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So, I'm still confused, why am I not getting it? There is not any kind of guarantee that the port is "well known" at all, that would only work if only 1 thing on the other side of the NAT would forward that port
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OK. Your client connects to the server to report itself, right? The port number that the client uses will NOT be the port number that it uses to listen for other clients connecting to it. The port number that the server sees the client using is only temporary and only good for that one connection. It the client contacts the server tomorrow to report itself, it'll probably use a different port number. You cannot use this port number for anything useful. The only important part is the server can learn the Internet visible IP address of the client. That's it.
Now, in order for the client to listen for connections from other clients, it has to open a specific "well-known" port to listen for connections on. What makes this "well-known"?? YOU! You pick the port number that your app is going to use to listen for connections. This is going to be a standardized port number that every client is going to listen on.
Great! Now you've got a port number. Well, your client software has a port number. The user's NAT doesn't know that. So, the NAT has to be configured to forward incomming connections from the public internet to a specific IP and port number on the inside of the network, preferrably the client machine(!), and the port that the client is listening on.
Now, back up a second. The NAT can forward any port number exposed to the public internet to any port number on the inside network. So, you can easily have the NAT listening for a different port number on the public internet, but usually you don't want to do this. Normally, you'd just use the same port number that the client is listening for. Just keep it in mind for troubleshooting purposes.
So, now you've got the NAT listening on your well-known port and forward traffic on that port to the client inside the network. Now, another client starts up, contacts the server to report in and get the list of currently connected clients. Your new client attempts to connect to another client, using the IP address it got from the server and the well-known port number. That request is going to get forwarded to the first client by the NAT, and any response sent back to the new client.
Now, if your clients are only supporting a single connection, you can leave it at this. If your clients are going to be able to talk to multiple clients simultaneously, you'd use this initial connection to negotiate with the calling client a second connection on a different port. The new clinet would have to start a listener on a random port, telling the first client which port that is on the initial connection port. The first client would then open a connection back to the new client to start the communication channel the two clients are going to use to chat with. The initial connection can then be closed down and the first client can then reuse the same port to listen for another connection request...
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Ok, but what will happen if more than 1 client is on the same NAT?
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With the communication model you've chosen, you can only have 1 client behind the NAT, unless the NAT sits behind more than 1 IP address.
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Ok, but that is bad.. and bittorrent definitely allows this, and they'd basically have the same problem. Now, I know they send the port number to the server as part of the protocol. Is that why it works? Allowing the port to be manually configured and then send it to the server? Because that's definitely something I could do
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Guy, read up on how servers work first. Just keep 2 things in mind. Your "client" software is really two pieces - a client AND a server listening for connections. The server side is ALWAYS listening on a well-known port and this channel is used to setup negotiated ports for further communication. All you have to do to find this out is run other client/server software under a network sniffer to find this out. But, that itself requires an in-depth knowledge of how TCP/IP and client/server communication works.
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Am I really that stupid? I know how it works normally, I'm just getting thrown off by NAT..
If one could forward the same port twice then I wouldn't have any problem but you can't - making the port unknown.
So in order to hack around that, I thought I could bind to the port (the one that will be listening later) for the outgoing connection so that the server could read the correct port. The documentation of bind suggests that that is possible, but it's not (but from what I know from TCP/IP, it should be). So the only way that I see to get around it is to manually configure the ports for all clients behind the same NAT and send the port number to the server.
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yea well ok I guess I deserved that, thanks for your help, I'll go re-read my books on networking
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Dave Kreskowiak wrote: The new clinet would have to start a listener on a random port, telling the first client which port that is on the initial connection port. The first client would then open a connection back to the new client to start the communication channel
Won't NAT block that, if you have only setup port forwarding for the well known client port?
Great response, BTW. +5.
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Well, you'd have to tell the NAT to allow that port also. I think they use UPnP for that if I'm not mistaken.
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Ok, well, I think I've heard enough - I will just accept that it sometimes fails. Not ideal, but, well, NAT is not ideal itself (crazy hack..)
Thanks everyone 
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Hi All, I am kind of new in programming in c#. Recently I am working on a window base application. The Application has a dataGridView and it fills with data which comes from many diffrent source. The dataGridView display all the data in which order they come.
My Question is how do I print only the data i wanto print. for examlple, the dataGridView has 10 rows. within 10 rows there are three kind of data. I only want to print one kind of three.
Printing fucntion look for only those row that has same data value.
any rows that matches with my selected row should go to print rest should be ignored.
any help.
Thanks much in advance.
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You can find an enlightening MSDN forum post here. That's got many different code snippets on, and a link to an interesting article as well. I would assume that if you only want to print certain rows, then you can go to the piece of code which iterates through each row and check if you want that row. If you do, then draw it; if not, continue to the next iteration
Between the idea
And the reality
Between the motion
And the act
Falls the Shadow
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Take a look at DataGridView Printing by Selecting Columns and Rows[^].
It doesn't do precisely what you want, but it is near enough that you ought to be able to adapt it.
If you try to do that and get stuck, please make a new thread with the code you are stuck on, and exactly what the problem is.
As an alternative, you could give the user somewhere to enter the selection criteria, TextBoxes or CheckBoxes for example, and then only load that data into your grid. There are lots of articles on printing datagridviews.
Henry Minute
Do not read medical books! You could die of a misprint. - Mark Twain
Girl: (staring) "Why do you need an icy cucumber?"
“I want to report a fraud. The government is lying to us all.”
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Is there an API available that can control the sound card bass and treble?
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IT probably depends on your sound card but not sure
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Their is no direct API to provide equalization. What you have to do is write a DirectX filter and use it control treble/bass.
Check the DirectX API for more information.
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Your sound card manufacturer may provide one, but there is no general standard for this as far as I know.
You can create your own audio player and provide your own DSP in that. Search for Audio DSP C# and you should find something to help you get started.
DaveBTW, in software, hope and pray is not a viable strategy. (Luc Pattyn)
Visual Basic is not used by normal people so we're not covering it here. (Uncyclopedia)
Why are you using VB6? Do you hate yourself? (Christian Graus)
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Take a look at Bass.Net[^]. I haven't tried it myself, but I've run across a few things that use it. It seems to have a pretty good reputation from what I can tell.
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hi,
how can i make a differnt view in a form?
i tried to use MDI form to open each time a different form(new controls and view), but i couln'd open it in my designated area (a panel in the right side of the screen), it opens anywhere on the form.
is it the right way to do what i want?
many thanks,
Samy
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The answer to this question depends on what you want to happen.
Do you want each view that you open, to remain open when you make a new selection in the TreeView? So that there could be many views open.
Do you want each view that you open, to be the only view open?
Henry Minute
Do not read medical books! You could die of a misprint. - Mark Twain
Girl: (staring) "Why do you need an icy cucumber?"
“I want to report a fraud. The government is lying to us all.”
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yes - i need that each node will show one view.
each time there will be only one view available.
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In that case you do not need MDI (Multiple Document Interface).
Either design your different views as UserControls or as a Form. The process is pretty much the same.
Then in the code that you use when a node is selected (I don't know if you are using BeforeSelect/AfterSelect or something else), your code should look something like this.
I am assuming the Panel on the right is called displayPanel, and that you decide to use names for your views like WhateverView
this.displayPanel.Controls.Clear();
WhateverView newView = new WhateverView();
newView.Dock = DockStyles.Fill;
this.displayPanel.Controls.Add(newView);
This code should work for a single type of 'view'
If you are using Form descendants you might also like to set their FormBorderStyle to FormBorderStyle.None
Just to see if it works as you would like, decide on one node, for now, and therefore one 'view' and use the code above to test it.
Of course you will need a way to decide which type of view to use depending on the node selected, assuming that you have more than one. For that you might want to investigate the Factory Pattern
Put (very) simply, this is a class that will return the right sort of 'view' when asked.
e.g the line WhateverView newView = ViewFactory.GetView("mp3"); will give you a view for looking at MP3 stuff.
the line WhateverView newView = ViewFactory.GetView("folder"); will give you a view for looking at Folder stuff.
There are many ways to select which type of 'view', this is just one of them.
Henry Minute
Do not read medical books! You could die of a misprint. - Mark Twain
Girl: (staring) "Why do you need an icy cucumber?"
“I want to report a fraud. The government is lying to us all.”
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great it work.
but i could only do the WhateverView as usercontrol. if i try to do it with a form - the "this.displayPanel.Controls.Add(newView) " cant work.
many thanks!
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