Hello! I'm prepping for a simple work project and am trying to familiarize myself with the basics of socket programming in a Unix dev environment. At this point, I have some basic server side code setup to listen for incoming TCP connection requests from clients after the parent socket has been created and is set to listen...

int sockfd, newfd;    
unsigned int len;
socklen_t sin_size;
char msg[]="Test message sent";
char buf[MAXLEN];
int st, rv;
struct addrinfo hints, *serverinfo, *p; 
struct sockaddr_storage client;
char ip[INET6_ADDRSTRLEN];

.
. //parent socket creation and listen code omitted for simplicity
.

//wait for connection requests from clients
while(1)
  {
    //Returns the socketID and address of client connecting to socket
    if( ( newfd = accept(sockfd, (struct sockaddr *)&client, &len) ) == -1 ){
      perror("Accept");
      exit(-1);
    }

    if( (rv = recv(newfd, buf, MAXLEN-1, 0 )) == -1) {
      perror("Recv");
      exit(-1);
    }
    struct sockaddr_in *clientAddr = ( struct sockaddr_in *) get_in_addr((struct sockaddr *)&client);
    inet_ntop(client.ss_family, clientAddr, ip, sizeof ip);
    printf("Receive from %s: query type is %s\n", ip, buf);

    if( ( st = send(newfd, msg, strlen(msg), 0)) == -1 ) {
      perror("Send");
      exit(-1);
    }

    //ntohs is used to avoid big-endian and little endian compatibility issues
    printf("Send %d byte to port %d\n", ntohs(clientAddr->sin_port) );

    close(newfd);
  }

}

I found the get_in_addr function online and placed it at the top of my code and use it to obtain the IP address of the client connecting...

// get sockaddr, IPv4 or IPv6:
void *get_in_addr(struct sockaddr *sa)
{
  if (sa->sa_family == AF_INET) {
    return &(((struct sockaddr_in*)sa)->sin_addr);
  }

  return &(((struct sockaddr_in6*)sa)->sin6_addr);
}

but the function always returns the IPv6 IP address since thats what the sa_family property is set as.

My question is, is the IPv4 IP address stored anywhere in the data I'm using and, if so, how can I access it?

Thanks so much in advance for all your help!