Rapid Spanning-tree TC

[kevin.r719]

I'm having trouble understanding why RSTP behaves the way it does. I've been reading through http://www.cisco.com/c/en/us/support/docs/lan-switching/spanning-tree-protocol/24062-146.html

and it says when a TC occurs it "clears mac addresses on all ports EXCEPT the one that received the TCN"

Isn't this backwards? if we have switches hooked up like this where C is the root, A -> B -> C 

1) B generates TCN and forwards it to C

2) what does B do with the stuff it's learned from A? clear the MAC table? or wait for timeout?

3) since C learned A's macs from B, doesn't that mean it will still retain the MACs learned from A? 

if C lead to another branch, say D, D would receive the TCN from C and still keep A's macs in it's table correct? because that's the port it learned it from

[esa_fresa]

The Cisco doc has it correct.

1. In your example of A->B->C, I'm assuming that A also has a physical link to C (the classic triangle looking configuration) and that this A-C link went down, causing the A-B link to go from blocking to forwarding on B's side and so causing B to mark the topology change bit on hellos sent to C.

2. B will start using the information from A immediately. The port was previously blocking and doesn't have any MAC addresses learned.

3. Prior to the A-C link going down, C was learning A's macs from A, not B. Remember that the A-B port was blocking on B's end, so none of these macs would have been propagated. What C will do is exactly what you stated, "clears mac addresses on all ports EXCEPT the one that received the TCN". C will clear the macs that it learned previously on its A-C link just like we want it to (well, now it doesn't matter because that link is down, but the theory is still good)

4. If D learned the A macs via the C-D link, then that's correct, it will keep them in the table, and that's what we want. Whether the macs go A>C>D or A>B>C>D, the D switch is still receiving the A macs from switch C.

I hope this clears things up!