Why does Guid.ToByteArray() order the bytes the way it does?

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If you read the Examples section from the GUID constructor, you’ll find your answer:

Guid(1,2,3,new byte[]{0,1,2,3,4,5,6,7}) creates a Guid that corresponds to "00000001-0002-0003-0001-020304050607".

a is a 32-bit integer, b is a 16-bit integer, c is a 16-bit integer, and d is simply 8 bytes.

Because a, b, and c are integer types rather than raw bytes, they are subject to endian ordering when choosing how to display them. The RFC for GUID’s (RFC4122) states that they should be presented in big endian format.

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