Isn’t on big endian machines UTF-8's byte order different than on little endian machines? So why then doesn’t UTF-8 require a BOM?

Question

UTF-8 can contain a BOM. However, it makes no difference as to the endianness of the byte stream. UTF-8 always has the same byte order.

Answer 1

To answer c): UTF-16 and UTF-32 represent characters as 16-bit or 32-bit words, so they are not byte-oriented.

For UTF-8, the smallest unit is a byte, thus it is byte-oriented. The alogrithm reads or writes one byte at a time. A byte is represented the same way on all machines.

For UTF-16, the smallest unit is a 16-bit word, and for UTF-32, the smallest unit is a 32-bit word. The algorithm reads or writes one word at a time (2 bytes, or 4 bytes). The order of the bytes in each word is different on big-endian and little-endian machines.

Answer 2

The byte order is different on big endian vs little endian machines for words/integers larger than a byte.

e.g. on a big-endian machine a short integer of 2 bytes stores the 8 most significant bits in the first byte, the 8 least significant bits in the second byte. On a little-endian machine the 8 most significant bits will the second byte, the 8 least significant bits in the first byte.

So, if you write the memory content of such a short int directly to a file/network, the byte ordering within the short int will be different depending on the endianness.

UTF-8 is byte oriented, so there's not an issue regarding endianness. the first byte is always the first byte, the second byte is always the second byte etc. regardless of endianness.