C语言 C 语言中的 IPv6 解析
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/2962664/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
IPv6 parsing in C
提问by The Stig
I wanted to know how I can parse an IPv6 address in C and convert it to a 128 bit value?
我想知道如何在 C 中解析 IPv6 地址并将其转换为 128 位值？
So a hex address like 1:22:333:aaaa:b:c:d:eneeds to be converted to its 128 bit equivalent binary. The problem is the IP address could be of the type ::2and its variant since they are valid IPv6 address.
因此1:22:333:aaaa:b:c:d:e需要将十六进制地址转换为其 128 位等效二进制。问题是 IP 地址可能属于该类型::2及其变体，因为它们是有效的 IPv6 地址。
The input is from the keyboard and hence is in ASCII format.
输入来自键盘，因此采用 ASCII 格式。
回答by dreamlax
You can use POSIX inet_ptonto convert a string to a struct in6_addr.
您可以使用 POSIXinet_pton将字符串转换为struct in6_addr.
#include <arpa/inet.h>
...
const char *ip6str = "::2";
struct in6_addr result;
if (inet_pton(AF_INET6, ip6str, &result) == 1) // success!
{
//successfully parsed string into "result"
}
else
{
//failed, perhaps not a valid representation of IPv6?
}
回答by Thanatos
getaddrinfo()can understand IPv6 addresses. Pass AF_INET6 to it in the hints, as well as AI_NUMERICHOST (to prevent a DNS lookup). Linux has it, Windows has it as of Windows XP.
getaddrinfo()可以理解 IPv6 地址。在提示中将 AF_INET6 以及 AI_NUMERICHOST 传递给它（以防止 DNS 查找）。Linux 拥有它，Windows 从 Windows XP 开始拥有它。
回答by Ales Teska
You can use getaddrinfo()POSIX function. It is more flexible than inet_pton(), for example it automatically detects IPv4 and IPv6 address formats, it can resolve even hostnames (using DNS resolving) and port/service names (using /etc/services).
您可以使用getaddrinfo()POSIX 函数。它比 更灵活inet_pton()，例如它会自动检测 IPv4 和 IPv6 地址格式，甚至可以解析主机名（使用 DNS 解析）和端口/服务名称（使用/etc/services）。
#include <sys/types.h>
#include <netdb.h>
#include <netdb.h>
....
const char *ip6str = "::2";
struct sockaddr_storage result;
socklen_t result_len;
struct addrinfo *res = NULL;
struct addrinfo hints;
memset(&hints, 0, sizeof(struct addrinfo));
hints.ai_family = PF_UNSPEC;
hints.ai_socktype = SOCK_STREAM;
hints.ai_flags = AI_DEFAULT | AI_NUMERICHOST | AI_NUMERICSERV;
rc = getaddrinfo(ip6str, NULL, &hints, &res);
if (rc != 0)
{
fprintf(stderr, "Failure to parse host '%s': %s (%d)", ip6str, gai_strerror(rc), rc);
return -1;
}
if (res == NULL)
{
// Failure to resolve 'ip6str'
fprintf(stderr, "No host found for '%s'", ip6str);
return -1;
}
// We use the first returned entry
result_len = res->ai_addrlen;
memcpy(&result, res->ai_addr, res->ai_addrlen);
freeaddrinfo(res);
The IPv6 address is stored in the struct sockaddr_storage resultvariable.
IPv6 地址存储在struct sockaddr_storage result变量中。
if (result.ss_family == AF_INET6) // Ensure that we deal with IPv6
{
struct sockaddr_in6 * sa6 = (struct sockaddr_in6 *) &result;
struct in6_addr * in6 = &sa6->sin6_addr;
in6->s6_addr[0]; // This is a first byte of the IPv6
in6->s6_addr[15]; // This is a last byte of the IPv6
}
回答by YeenFei
To parse IPv6 in C, you need to build yourself a utility function, which tokenized string (colon for hex blocks, and forward-slash for subnet bits).
要在 C 中解析 IPv6，您需要自己构建一个实用程序函数，该函数将字符串标记化（十六进制块的冒号，子网位的正斜杠）。
- Tokenize raw IPv6 string into smaller substring.
- Convert non-empty substring into hex blocks. (ASCII to decimal conversion)
- Expand hex block into 2-bytes by padding zero in front. (only leading zeroes get trimmed)
- Complete IPv6 should have 8 hex blocks, calculate missing hex-block(s). (zeroes grouping can happen only once)
- Reinsert missing hex-block. (use index of the empty substring)
- 将原始 IPv6 字符串标记为更小的子字符串。
- 将非空子字符串转换为十六进制块。（ASCII 到十进制转换）
- 通过在前面填充零将十六进制块扩展为 2 个字节。（只有前导零被修剪）
- 完整的 IPv6 应该有 8 个十六进制块，计算丢失的十六进制块。（零分组只能发生一次）
- 重新插入丢失的六角块。（使用空子串的索引）
回答by Fernando Gonzalez Sanchez
Rosetta has samples in several languages: https://rosettacode.org/wiki/Parse_an_IP_Address
Rosetta 有多种语言的样本：https: //rosettacode.org/wiki/Parse_an_IP_Address
回答by Patryk
if you can use boost, something like this should work:
如果你可以使用boost，这样的事情应该可以工作：
#include<boost/asio.hpp>
using boost::asio::ip;
bool parseIpv6String(std::string ipv6_string, char* dest){
try{
address_v6 addr = address_v6::from_string(ipv6_string);
memcpy(dest,addr.to_bytes().data(), 16);
}catch(...){
return false;
}
return true;
}
It is a bit more portable than for example POSIX specific functions.
它比例如 POSIX 特定功能更便携。
回答by Ken Bloom
In Windows, you can use WSAStringToAddress, which is available since Windows 2000.
在 Windows 中，您可以使用WSAStringToAddress自 Windows 2000 起可用的 。
