3

我需要弄清楚，ip是否属于ip mask。例如：

ip = 192.168.0.1 掩码 = 192.168.0.1/24。

我找到了将 ip 转换为掩码的函数：

inet_cidrtoaddr(int cidr, struct in_addr *addr)
{
        int ocets;

        if (cidr < 0 || cidr > 32) {
                errno = EINVAL;
                return -1;
        }
        ocets = (cidr + 7) / 8;

        addr->s_addr = 0;
        if (ocets > 0) {
                memset(&addr->s_addr, 255, (size_t)ocets - 1);
                memset((unsigned char *)&addr->s_addr + (ocets - 1),
                       (256 - (1 << (32 - cidr) % 8)), 1);
        }

        return 0;
}

我如何比较 ip 和 cidr 范围？

4

4 回答 4

9

因此，将 Olis 的答案放入代码中：

// Check if 192.168.0.1 is inside 192.168.0.0/24
in_addr ip, net, netmask;
inet_aton("192.168.0.1", &ip);
inet_aton("192.168.0.0", &net);

他说：

inet_cidrtoaddr(24, &netmask);
bool is_inside = ((ip.s_addr & netmask.s_addr) == (net.s_addr & netmask.s_addr));

我更喜欢addr4_match方法：

bool cidr_match(const in_addr &addr, const in_addr &net, uint8_t bits) {
  if (bits == 0) {
    // C99 6.5.7 (3): u32 << 32 is undefined behaviour
    return true;
  }
  return !((addr.s_addr ^ net.s_addr) & htonl(0xFFFFFFFFu << (32 - bits)));
}
bool is_inside = cidr_match(ip, net, 24);

我尝试了一堆不同的输入：https ://gist.github.com/duedal/b83303b4988a4afb2a75

如果有人发现这也需要 IPv6 解决方案：

bool cidr6_match(const in6_addr &address, const in6_addr &network, uint8_t bits) {
#ifdef LINUX
  const uint32_t *a = address.s6_addr32;
  const uint32_t *n = network.s6_addr32;
#else
  const uint32_t *a = address.__u6_addr.__u6_addr32;
  const uint32_t *n = network.__u6_addr.__u6_addr32;
#endif
  int bits_whole, bits_incomplete;
  bits_whole = bits >> 5;         // number of whole u32
  bits_incomplete = bits & 0x1F;  // number of bits in incomplete u32
  if (bits_whole) {
    if (memcmp(a, n, bits_whole << 2)) {
      return false;
    }
  }
  if (bits_incomplete) {
    uint32_t mask = htonl((0xFFFFFFFFu) << (32 - bits_incomplete));
    if ((a[bits_whole] ^ n[bits_whole]) & mask) {
      return false;
    }
  }
  return true;
}

检查 2001:db8::ff00:42:8329 是否存在于 2001:db8/32 中。注意 inet_net_pton 非常挑剔，它是 2001:db8/32 而不是 2001:db8::/32。但是 2001:db8::/48 是完全有效的（也称为 2001:db8:0/48）。

in6_addr ip6, net6, net6_48;
memset(&net6, 0, sizeof(net6));
memset(&net6_48, 0, sizeof(net6_48)); 
assert(inet_pton(AF_INET6, "2001:db8::ff00:42:8329", &ip6));

int bits = inet_net_pton(AF_INET6, "2001:db8/32", &net6, sizeof(net6));
assert((bits != -1));  // assert that inet_net_pton understood us
bool is_inside6 = cidr6_match(ip6, net6, bits);

int bits_48 = inet_net_pton(AF_INET6, "2001:db8::/48", &net6_48, sizeof(net6_48));
assert((bits_48 == 48));
bool is_inside6_48 = cidr6_match(ip6, net6_48, bits_48);
于 2014-08-13T15:58:18.733 回答
2

这个函数计算一个网络掩码（例如，255.255.255.128 形式的东西）。因此，要检查指定的 IP 地址是否属于指定的子网，只需将掩码应用于 CIDR 地址和 IP 地址（使用按位 AND执行此操作）。如果结果相同，则 IP 地址有效。

于 2011-08-27T11:06:00.273 回答
2

如果您有 IP 地址、网络地址和网络掩码，那么您可以使用如下函数：

bool
is_in_net (
        const struct in_addr*   addr,     /* host byte order */
        const struct in_addr*   netaddr,
        const struct in_addr*   netmask
        )
{
   if ((addr->s_addr & netmask->s_addr) == (netaddr->s_addr & netmask->s_addr))
      return true;
   return false;
}
于 2011-08-27T11:20:12.217 回答
0

尝试这个：

const std::uint32_t CIDR_PREFIXES[33] = {
    [0] = htonl(0),
    [1] = htonl(0x80000000),
    [2] = htonl(0xC0000000),
    [3] = htonl(0xE0000000),
    [4] = htonl(0xF0000000),
    [5] = htonl(0xF8000000),
    [6] = htonl(0xFC000000),
    [7] = htonl(0xFE000000),
    [8] = htonl(0xFF000000),
    [9] = htonl(0xFF800000),
    [10] = htonl(0xFFC00000),
    [11] = htonl(0xFFE00000),
    [12] = htonl(0xFFF00000),
    [13] = htonl(0xFFF80000),
    [14] = htonl(0xFFFC0000),
    [15] = htonl(0xFFFE0000),
    [16] = htonl(0xFFFF0000),
    [17] = htonl(0xFFFF8000),
    [18] = htonl(0xFFFFC000),
    [19] = htonl(0xFFFFE000),
    [20] = htonl(0xFFFFF000),
    [21] = htonl(0xFFFFF800),
    [22] = htonl(0xFFFFFC00),
    [23] = htonl(0xFFFFFE00),
    [24] = htonl(0xFFFFFF00),
    [25] = htonl(0xFFFFFF80),
    [26] = htonl(0xFFFFFFC0),
    [27] = htonl(0xFFFFFFE0),
    [28] = htonl(0xFFFFFFF0),
    [29] = htonl(0xFFFFFFF8),
    [30] = htonl(0xFFFFFFFC),
    [31] = htonl(0xFFFFFFFE),
    [32] = htonl(0xFFFFFFFF),
};

bool matchCIDR(const in_addr &network, const in_addr &addr, unsigned prefix)
{
    if (prefix > 32)
        return false;

    std::uint32_t cidr = CIDR_PREFIXES[prefix];
    std::uint32_t hNetwork = network.s_addr;
    std::uint32_t hAddr = addr.s_addr;

    return hNetwork == (hAddr & cidr);
}

虽然在 C++ 中，应该很容易使它成为 C

于 2019-07-30T07:40:10.547 回答