linux - [Solved-5 Solutions] How to get the ip address of the machine in Linux ? - ubuntu - red hat - debian - linux server - linux pc
Linux - Problem :
How to get the ip address of the machine ?
Linux - Solution 1:
#include <stdio.h>
#include <sys/types.h>
#include <ifaddrs.h>
#include <netinet/in.h>
#include <string.h>
#include <arpa/inet.h>
int main (int argc, const char * argv[]) {
struct ifaddrs * ifAddrStruct=NULL;
struct ifaddrs * ifa=NULL;
void * tmpAddrPtr=NULL;
getifaddrs(&ifAddrStruct);
for (ifa = ifAddrStruct; ifa != NULL; ifa = ifa->ifa_next) {
if (!ifa->ifa_addr) {
continue;
}
if (ifa->ifa_addr->sa_family == AF_INET) { // check it is IP4
// is a valid IP4 Address
tmpAddrPtr=&((struct sockaddr_in *)ifa->ifa_addr)->sin_addr;
char addressBuffer[INET_ADDRSTRLEN];
inet_ntop(AF_INET, tmpAddrPtr, addressBuffer, INET_ADDRSTRLEN);
printf("%s IP Address %s\n", ifa->ifa_name, addressBuffer);
} else if (ifa->ifa_addr->sa_family == AF_INET6) { // check it is IP6
// is a valid IP6 Address
tmpAddrPtr=&((struct sockaddr_in6 *)ifa->ifa_addr)->sin6_addr;
char addressBuffer[INET6_ADDRSTRLEN];
inet_ntop(AF_INET6, tmpAddrPtr, addressBuffer, INET6_ADDRSTRLEN);
printf("%s IP Address %s\n", ifa->ifa_name, addressBuffer);
}
}
if (ifAddrStruct!=NULL) freeifaddrs(ifAddrStruct);
return 0;
}
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Linux - Solution 2:
- Create a socket.
- Perform ioctl(
, SIOCGIFCONF, (struct ifconf)&buffer);
Read /usr/include/linux/if.h for information on the ifconf and ifreq structures. This should give you the IP address of each interface on the system. Also read /usr/include/linux/sockios.h for additional ioctls.
Linux - Solution 3:
- It uses the local routing table to find the IP address of the ethernet interface that would be used for a connection to a specific external host.
- By using a connected UDP socket, you can get the information without actually sending any packets.
- The approach requires that you choose a specific external host.
void GetPrimaryIp(char* buffer, size_t buflen)
{
assert(buflen >= 16);
int sock = socket(AF_INET, SOCK_DGRAM, 0);
assert(sock != -1);
const char* kGoogleDnsIp = "8.8.8.8";
uint16_t kDnsPort = 53;
struct sockaddr_in serv;
memset(&serv, 0, sizeof(serv));
serv.sin_family = AF_INET;
serv.sin_addr.s_addr = inet_addr(kGoogleDnsIp);
serv.sin_port = htons(kDnsPort);
int err = connect(sock, (const sockaddr*) &serv, sizeof(serv));
assert(err != -1);
sockaddr_in name;
socklen_t namelen = sizeof(name);
err = getsockname(sock, (sockaddr*) &name, &namelen);
assert(err != -1);
const char* p = inet_ntop(AF_INET, &name.sin_addr, buffer, buflen);
assert(p);
close(sock);
}
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Linux - Solution 4:
As you have found out there is no such thing as a single "local IP address". Here's how to find out the local address that can be sent out to a specific host.
- Create a UDP socket
- Connect the socket to an outside address (the host that will eventually receive the local address)
- Use getsockname to get the local address
Linux - Solution 5:
You can try this:
// ifconfig | perl -ne 'print "$1\n" if /inet addr:([\d.]+)/'
#include <stdlib.h>
int main() {
setenv("LANG","C",1);
FILE * fp = popen("ifconfig", "r");
if (fp) {
char *p=NULL, *e; size_t n;
while ((getline(&p, &n, fp) > 0) && p) {
if (p = strstr(p, "inet ")) {
p+=5;
if (p = strchr(p, ':')) {
++p;
if (e = strchr(p, ' ')) {
*e='\0';
printf("%s\n", p);
}
}
}
}
}
pclose(fp);
return 0;
}