Question

I am converting a boost::asio::ip::address_v6 IP (say 1456:94ce:2567:a4ef:1356:94de:2967:a4e8) first into 16 bytes unsigned char array by doing the following :

auto ip = boost::asio::ip::address_v6::from_string("1456:94ce:2567:a4ef:1356:94de:2967:a4e8");

auto v6Bytes = ip.boost::asio::ip::address_v6::to_bytes();

Now my next objective is to use 8 bytes from the byte array and convert them into uint64_t (say I get num1). Similarly, using the next 8 bytes from the array, I want to generate another uint64_t (say num2). What logic can I use here for conversion?

Also, once I get num1 and num2, I want to use them and convert back to a

std::array<unsigned char, 16>

What logic can I use here?

Answer 1

唯一受支持的方法是使用std::memcpy。将前八个字节复制到一个变量中，然后将其他八个字节复制到第二个变量中。

这可以通过&运算符的地址和指针算法轻松实现：

std::uint64_t part1, part2;
std::memcpy(&part1, v6Bytes.data(), 8);
std::memcpy(&part2, v6Bytes.data() + 8, 8);

复制相反的方法以将数据返回阵列。

Answer 2

我认为原始数组是网络字节顺序（bigendian），并执行以下操作：

auto ip = boost::asio::ip::address_v6::from_string("1456:94ce:2567:a4ef:1356:94de:2967:a4e8");

auto v6Bytes = ip.boost::asio::ip::address_v6::to_bytes();

std::uint64_t num1;
std::uint64_t num2;

std::copy(std::begin(v6Bytes), std::begin(v6Bytes) + std::size(v6Bytes) / 2, (unsigned char*)&num1);
std::copy(std::begin(v6Bytes) + std::size(v6Bytes) / 2, std::end(v6Bytes), (unsigned char*)&num2);

// assume network byte order

boost::endian::big_to_native(num1);
boost::endian::big_to_native(num2);

// and back again

std::array<unsigned char, 16> bytes;

boost::endian::native_to_big(num1);
boost::endian::native_to_big(num2);

std::copy((unsigned char*)&num1, ((unsigned char*)&num1) + 8, bytes.data());
std::copy((unsigned char*)&num2, ((unsigned char*)&num2) + 8, bytes.data() + 8);

assert(bytes == v6Bytes);

How to convert boost::asio::ip::address_v6 IP into 2 uint64_t numbers and back from 2 uint64_t to v6 address?

2 个答案: