Why does an 8-bit field have endianness?

Question

See the definition of TCP header in /netinet/tcp.h:

struct tcphdr
  {
    u_int16_t th_sport;         /* source port */
    u_int16_t th_dport;         /* destin         


        

Answer 1

My reading of the comment is that the two one-byte fields together are construed as a two-byte value (or were -- seems like one byte is unused anyway). Rather than declare one two-byte value, they declare two one-byte values but reverse the order of declaration depending on endian-ness.

Answer 2

It may help to know that this is code is only run if "# ifdef __FAVOR_BSD". It is from /usr/include/netinet/tcp.h

# ifdef __FAVOR_BSD
typedef u_int32_t tcp_seq;
/*
 * TCP header.
 * Per RFC 793, September, 1981.
 */
struct tcphdr

Answer 3

Its possible that in this machine the endianess also refers to the bit order as well as the byte order. This wikipedia article mentions that this is sometimes the case.

Answer 4

This is compiler-dependent and non-portable. How bit fields are ordered is implementation dependent, it would be far better here to use an 8-bit field and shift/mask to obtain the subfields.

Answer 5

My understanding is that bit ordering and endianness are generally two different things. Structures with bit-fields are generally not portable across compilers/architectures. Sometimes ifdefs can be used to support different bit orderings. In this case the endianness is really irrelevant and it should be an ifdef about the bit ordering. The assumption that some endianesses have a certain bit-order may be true in some cases.

Answer 6

There are two kind of order. One is byte order, one is bitfield order. There is no standard order about the bitfield order in C language. It depends on the compiler. Typically, the order of bitfields are reversed between big and little endian.