如何更改IPV6地址的字节顺序（从网络到主机，反之亦然）？ [英] How can I change the byte order (from network to host, and vice versa) of an IPV6 address?

问题描述

我知道 ntoh {s，l} 和 hton {s，l} 2和4个字节的整数。
现在，我面临转换16个字节长的IPv6地址的问题。

I am aware of ntoh{s,l} and hton{s,l}, which work on integers of 2 and 4 bytes. Now, I am facing the problem to translate an IPv6 address, which is 16 bytes long.

是否有用于此目的的现成函数？

Is there a ready-made function for that purpose?

TIA，
Jir

TIA, Jir

推荐答案

我不是确保 ntoh 和 hton 在IPv6中相关。您不是本地的128位类型吗？

I'm not sure that ntoh and hton are relevant in IPv6. You don't have a native 128-bit type, do you?

根据 http://www.mail-archive.com/[e-mail]/msg00195.html ：

IPv6地址只要采用二进制形式（在
上是主机，路由器上，
等）。在其他地方，请参阅RFC
2553，第3.2节。

IPv6 addresses are expected to be representd in network byte order whenever they are in binary form (on the wire, on a host, in a router, etc). Among other places, see RFC 2553, section 3.2.

来自RFC 2553 ：

3.2 IPv6地址结构

3.2 IPv6 Address Structure

一个新的in6_addr结构包含一个IPv6地址，并且由于包含以下内容而被定义：

A new in6_addr structure holds a single IPv6 address and is defined as a result of including :

struct in6_addr {
    uint8_t  s6_addr[16];      /* IPv6 address */
};

此数据结构包含十六个8位元素的数组，这些元素组成一个128位IPv6地址。 IPv6地址以网络字节顺序存储。

This data structure contains an array of sixteen 8-bit elements, which make up one 128-bit IPv6 address. The IPv6 address is stored in network byte order.

上面的in6_addr结构通常是通过嵌入式联合实现的，该联合具有额外的字段，这些字段以类似于 struct in_addr的BSD实现。为简单起见，此处省略了这些其他实现细节。

The structure in6_addr above is usually implemented with an embedded union with extra fields that force the desired alignment level in a manner similar to BSD implementations of "struct in_addr". Those additional implementation details are omitted here for simplicity.

示例如下：

struct in6_addr {
    union {
        uint8_t  _S6_u8[16];
        uint32_t _S6_u32[4];
        uint64_t _S6_u64[2];
    } _S6_un;
};
#define s6_addr _S6_un._S6_u8

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