c++structstdtuple

Converting a struct to/from a std::tuple


Assuming I have a struct and a std::tuple with the same type layout:

struct MyStruct { int i; bool b; double d; }
using MyTuple = std::tuple<int,bool,double>;

Is there any standardized way to cast one to another?

I'm not sure if trivial memory copying can do the trick, as it is alignment and implementation dependent.


Solution

  • Unfortunately there is no automatic way to do that, BUT an alternative is adapt the struct to Boost.Fusion sequence. You do this once and for all for each new class.

    #include <boost/fusion/adapted/struct/adapt_struct.hpp>
    ...
    struct MyStruct { int i; bool b; double d; }
    
    BOOST_FUSION_ADAPT_STRUCT(
        MyStruct,
        (int, i)
        (bool, b)
        (double, d)
    )
    

    The use MyStruct as if it where a Fusion.Sequence (it fits generically almost everywhere you already use std::tuple<...>, if you make those functions generic.) As a bonus you will not need to copy your data members at all.

    If you really need to convert to std::tuple, after "Fusion-adapting" you can do this:

    #include <boost/fusion/adapted/std_tuple.hpp>
    #include <boost/fusion/algorithm/iteration/for_each.hpp>
    #include <boost/fusion/algorithm/transformation/zip.hpp>
    ...
    auto to_tuple(MyStruct const& ms){
       std::tuple<int, bool, double> ret;
       auto z = zip(ret, ms);
       boost::fusion::for_each(z, [](auto& ze){get<0>(ze) = get<1>(ze);});
       // or use boost::fusion::copy
       return ret;
    }
    

    The truth is that std::tuple is a half-backed feature. It is like having STD containers and no algorithms. Fortunatelly we have #include <boost/fusion/adapted/std_tuple.hpp> that allows us to do amazing things.

    Full code:

    By including the std_tuple.hpp header from Boost.Fusion std::tuple is automatically adapted to a Boost.Fusion sequence, thus the following is possible by using Boost.Fusion as a bridge between your struct and std::tuple:

    #include <iostream>
    #include <string>
    #include <tuple>
    
    #include <boost/fusion/adapted/struct/adapt_struct.hpp>
    #include <boost/fusion/algorithm/auxiliary/copy.hpp>
    #include <boost/fusion/adapted/std_tuple.hpp>
    
    struct foo
    {
      std::string a, b, c;
      int d, e, f;
    };
    
    BOOST_FUSION_ADAPT_STRUCT(
        foo,
        (std::string, a)
        (std::string, b)
        (std::string, c)
        (int, d)
        (int, e)
        (int, f)
    )
    
    template<std::size_t...Is, class Tup>
    foo to_foo_aux(std::index_sequence<Is...>, Tup&& tup) {
      using std::get;
      return {get<Is>(std::forward<Tup>(tup))...};
    }
    template<class Tup>
    foo to_foo(Tup&& tup) {
      using T=std::remove_reference_t<Tup>;
      return to_foo_aux(
        std::make_index_sequence<std::tuple_size<T>{}>{},
        std::forward<Tup>(tup)
      );
    }
    
    template<std::size_t...Is>
    auto to_tuple_aux( std::index_sequence<Is...>, foo const& f ) {
      using boost::fusion::at_c;
      return std::make_tuple(at_c<Is>(f)...);
    }
    auto to_tuple(foo const& f){
      using T=std::remove_reference_t<foo>;
      return to_tuple_aux(
        std::make_index_sequence<boost::fusion::result_of::size<foo>::type::value>{},
        f
      );    
    }
    
    int main(){
    
    
        foo f{ "Hello", "World", "!", 1, 2, 3 };
    
        std::tuple<std::string, std::string, std::string, int, int, int> dest = to_tuple(f);
        // boost::fusion::copy(f, dest); // also valid  but less general than constructor
    
        std::cout << std::get<0>(dest) << ' ' << std::get<1>(dest) << std::get<2>(dest) << std::endl;
        std::cout << at_c<0>(dest) << ' ' << at_c<1>(dest) << at_c<2>(dest) << std::endl; // same as above
    
        foo f2 = to_foo(dest);
    
        std::cout << at_c<0>(f2) << ' ' << at_c<1>(f2) << at_c<2>(f2) << std::endl;
    }
    

    I will not recommend reinterpret_cast<std::tuple<...>&>(mystructinstance.i) because that will result in negative votes and it is not portable.