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What's Socket Addressing ?
An Introduction to Socket Programming
What's Socket Addressing ?
A Socket Address is a host.port pair (communication is between host.port pairs -- one on the server, the other on the client). We know how to determine host numbers and service numbers so we're well on our way to filling out a structure were we specify those numbers. The structure is sockaddr_in, which has the address family is AF_INET as in this fragment:
int tcpopen(host,service)
char *service, *host;
{ int unit;
struct sockaddr_in sin;
struct servent *sp;
struct hostent *hp;
etc...
if ((sp=getservbyname(service,"tcp"))
== NULL) then error...
if ((hp=gethostbyname(host)) == NULL)
then error...
bzero((char *)&sin, sizeof(sin));
sin.sin_family=AF_INET;
bcopy(hp->h_addr,(char *)&sin.sin_addr,
hp->h_length);
sin.sin_port=sp->s_port;
etc...
The code fragment is filling in the IP address type AF_INET, port number and IP address in the Socket Address structure -- the address of the remote host.port where we want to connect to find a service.
There's a generic Socket Address structure, a sockaddr, used for communication in arbitrary domains. It has an address family field and an address (or data) field:
/* from: /usr/include/sys/socket.h */
struct sockaddr {
u_short sa_family; /*address family */
char sa_data[14];/*max 14 byte addr*/
};
The sockaddr_in structure is for
Internet Socket Addresses
address family AF_INET). An instance
of the generic socket address.
/* from: /usr/include/netinet/in.h */
struct sockaddr_in {
short sin_family; /* AF_INET */
u_short sin_port; /* service port */
struct in_addr sin_addr; /*host number */
char sin_zero[8]; /* not used */
};
The family defines the interpretation of the data. In other domains addressing will be different -- services in the UNIX domain are names (eg. /dev/printer). In the sockaddr_in structure we've got fields to specify a port and a host IP number (and 8 octets that aren't used at all!). That structure specifies one end of an IPC connection. Creating that structure and filling in the right numbers has been pretty easy so far.
(Continued on next question...)
Other Interview Questions
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