Filter interfaces depending on what is required in linux with C

Question

Filter interfaces depending on what is required in linux with C

Navigation

#1 by (1 votes)

0

and with help from here I was able to obtain a program that lists all the network interfaces, the program is as follows:

#define _GNU_SOURCE     /* To get defns of NI_MAXSERV and NI_MAXHOST */
#include <arpa/inet.h>
#include <sys/socket.h>
#include <netdb.h>
#include <ifaddrs.h>
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <linux/if_link.h>

int main(int argc, char *argv[])
{
struct ifaddrs *addrs, *tmp;
getifaddrs(&addrs);
tmp = addrs;

/*solo imprime el nombre de los puertos disponibles*/
while (tmp)
{
    if (tmp->ifa_addr && tmp->ifa_addr->sa_family == AF_PACKET)
        printf("%s\n", tmp->ifa_name);

    tmp = tmp->ifa_next;
}
freeifaddrs(addrs);
}

What I would miss is that now I ask for the similar ones and return in an array the interfaces that are similar (that is, they start with the same thing).

For example, it would be like asking:

Give me the interfaces that start with "eth". The result could be an array with the strings: eth0, eth1, eth3.

I understand that the interface that you deploy is done through the ifa_name , but as I do the comparison with everything that started with eth for example

lista c

asked by Edú Arias 01.12.2016 в 00:09

source

1 answer

Problem with geometry.spherical.computeDistanceBetween () Google Maps ItextSharp does not recognize the accents

score 1 · Answer 1

To achieve the goal, we will pass the prefix to compare your program as an argument in the command line. This being the case, the answer to your question is actually made up of 2 parts.

The first is

Read the command line arguments

When you write the main function of a C program, you usually see one of these two definitions:

int main(void)
int main(int argc, char **argv)

The second form receives the arguments that have been passed to the program from the command line, and the number of arguments specified (the arguments are separated by spaces).

Thus, the arguments of main are:

int argc - The number of arguments that have been passed to the program when it is executed. That value is at least 1 .
char **argv - It is a pointer to char * . Alternatively it can be: char *argv[] , which means an 'array of char * '. This is an array of pointers to strings ending in null.

Basic example:

For example, you could print the arguments that your program receives with this program:

#include <stdio.h>

int main(int argc, char **argv)
{
    for (int i = 0; i < argc; ++i)
    {
        printf("argv[%d]: %s\n", i, argv[i]);
    }
}

Using GCC 4.5 to compile this code into a file called args.c would default to an executable called a.out .

[jachguate@lilun c_code]$ gcc -std=c99 args.c

When executing it ...

[jachguate@lilun c_code]$ ./a.out hello there
argv[0]: ./a.out
argv[1]: hello
argv[2]: there

As you can see, in argv , the first element argv[0] is the name of the program itself (it is not something that is defined in a standard, but it is common that it works). The arguments you received are in argv[1] and following.

This is pretty rudimentary, but it works for simple cases (like this one).

The other part is:

Compare if a character string starts with another:

Once you have the argument, now basically we need to know if the name of the interface that we are processing within the cycle while , stored in ifa_name starts with said prefix , for that we can make use of of the strncmp function, which basically compares the first n bytes of two strings and returns 0 if the strings They are equal. The way to call it would be with an if, in a structure similar to this:

if strncmp(prefijo, cadena, strlen(prefijo)) == 0 {
    printf("cadena SI inicia con prefijo!");
}

Now let's put the pieces together

Putting this together in code is a task left to the reader, but basically what I would do is follow the following steps:

Check the number of arguments received. If no argument has been passed to the program, inform the user of the form of use and exit with error (the result of main could be 1 and not 0, for example)
Within the while, add with a logical operator && (and) the comparison condition with the prefix, at the end of the if already existing. So that only those that meet all the conditions are printed.