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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. If Utf-8 stored all code-points in a single byte, then it would make sense why endianness doesn’t play any role and thus why BOM isn’t required. But since code points 128 and above are stored using 2, 3 and up to 6 bytes, which means their byte order on big endian machines is different than on little endian machines, so how can we claim Utf-8 always has the same byte order? Thank you EDIT: UTF-8 is byte oriented I understand that if two byte UTF-8 character C consists of bytes B1 and B2 ( where B1 is first byte and B2 is last byte ), then with UTF-8 those two bytes are always written in the same order ( thus if this character is written to a file on little endian machine LEM , B1 will be first and B2 last. Similarly, if C is written to a file on big endian machine BEM , B1 will still be first and B2 still last). But what happens when C is written to file F on LEM , but we copy F to BEM and try to read it there? Since BEM automatically swaps bytes ( B1 is now last and B2 first byte ), how will app ( running on BEM ) reading F know whether F was created on BEM and thus order of two bytes wasn’t swapped or whether F was transferred from LEM , in which case BEM automatically swapped the bytes? I hope question made some sense EDIT 2: In response to your edit: big-endian machines do not swap bytes if you

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