netlink API

[Vinay Venkataraghavan <raghavanvinay@xxxxxxxxx>]

Hi folks,

It looks like the netlink interface has changed in the 2.6.X kernel. I am not sure what X is. But the old API used to be:



extern struct sock *netlink_kernel_create(int unit, void (*input)(struct sock *sk, int len));


The new API is as follows:


extern struct sock *netlink_kernel_create(struct net *net,
                                           int unit,unsigned int groups,
                                           void (*input)(struct sk_buff *skb),
                                           struct mutex *cb_mutex,
                                           struct module *module);

But I am having problems getting my user space application to communicate with my driver.

I am creating a netlink kernel socket in my driver as follows:


nls = netlink_kernel_create(&init_net, NETLINK_USERSOCK, 0,
                                my_input, 0, THIS_MODULE);


and in my user space program i am trying to open a socket and write to it as follows:

struct sockaddr_nl nlDest;

    sockfd = socket(AF_NETLINK, SOCK_RAW, NETLINK_USERSOCK);

    if(sockfd < 0) {
        perror("Unable to create netlink socket. \n");
    }
    else {
        memset(&nlDest, 0, sizeof(nlDest));
        nlDest.nl_family = AF_NETLINK;
        nlDest.nl_pid = getpid();
        nlDest.nl_groups = 0;
        if (bind(sockfd, (struct sockaddr *)&nlDest, sizeof(nlDest)) < 0)
        {
            perror("could not bind netlink socket.\n");
            close(sockfd);
            exit(0);
        }
    }

But when I try to send a message using:

rc = sendmsg(sockfd, &msg, MSG_DONTWAIT);

I get an error with errno = 22. This mean Invalid argument. 

msg here is of type struct msghdr.

A couple of other things that would help is:

- how do I know that the netlink socket in the kernel has been created?
- secondly how do I know that the message will be appropriately routed to my function that I have registered with the socket.


Any help in this matter would be greatly appreciated. 

Thanks,
Vinay


      
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