very basic question about MAC addresses - globally unique identifying bit

Sep 01, 2009 3 Replies

I'm confused, and would appreciate some guidance. I'm reading "The All New Switch Book" (Seifert & Edwards), and I'm on page 25, where they say that, before the IEEE took over, Xerox issued some OUIs that had their second bit equal to 1. But the examples given is that of



3COM, with 02-60-8C-xx-yy-zz, and DEC, with AA-AA-03-xx-yy-zz. But when I translate those to binary, I get this -


02-60-8C = 0000 0010 0110 0000 1000 1110 AA-AA-03 = 1010 1010 1010 1010 0000 0011



Neither of these have "1" as the 2nd bit. So I'm misunderstanding either the text or the proper way to convert Hex addresses to binary. What am I not getting?



thanks



Mark


< I'm confused, and would appreciate some guidance. I'm reading "The < All New Switch Book" (Seifert & Edwards), and I'm on page 25, where < they say that, before the IEEE took over, Xerox issued some OUIs that < had their second bit equal to 1. But the examples given is that of < 3COM, with 02-60-8C-xx-yy-zz, and DEC, with AA-AA-03-xx-yy-zz. But < when I translate those to binary, I get this - < 02-60-8C = 0000 0010 0110 0000 1000 1110 < AA-AA-03 = 1010 1010 1010 1010 0000 0011 < Neither of these have "1" as the 2nd bit. So I'm misunderstanding < either the text or the proper way to convert Hex addresses to < binary. What am I not getting?

Second from the right. Though most CS people would number starting from 0 at the right, or 0 at the left...

-- glen

In case Glen's answer was too cryptic, they are referring to the bit order as it is transferred across the wire over an Ethernet interface.

Over Ethernet, each byte (octet) is transmitted least significant bit first. So in both of those examples, the second bit over the wire is in fact set to 1.

Bert

That makes it crystal clear. Thank you both for the info.

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