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;         /* destination port */
    tcp_seq th_seq;             /* sequence number */
    tcp_seq th_ack;             /* acknowledgement number */
#  if __BYTE_ORDER == __LITTLE_ENDIAN
    u_int8_t th_x2:4;           /* (unused) */
    u_int8_t th_off:4;          /* data offset */
#  endif
#  if __BYTE_ORDER == __BIG_ENDIAN
    u_int8_t th_off:4;          /* data offset */
    u_int8_t th_x2:4;           /* (unused) */
#  endif
    u_int8_t th_flags;
#  define TH_FIN        0x01
#  define TH_SYN        0x02
#  define TH_RST        0x04
#  define TH_PUSH       0x08
#  define TH_ACK        0x10
#  define TH_URG        0x20
    u_int16_t th_win;           /* window */
    u_int16_t th_sum;           /* checksum */
    u_int16_t th_urp;           /* urgent pointer */
};

Why does the 8-bit field have a different order in endianness? I thought only 16-bit and 32-bit fields mattered with byte order, and you could convert between endians with ntohs and ntohl, respectively. What would the function be for handling 8-bit things? If there is none, it seems that a TCP using this header on a little endian machine would not work with a TCP on a big endian machine.

Answer 1

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.