CODEWITHSUNDEEP
c++boost-asio
Sanjana Gupta
Sanjana Gupta

Reputation: 43

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

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?

Upvotes: 0

Views: 649

Answers (2)

Galik
Galik

Reputation: 48605

I would assume the original array is in network byte order (bigendian) and do something like this:

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);

Upvotes: 0

Some programmer dude
Some programmer dude

Reputation: 409136

The only supported way is to use std::memcpy. Copy the first eight bytes into one variable, and then the other eight bytes into the second.

This can easily be accomplished with the address-of & operator and pointer arithmetic:

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

Copy the opposite way for getting the data back into the array.

Upvotes: 1

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