给文本字符串是否是用标准C套接字API的IPv6地址或IPv4地址 [英] Tell whether a text string is an IPv6 address or IPv4 address using standard C sockets API

问题描述

我有哪里外部组件通过我这包含一个IP地址的字符串的程序。那么我就需要把它变成一个URI。对于IPv4这是容易的;我prePEND的http：//和追加/。然而，对IPv6的我还需要围绕着它在方括号[]。

I have a program where an external component passes me a string which contains an IP address. I then need to turn it into a URI. For IPv4 this is easy; I prepend http:// and append /. However, for IPv6 I need to also surround it in brackets [].

有没有一个标准套接字API调用来确定地址的地址族？

Is there a standard sockets API call to determine the address family of the address?

推荐答案

使用的getaddrinfo （ ），并设置提示标志AI_NUMERICHOST，家庭AF_UNSPEC，在从的getaddrinfo全成归来，所产生的结构addrinfo中.ai_family成员将不是AF_INET或AF_INET6。

Use getaddrinfo() and set the hint flag AI_NUMERICHOST, family to AF_UNSPEC, upon successfull return from getaddrinfo, the resulting struct addrinfo .ai_family member will be either AF_INET or 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

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