c++ - 将C++ IPv6字符串表示形式转换为boost::multiprecision::uint128_t

Question

我需要将IPv6字符串地址转换为boost::multiprecision::uint128_t
对于IPv4，我使用以下算法:

uint32_t byte1 = 0, byte2 = 0, byte3 = 0, byte4 = 0;
sscanf(ipAddress, "%3d.%3d.%3d.%3d", &byte1, &byte2, &byte3, &byte4);
uint32_t ip = (byte1 << 24) | (byte2 << 16) | (byte3 << 8) | (byte4);

对于IPv6，我该怎么做？

Answer 1

使用Wikipedia的示例:



还使用Boost Asio的address_v6实现而不是1970年代的解析:

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#include <boost/asio/ip/address_v6.hpp>
#include <boost/multiprecision/cpp_int.hpp>
#include <cstdio>

using boost::asio::ip::address_v6;
using boost::multiprecision::int128_t;

int main() {
    auto v6 = address_v6::from_string("2001:0DBB:AC10:FE01::");

    int128_t val {};
    for (uint8_t b : v6.to_bytes())
        (val <<= 8) |= b;

    std::cout << std::hex << std::showbase << val << std::endl;
}

版画

0x20010dbbac10fe010000000000000000