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?

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

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 endianess. UTF-8 is byte oriented, so there's not an issue regarding endianess.

The first byte is always the first byte, the second byte is always the second byte etc. Regardless of endianess.

Could you see my edit, since there's still something I don't quite understand – user437291 Sep 30 '10 at 19:53 2 Neither BEM nor LEM swap any bytes when you deal with bytes. They'll be swapped if you read more than 1 byte as a larger type, e.g. 2 bytes as a short or 4 bytes as an int then you have to care about which byte goes where within the integer – nos Sep 30 '10 at 20:20 Bytes are not "automatically" swapped at all. Depending on endianness they have different meaning (if part of a larger integer), but there is no swapping.

– Ben Voigt Oct 1 '10 at 18:59.

Check out the article on wikipedia: en.wikipedia.org/wiki/Big_endian.

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