Socket Destination Port Problem

This is a discussion on Socket Destination Port Problem within the Networking/Device Communication forums, part of the General Programming Boards category; Hi everybody, I have an application where two different socket connections run in order to communicate with a machine. Over ...

  1. #1
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    Socket Destination Port Problem

    Hi everybody,

    I have an application where two different socket connections run in order to communicate with a machine. Over the first one (UDP) I send a request packet to ask for data and over the other one connection (TCP) I receive the data.
    I repeat this procedure two times that is, I request a block of data and when it's finished I request a second one.By the first time it works without problems but by the second time I discovered with wireshark that the request packet is sent to another port (not over the original port). I know I haven't made changes between the first request call and the second one because I have a function where I initialize all connections at the beginning and it works for the first transmission.

    Could the operative system change the destination port?

    Can anyone help me?

    Thank you very much

  2. #2
    CSharpener vart's Avatar
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    no, OS will not change port for you

    why not to use same connection?
    The first 90% of a project takes 90% of the time,
    the last 10% takes the other 90% of the time.

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    Hi,

    I must use different connections (design requirements...)

  4. #4
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    try to post some code (a relevant minimal example), there is surely something that changes between the two requests.

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    Hi again,

    Now I can describe the problem better

    I have a tcp server that wait for data in the recv function after accept. I receive data buffer of 1500 bytes but I don't receive only one buffer. I receive more. The first time the socket receives something port=2000 (so it's right) but the second time and successively port=49154. Something is wrong with the receive function...

    Code:
    AcceptSocket = accept(socketInTCP, NULL, NULL);		 
    		 if (AcceptSocket == INVALID_SOCKET)
    		 {	 
    			
    			 printf("Server: WSA:\n");
    			
    		 }
    		 else
    		 {   
    			 printf("Server: Client Connected!\n");
    			 //Receive Packages from Client
    			 while(run_tcpserver)
    			 {  	bytesRecv = recv(AcceptSocket, recvbuf, sizeof(recvbuf), 0);
    				if (bytesRecv > 0)
    				{if(ctr1==1)
    					{	
    						WaitForSingleObject(mutex_req,INFINITE);
    						//The first time it receives port = 2000
    					}
    				 
                                     //the second time and successively port =49154
                                     ctr1=0;
    				 ofstream outfile(outputfile, ios::out |ios::ate | ios::binary);//open
    				 rcv_data=rcv_data+bytesRecv;
    				 cout << "\xd" << rcv_data;
    				 
    				 outfile.write (var, sizeof(var));
    				 outfile.close();
    				 memset(recvbuf,0,sizeof(recvbuf));
    				 
                                     if(rcv_data==(length_dummy*4))
    					{
    						ReleaseMutex(mutex_req);
    						
    					}
    have anyone any ideas??

  6. #6
    CSharpener vart's Avatar
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    I do not see where you do anything with the recv_buf
    The first 90% of a project takes 90% of the time,
    the last 10% takes the other 90% of the time.

  7. #7
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    I don't know. but the destination port is wrong after successively recv...

    that is my question....

  8. #8
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    now I have this code

    Code:
    while(AcceptSocket==SOCKET_ERROR)
    {
       Acceptsocket=accept(socketInTCP,NULL;NULL)
    }
    socketInTCP=AcceptSocket
    while(run_tcpserver)
    		{  //comprobar length_dummy, abrir fichero, blucle, cerra fichero
    			bytesRecv = recv(socketInTCP, recvbuf, sizeof(recvbuf), 0);
    			if (bytesRecv > 0)
    				{if(ctr1==1)
    					{	
    						
    						p=ntohs(clientService.sin_port);
    						cout << "port the first time it receives:" << p << "\n";
    						m=0;
    						
    						
    					}
    					ctr1=0;
    					port_array[m]=ntohs(clientService.sin_port);
    					m++;
    
    					ofstream outfile(outputfile, ios::out |ios::ate | ios::binary);//open
    					rcv_data=rcv_data+bytesRecv;
    					cout << "\xd" << rcv_data;
       					
    
    				 outfile.write (var, sizeof(var));
    				 outfile.close();
    				 memset(recvbuf,0,sizeof(recvbuf));
    				 if(rcv_data==(length_dummy*4))
    					{
    						
    						ReleaseMutex(req_mutex);
    					}
    			 
    			}
    	}
    the first time p=2000

    and

    the second time

    port_array[0]=2000
    but
    port_array[1]=49154.....

  9. #9
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    I don't understand why after the TCP connection the destination port of my UDP connection is changed...but I have made debugging and I know that it occurs...

  10. #10
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    Sorry, I wrote something wrong,

    the first time it receives port=2000

    but the second time=49154 and not 2000

  11. #11
    CSharpener vart's Avatar
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    Quote Originally Posted by radeberger View Post
    I don't understand why after the TCP connection the destination port of my UDP connection is changed...but I have made debugging and I know that it occurs...
    It seems somethere you have a memory overrun
    The first 90% of a project takes 90% of the time,
    the last 10% takes the other 90% of the time.

  12. #12
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    I think it's a synchronization problem because I use a mutex for the screen and I think something is wrong but I can't see it....

  13. #13
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    what would you recommend me to synchronize the cout on the screen? a mutex or a semaphore?

  14. #14
    CSharpener vart's Avatar
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    Quote Originally Posted by radeberger View Post
    what would you recommend me to synchronize the cout on the screen? a mutex or a semaphore?
    what do you want to synchronize cout with?

    If you are using MS multi-threaded library - cout is alrady synchronized
    The first 90% of a project takes 90% of the time,
    the last 10% takes the other 90% of the time.

  15. #15
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    I don't want that two threads print out something on the screen at the same time

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