"Priscilla Oppenheimer" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> At 12:41 PM 11/1/00, Jim Erickson wrote:
> >"Aaron" <[EMAIL PROTECTED]> wrote in message
> >[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> > > Hi, everyone!
> > >
> > > I have a question about the MAC layer address, and I use the Ethernet
for
> > > making an example.
> > >
> > > I have a question about the following whether it is right:
> > > when we get a MAC address, such as 0030.b6f7.3000 (Cisco),
> > > 1. Whether the I/G and U/L bit are already set to zero?
> > > 2. When a multicast packet shoud be sent to this address, the
destination
> > > address in the MAC packet header should be set to 0130.b6f7.3000?
> >
> >Wouldn't a multicast packet be sent to the MAC address of the multicast
> >group, as translated from the multicast IP address, not to the individual
> >host's MAC address (or an alteration thereof)? If so, the first three
bytes
> >are 0100.5e, with the remainder determined by the translation of the
> >multicast IP address. Then, if it was, say, an all routers multicast, the
> >destination MAC would be 0100.5e00.0002
>
> What if it's not IP? &;-) Seriously, you're up a layer from the original
> question. He was just trying to understand the MAC-layer multicast bit.
> Notice that the multicast bit is set in your example. The first byte is 01
> in HEX, or 0000 0001 in binary. The first bit transmitted will be a one.
> It's on the right because we're using canonical Ethernet in the example.
You're right, that's exactly what I was missing. I just got through with
BCMSN, so I've got _IP_ multicast on the brain, and that's what I was
assuming. Thanks for the clarification.
---JRE---
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