如何映射IPv4的IPv6地址转换为IPv4（字符串格式）？ [英] How to convert IPv4-mapped-IPv6 address to IPv4 (string format)?

问题描述

我有一个结构sockaddr 包含与IPv4映射的IPv6地址结构 :: FFFF：10.0.0.1 。我想获得一个字符串只IPv4版本（在这种情况下， 10.0.0.1 ）的C语言编程。我如何去实现呢？

I have a struct sockaddr structure containing an IPv4-mapped-IPv6 address like ::ffff:10.0.0.1. I want to obtain only the IPv4 version of it in a string (in this case, 10.0.0.1) in C programming language. How do I go about achieving it?

推荐答案

由于您的结构中包含IPv6地址，我会假设你有一个结构sockaddr * 指针（让我们将其命名为 addrPtr ）指向结构体sockaddr_in6 结构。

As your structure contains an IPV6 address, I'll assume your have a struct sockaddr * pointer (let's name it addrPtr) pointing to a struct sockaddr_in6 structure.

您可以轻松地获得该地址字节。

You can get the address bytes easily.

const uint8_t *bytes = ((const struct sockaddr_in6 *)addrPtr)->sin6_addr.s6_addr;

然后加入12指针，因为12第一个字节是不感兴趣（10 0×00 ，然后2 0xFF的 ）。只有4最后的母校。

Then add 12 to the pointer because the 12 first bytes are not interesting (10 0x00, then 2 0xff). Only the 4 last ones mater.

bytes += 12;

现在，我们可以使用这些四个字节做任何我们想做的。例如，我们可以将它们存储到一个IPv4的 struct in_addr，这个地址

Now, we can use those four bytes to do whatever we want. For example, we might store them into a IPv4 struct in_addr address.

struct in_addr addr = { *(const in_addr_t *)bytes };

然后，我们可以使用 inet_ntop 得到一个字符串（在＆LT宣布; ARPA / inet.h＆GT; ）。

Then we can get a string using inet_ntop (declared in <arpa/inet.h>).

char buffer[16]; // 16 characters at max: "xxx.xxx.xxx.xxx" + NULL terminator
const char *string = inet_ntop(AF_INET, &addr, buffer, sizeof(buffer));

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