the size of the bitfield – and no fair cheating and making the bitfield size and the loop length a macro. 😉

Answer 1

You cannot determine the size of bit-fields in C. You can, however, find out the size in bits of other types by using the value of CHAR_BIT, found in <limits.h>. The size in bits is simply CHAR_BIT * sizeof(type).

Do not assume that a C byte is an octet, it is at least 8 bit. There are actual machines with 16 or even 32 bit bytes.

Concerning your edit:
I would say a bit-field int a: n; has a size of n bits by definition. The extra padding bits when put in a struct belong to the struct and not to the bit-field.

My advice: Don’t use bit-fields but use (arrays of) unsigned char and work with bitmasks. That way a lot of behaviour (overflow, no padding) is well defined.