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Socket programming

 
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# 8  
Old 03-11-2010
Yes. Have your program open a socket in listening mode, and have it fork() on accept(). The child inherits the new socket, and can transfer (read and write) data through it. Meanwhile the parent can still accept new connections on the same port, and spawn additional child processes. All the while they use the same TCP or UDP port. That's the same mechanism that Apache uses if you've configured it in prefork mode.

To illustrate:
Code:
# netstat -antp | grep ':80'
tcp        0      0 1.2.3.4:80        0.0.0.0:*               LISTEN      534/httpd2-prefork
tcp        0      0 1.2.3.5:80        0.0.0.0:*               LISTEN      534/httpd2-prefork

From a different machine:
Code:
$ telnet 1.2.3.4 80
Trying 1.2.3.4...
Connected to 1.2.3.4
Escape character is '^]'.
HEAD / HTTP/1.1
Host: 1.2.3.4
Connection: keep-alive
Accept: */*

HTTP/1.1 403 Forbidden
Date: Thu, 11 Mar 2010 07:44:44 GMT
Server: Apache/2.2.3 (Linux/SUSE)
Vary: accept-language,accept-charset
Accept-Ranges: bytes
Connection: close
Content-Type: text/html; charset=iso-8859-1
Content-Language: en

The keep-alive directive will keep the connection open. Meanwhile, we take another look at netstat:
Code:
# netstat -antp | grep ':80'
tcp        0      0 1.2.3.4:80        0.0.0.0:*               LISTEN      534/httpd2-prefork
tcp        0      0 1.2.3.5:80        0.0.0.0:*               LISTEN      534/httpd2-prefork
tcp        0      0 1.2.3.4:80        1.2.9.8:59388           ESTABLISHED 1033/httpd2-prefork

OMG, 2 processes on the same Port, one listening the other communicating. But wait, it gets even worse if we open 2 telnet sessions:
Code:
# netstat -antp | grep ':80'
tcp        0      0 1.2.3.4:80        0.0.0.0:*               LISTEN      534/httpd2-prefork
tcp        0      0 1.2.3.5:80        0.0.0.0:*               LISTEN      534/httpd2-prefork
tcp        0      0 1.2.3.4:80        1.2.9.8:39364           ESTABLISHED 22914/httpd2-prefor
tcp        0      0 1.2.3.4:80        1.2.9.8:39365           ESTABLISHED 26486/httpd2-prefor

Now it's even 3 processes on the same port.

You see, it is possible, as long as you have a central instance (in this case the main Apache process) handling incoming requests and delegating them to other processes.
 
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