Bridging Ethernet to Token Ring and endianness

[dbrb2]

Hi,

I've been thinking about bridging ethernet and token ring. The big problem everyone mentions is that, big vs little endian. For isnatance, if an ethernet ARP request is issued and bridged to the token ring network, then the reply MAC address is not reordered as it is not in layer 2, and the requestor has an incorect ARP cache entry.

As far as I can see, this should not be a problem if all the data was either big endian or little endian - you could just blindly reorder every byte passing thrugh the bridge without caring what layer it was in.

The fact that it is a problem presumably means that there is a mix of big and little endian in any packet, and so just blindly reordering eveything would create as many problems as it solved?

Is only part of a token ring packet big-endian?

Cheers,


Ben

 

[bcastner]

Use XP and its native bridging, or purchase a hardware ethernet to Token Ring bridge.

There is no issue with a mixed-media router in either case, or little-endian or big-endian should not be your concern.

Let the mixed-media routers handle the chore.

 

[dbrb2]

Hi,

I'm not rally worried about it - just interesed. If I were to use a mixed media router, why can't it just blindly reorder every byte that passes through? Why des it need to selectively
reorder certain fields only? Are there some fields that are big endian and some that are little (perhaps in the data fields...?)

Cheers,

Ben

 

[TimRegester]

Try and get hold of a switch that can do both Token Ring and Ethernet. This will do the mac translation for you. If it has layer 3 capability it will also do the ARP resolution.

Suggestions are

3Com Corebuilder 2500, 6000 and possibly 5000. All should be available cheaply.

Nortel also made switches that had TR and Ethernet blades.

You could also use a router.

 

[dbrb2]

Thanks - it was really only a hypothetical question.

I think the answer must be that the bridge/router can only reorder headrers it knows about. Otherwise it would not know whether a given 8 bits was a byte that needed reordering, or half a word which needed reordering, but as a block of 16 rather than 8.

This would suggest there will always be some un-reordered data - is this dealt with by software...?