使用标准C套接字API判断文本字符串是IPv6地址还是IPv4地址

Question

我有一个程序,其中一个外部组件传递给我一个包含IP地址的字符串.然后我需要把它变成一个URI.对于IPv4来说这很容易; 我在前面加上http://并追加/.但是,对于IPv6,我还需要用括号[]括起来.

是否有标准的套接字API调用来确定地址的地址族？

Answer 1

的种类.您可以inet_pton()尝试首先将字符串解析为IPv4(AF_INET)然后解析IPv6(AF_INET6).返回代码将告诉您函数是否成功,因此字符串包含尝试类型的地址.

例如:

#include <arpa/inet.h>
#include <stdio.h>

static int
ip_version(const char *src) {
    char buf[16];
    if (inet_pton(AF_INET, src, buf)) {
        return 4;
    } else if (inet_pton(AF_INET6, src, buf)) {
        return 6;
    }
    return -1;
}

int
main(int argc, char *argv[]) {
    for (int i = 1; i < argc; ++i) {
        printf("%s\t%d\n", argv[i], ip_version(argv[i]));
    }

    return 0;
}

Answer 2

使用getaddrinfo()并将提示标志AI_NUMERICHOST,family设置为AF_UNSPEC,从getaddrinfo成功返回后,生成的struct addrinfo .ai_family成员将是AF_INET或AF_INET6.

编辑,小例子

#include <sys/types.h>
#include <stdio.h>
#include <string.h>
#include <sys/socket.h>
#include <netdb.h>

int main(int argc, char *argv[])
{
    struct addrinfo hint, *res = NULL;
    int ret;

    memset(&hint, '\0', sizeof hint);

    hint.ai_family = PF_UNSPEC;
    hint.ai_flags = AI_NUMERICHOST;

    ret = getaddrinfo(argv[1], NULL, &hint, &res);
    if (ret) {
        puts("Invalid address");
        puts(gai_strerror(ret));
        return 1;
    }
    if(res->ai_family == AF_INET) {
        printf("%s is an ipv4 address\n",argv[1]);
    } else if (res->ai_family == AF_INET6) {
        printf("%s is an ipv6 address\n",argv[1]);
    } else {
        printf("%s is an is unknown address format %d\n",argv[1],res->ai_family);
    }

   freeaddrinfo(res);
   return 0;
}

$ ./a.out 127.0.0.1
127.0.0.1 is an ipv4 address
$ ./a.out ff01::01
ff01::01 is an ipv6 address