如何比较IPv6大于/小于C [英] How to compare an IPv6 is greater/less than in C

问题描述

使用 inet_pton 转换IPv4以比较它是否在IPv4范围内非常简单。但是，我不确定如何使用 inet_pton 和 in6_addr 来查看它是否小于另一个IP。这就是我的想法：

Converting an IPv4 with inet_pton to compare if it's within a range of IPv4 is fairly simple. However, I'm not sure how to use inet_pton and in6_addr and see if it's less/greater than another IP. Here's what I was thinking:

#include <arpa/inet.h>

  ...

const char *ip6str = "0:0:0:0:0:ffff:c0a8:3";
const char *first = "0:0:0:0:0:ffff:c0a8:1";
const char *last = "0:0:0:0:0:ffff:c0a8:5";

struct in6_addr result, resfirst, restlast;
uint8_t ipv6[16]; // perhaps to hold the result?

inet_pton(AF_INET6, first, &resfirst);
inet_pton(AF_INET6, last, &reslast);
inet_pton(AF_INET6, ip6str, &result);

//assuming inet_pton succeed
if(result.s6_addr >= resfirst.s6_addr && result.s6_addr <= reslast.s6_addr)
    //within range

推荐答案

您可以使用 memcmp ，因为它们以网络字节顺序存储（aka大字节序）。

You can use memcmp for this since they're stored in network byte order (aka big endian).

if (memcmp(&result, &resfirst, sizeof(result)) > 0 && memcmp(&result, &reslast, sizeof(result)) < 0)

我想您可能是说> = 的意思，而且可能是< = 。

I think you might've meant >= though and possibly <= as well.

实际上，对于IPv4，您也必须这样做，至少在很少的情况下字节序机器。

In fact, you'd have to do it this way for IPv4 as well, at least on little endian machines.

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