I have some problems for understand RSTP. I read IEEE Std 802.1D™- 2004 and have some questions.
In 17.3 I see this notes: Each Edge Port transitions directly to the Forwarding Port State, since there is no possibility of it participating in a loop.
But I don't see it in state machine schemes.
17.30 Port State Transition state machine doesn't have direct way for transit from Discarding state to Forwarding state with out Learning state.
and in 17.29.3 Designated Port states I don't see this behavior for Edge port.
port with true operEdge first go to DESIGNATED_LEARN and then go to DESIGNATED_FORWARD.
Can you help me understand this moments?
Please refer to Figure 17-22─Designated port role transtions in IEEE Std 802.1D™- 2004; 17.29.3 Designated Port states at page 176 as shown below.
In the case of an edge port (i.e,., operEdge == TRUE), when the link is up, the port state will transit from DESIGNATED_DISCARD to DESIGNATED_LEARN as the RED line and the condition 1. shown in below figure, and then it will move to DESIGNATED_FORWARD in the blink of an eye as the condition 2 and the BLUE line. It skips the fdWhile (forward delay) time.
Therefore, the note says that Each Edge Port transitions directly to the Forwarding Port State, since there is no possibility of it participating in a loop.
Related
The RFP for TFTP says that TID's in most circumstances:
should be randomly chosen, so that the probability that the same
number is chosen twice in immediate succession is very low.
The thing is, these "TID"s are also used as UDP port numbers. But a typical network interface cannot just be dedicated for TFTP use. Some ports are liable to be in use, and others should essentially be "reserved" for specific applications. I'm not even sure where a program could go to look up this information at runtime.
So how is a TFTP implementation supposed to deal with this?
Since the host selecting the TID/port is the one opening it and telling the other party which one it's opened, you can simply try to open the port; if it's already in use or otherwise unavailable, this will fail, and you can re-try with a different port. (Note that since UDP and TCP are difference protocols, a TCP application and a UDP application can both be using the "same" port, since they are not, in fact, the same at all!) Do this in a simple loop until you find a "good" one. (Probably best to define a maximum number of tries and simply fail the connection if that's met before a good port is found, as this could be a sign of other issues that prohibit this from working at all.)
Stick to the ephemeral port range to play nice with best practices, although note that different systems define different ranges for this purpose. You could pick the range suitable to your system, or simply try to use a port above the "well-known" port range (i.e. above 1024); this may not give you an "ephemeral port" per se for your system, but so long as you can open it it should work fine.
I newly wrote a simple chat application, but I didn't really understand the background of ICE Candidates.
When the peer create a connection they get ICE Candidates and they exchange them and set
them finally to the peerconnection.
So my question is, where do the ICE Candidates come from and how are they used and are they all really used ?
I have noticed that my colleague got less candidates when he executes the application on his machine, what could be the reason for different amount of Candidates ?
the answer from #Ichigo is correct, but it is a litte bit bigger. Every ICE contains 'a node' of your network, until it has reached the outside. By this you send these ICE's to the other peer, so they know through what connection points they can reach you.
See it as a large building: one is in the building, and needs to tell the other (who is not familiar) how to walk through it. Same here, if I have a lot of network devices, the incoming connection somehow needs to find the right way to my computer.
By providing all nodes, the RTC connection finds the shortest route itself. So when you would connect to the computer next to you, which is connected to the same router/switch/whatever, it uses all ICE's and determine the shortest, and that is directly through that point. That your collegue got less ICE candidates has to do with the ammount of devices it has to go through.
Please note that every network adapter inside your computer which has an IP adress (I have a vEthernet switch from hyper-v) it also creates an ICE for it.
ICE stands for Interactive Connectivity Establishment , its a techniques used in NAT( network address translator ) for establishing communication for VOIP, peer-peer, instant-messaging, and other kind of interactive media.
Typically ice candidate provides the information about the ipaddress and port from where the data is going to be exchanged.
It's format is something like follows
a=candidate:1 1 UDP 2130706431 192.168.1.102 1816 typ host
here UDP specifies the protocol to be used, the typ host specifies which type of ice candidates it is, host means the candidates is generated within the firewall.
If you use wireshark to monitor the traffic then you can see the ports that are used for data transfer are same as the one present in ice-candidates.
Another type is relay , which denotes this candidates can be used when communication is to be done outside the firewall.
It may contain more information depending on browser you are using.
Many time i have seen 8-12 ice-candidates are generated by browser.
Ichigo has a good answer, but doesn't emphasise how each candidate is used. I think MarijnS95's answer is plain wrong:
Every ICE contains 'a node' of your network, until it has reached the outside
By providing all nodes, the RTC connection finds the shortest route itself.
First, he means ICE candidate, but that part is fine. Maybe I'm misinterpreting him, but by saying 'until it has reached the outside', he makes it seem like a client (the initiating peer) is the inner most layer of an onion, and suggests the ICE candidate helps you peel the layers until you get to the 'internet', where can get to the responding peer, perhaps peeling another onion to get to it. This is just not true. If an initiating peer fails to reach a responding peer through the transport address, it discards this candidate and will try a different candidate. It does not store any nodes anywhere in the candidate. The ICE candidates are generated before any communication with the responding peer. An ice candidate does not help you peel the proverbial NAT onion. Also regarding the second quote I made from his answer, he makes it seem like ICE is used in a shortest path algorithm, where 'shortest' does not show up in the ICE RFC at all.
From RFC8445 terminology list:
ICE allows the agents to discover enough information
about their topologies to potentially find one or more paths by which
they can establish a data session.
The purpose of ICE is to discover which pairs of addresses will work. The way that ICE does this is to systematically try all possible pairs (in a carefully sorted order) until it finds one or more that work.
Candidate, Candidate Information: A transport address that is a
potential point of contact for receipt of data. Candidates also
have properties -- their type (server reflexive, relayed, or
host), priority, foundation, and base.
Transport Address: The combination of an IP address and the
transport protocol (such as UDP or TCP) port.
So there you have it, (ICE) Candidate was defined (an IP address and port that could potentially be an address that receives data, which might not work), and the selection process was explained (the first transport address pair that works). Note, it is not a list of nodes or onion peels.
Different users may have different ice candidates because of the process of "gathering candidates". There are different types of candidates, and some are obtained from the local interface. If you have an extra virtual interface on your device, then an extra ICE will be generated (I did not test this!). If you want to know how ICE candidates are 'gathered', read the 2.1. Gathering Candidates
I have a weird problem. I have a successfully working C++ (boost asio) P2P application which works on most of the NAT. The problem is when I give the initial start port number as 1000 it checks if 1000 is free else increment by one and chooses a port and starts handshaking. But when I have 10000, 20000, or any other huge port number the hole punching doesn't work on port restricted cone NAT.
How is that possible? I am pretty sure it nothing to do with the code. and recently it doesn't work on one of my friends' full cone NAT as well, but it has worked in many other full cone NATs. What could be the reason? Is there something I am missing about how a NAT behaves?
In many NAT implementations, there are protection rules in place which prevent one host from tying up a large percentage of ports on the WAN interface, e.g. like described here.
Depending on the router, the NAT table entries have different lifetimes, and there are always limits on how many ports can be allocated to a single client (I've seen numbers from 128 to 4096).
So I think when you get to the point where you need to use high ports, the NAT table for your source IP address is already full (or almost full) with entries from old connections, or connections from other apps, so the router either decides to decline or can't fit the new NAT entry for your port.
However, to be sure, I would try to repeat that on a controlled environment collecting Wireshark dumps on both sides of the NAT and analyze the packets. If possible, it would also be helpful to enable router logs and peek into them.
I understand this is not a "magic bullet", but hope it somehow helps you.
Don't try to choose the port number yourself. The operating system can do this faster and better than your code can.
Bind your socket to port 0 and let the OS choose an available port number for you. You didn't specify what programming language, but it usually involves a call to getsockname() after the bind() call is made to discover what local port is going to be used. Java and .NET have equivalent APIs for doing the same thing.
Then follow all the other steps here:
https://stackoverflow.com/a/8524609/104458
Not sure if this'll help but have you tried having one instance of the client application starting at 1001 and the other starting at 1000, then both increment by 1.
While the 1000 will fail on client B, client A has already tried 1001 and so punched that hole, so hopefully it'll work, right? In theory, it sounds OK in my head.
I am trying to get a list of all IP's on a LAN network. The reason for this is I am writing an application that uses the STAR TSP100LAN receipt printer.
The process for obtaining the IP address of the printer is quite cumbersome for the end user. It involves turning the printer off, holding the paper feed button, turning the printer back on, waiting 15 seconds for the printer to obtain an IP address through DHCP and then finally spitting out a receipt with this information on it.
Seeing as the printer is not Bonjour enabled is it possible to get the IP address through other methods?
Any help is much appreciated! I hope this isn't a repeat question, but through my searching I can't seem to find a solution!
UPDATE: Ok after a bit of thinking I have come up with a pseudo-solution:
Determine the iPad's current IPAddress through NSHost.
Strip the last quadrant from the IPAddress
Using the stripped string as a prefix, iterate 1-255 for the last quadrant
Each iteration, attempt to open a port to the given address using the printer's sdk
If I get a valid response, I know that the IP is a printer
If not I exclude the IP from the available printers list.
So far this has been working, I set a timeout of 5 milleseconds for each port open attempt. But have found that this can return some null results despite there actually being a printer on the network with an IP Address assigned.
Perhaps if I get a null result the first time I should increase the timeout to 15 milleseconds for a second attempt at searching.
Your approach of polling the local /8 subnet is probably the best you can do. I can't find any API to get more detailed information about the network interface (i.e. subnet mask) in iOS. (Although using the subnet mask would be a more correct way to determine the range to iterate, if you could get it.)
As you've seen, 5ms is a pretty tight interval; In my experience, 15ms can STILL be pretty tight for a TCP connection over WiFi. As a next step, I would suggest parallelizing the polling of the range, and thereby enabling you to extend the interval you're willing to wait. The simplest way would probably be to use GCD. You could also start this polling process in the background before the user explicitly needs the printer, which might improve the user-perceived responsiveness of your app.
Alternately, you could use the CFSocket API to open all these connections (CFSocketCreate, CFSocketConnectToAddress, and friends) and get parallelism by servicing them all on the main thread with callbacks/the runloop. Then, as those callbacks come in, make a note of which addresses answer on the given port. Unless the printer isn't using TCP for some reason, this should be workable. Once you know which addresses answer on that port at all, you can iterate over that (hopefully much smaller) list connecting with the printer SDK itself. This approach will give you even more (and way more elegant) parallelism than spawning a huge number of GCD threads, but can be difficult to get your head around if you've not worked with runloops before.
Hope that helps.
You can quickly winnow the list down from 255 to a smaller number by pinging the broadcast address then looking into your arp cache.
Only works for hosts that respond to broadcast pings.
If I understand the parallel port right, sending data from (D0 to D7) simultaneous, but that it can control the sticks individually?
example:
D0 = Input
D1 = Input
D2 = Output
...
...
...
D7 = Input
would it work?
what I want to do is to both send and receive data simultaneously.
Data wires (D0-D7) are being read or set simultaneously. For various tecniques for bidirectional I/O read the attached articles:
Standard parallel port: http://www.beyondlogic.org/spp/parallel.htm
EPP: http://www.beyondlogic.org/epp/epp.htm
ECP: http://www.beyondlogic.org/ecp/ecp.htm
This site is a good source for programming the parallel port.
The basic idea is that you need a DLL, add-on or library that allows you to access the I/O Ports of the PC. For Windows XP on up you need a specific driver that will allow you do this as the OS doesn't offer access out of the box.
The Parallel port will generally reside at one of three address 278,378, 3BC. This port. have the bytes you are reading or writing.
The (base)+1 port allows access to the status bytes. IE. 279,379, 3BD
The (base)+2 port allows access to the control bytes. IE. 27A,37A,3BE
The parallel port documentation will tell not only how to implement the common modes (like bi-directional) but how to control the port at the byte level to implement your own custom.
Back in the day there was only the standard mode available. You pump out your bytes at the (base) port. Some application, like mine, manipulated individual bits of that ports as form of cheap Digital I/O Controller. We did use the status and control bytes as additional inputs and outputs. There were commands you can send to the Parallel Port chip configure the modes precisely.
Today there are are hundreds of sites with example of using the Parallel Port to solve all kinds of problems. I would be surprised that one of doesn't have something you can use for you specific application.
Again the book I recommend starting with is Parallel Port complete. It tells just about everything you need to start with. If your application is too esoteric for that book it will give a springboard from which you can find the exact setup you need.
Of course by sending a number that has just the required bit set (2n) you'd get the expected result.
I'm not sure about bidirectional access though. I guess it's achieved by using control pins along with the data pins but that's just a guess.
Parallel ports doing EPP or ECP only allow D0-D7 to be all input or all output. Trying to do otherwise may fry your hardware.
See http://www.nor-tech.com/solutions/dox/ieee1284_parallel_ports.pdf, page 6.
However, parallel ports have several control lines that may be useful if you only need a small amount of input/output in the "other" direction.
I believe its bit 5 in the port's control register (base address + 2) that switches direction. (no hardware line attached)