What first confused me about endianness is that it is about byte order, not bit order. The latter would have seemed more logical, or is this just me?
replies(5):
Bits are typically not addressable, therefore do not have endiannness.
Bits are manipulated by special instructions, and those instructions are tied to arithmetic identities, due to the bits being interpreted as a binary number: like that a shift left is multiplication by 2.
In many instruction sets, the shift is a positive amount, and whether it is left or right is a different instruction. If it were the case that shifting one way is positive and the other way negative, then you have a kind of endiannness in that one machine uses positive for multiplication by powers of two, whereas another one for division. That would not result in an incompatible storage format though.
When data is transferred between machines as a sequence of bytes, there is a bit order in question, but it is taken care of by the compatibility of the data links.
Classic Ethernet is little endian at the bit level: the baseband pulses that represent the bit of a byte are sent into coax cable least-significant-bit first. RS-232 serial communication is the same: least significant bit first.
I think I²C is an example of a data link / physical protocol that is most-significant-bit first. So if you somehow hooked up an RS-232 end to I²C and got the communication to work, the bytes would be reversed.
We rarely, if ever, see bit endian effects because nobody does that --- transmit bytes between incompatible data links. If won't work for other reasons, like different framing bits, signaling conventions, voltages, speeds, synchronization methods, checksums, ...
Endianness of bits shows up in some data formats which pack individual bitfields of variable length.
Bitfields in C structures reveal bit endianness to some extent. What typically happens is that on a big endian target, bit fields are packed into the most significant bit of the underlying "cell" first. E.g.
struct { unsigned a : 1, b : 1 };
So in a certain sense, the allocation of members in C, as such, is little endian: the earlier struct members go to the lowest address, regardless of machine endian. It is probably because of that the bit order follows. Since putting bitfield a at the lowest address, as mandated by C field layout order, means that it has to go into the first byte, and that first byte is the most significant byte under big endian, it makes sense that the bit goes into the most significant bit position, for consistency.
That way we only have two possibilities to deal with for, say, a memory mapped status register:
struct port_status_word {
#if HAVE_BIG_ENDIAN
unsigned transmit_ready : 1;
unsigned data_received : 1;
unsigned carrier_present : 1;
// [ ... 29 more]
#else
// [ ... 29 more]
unsigned carrier_present : 1;
unsigned data_received : 1;
unsigned transmit_ready : 1;
#endif
};