struct to/from std::tuple conversion

Question

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

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

Is there any standartized way to cast one to another?

PS I know that trivial memory copying can do the trick, but it is alignment and implementation dependent

Answer 1

We can use structured bindings to convert a struct into a tuple with a bit of work.

Struct-to-tuple is very awkward.

template<std::size_t N>
struct to_tuple_t;

template<>
struct to_tuple_t<3> {
  template<class S>
  auto operator()(S&& s)const {
    auto[e0,e1,e2]=std::forward<S>(s);
    return std::make_tuple(e0, e1, e2);
  }
};

Now, write a to_tuple_t for each size you want to support. This gets tedious. Sadly I know of no way to introduce a parameter pack there.

template<std::size_t N, class S>
auto to_tuple(S&& s) {
  return to_tuple_t<N>{}(std::forward<S>(s));
}

I know of no way to calculate the value of N required either. So you'd have to type the 3 in auto t = to_tuple<3>(my_struct); when you call it.

I am not a master of structured bindings. There is probably a && or & or a decltype that would permit perfect forwarding on these lines:

    auto[e0,e1,e2]=std::forward<S>(s);
    return std::make_tuple(e0, e1, e2);

but without a compiler to play with, I'll be conservative and make redundant copies.

---

Converting a tuple into a struct is easy:

template<class S, std::size_t...Is, class Tup>
S to_struct( std::index_sequence<Is...>, Tup&& tup ) {
  using std::get;
  return {get<Is>(std::forward<Tup>(tup))...};
}
template<class S, class Tup>
S to_struct( Tup&&tup ) {
  using T=std::remove_reference_t<Tup>;

  return to_struct(
    std::make_index_sequence<std::tuple_size<T>{}>{},
    std::forward<Tup>(tup)
  );
}

SFINAE support based off tuple_size might be good for to_struct .

The above code works with all tuple-likes, like std::pair , std::array , and anything you custom-code to support structured bindings ( tuple_size and get<I> ).

---

Amusingly,

std::array<int, 3> arr{1,2,3};
auto t = to_tuple<3>(arr);

works and returns a tuple with 3 elements, as to_tuple is based on structured bindings, which work with tuple-likes as input.

to_array is another possibility in this family.

Answer 2

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.

Answer 3

Is there any standartized way to cast one to another?

There is no way to "cast" the one to the other.

The easiest may be to use a std::tie to pack the tuple out into the struct ;

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

auto t = std::make_tuple(42, true, 5.1);
MyStruct s;

std::tie(s.i, s.b, s.d) = t;

Demo .

You can further wrap this up in higher level macros or "generator" ( make style) functions, eg;

std::tuple<int, bool, double> from_struct(MyStruct const& src)
{
  return std::make_tuple(src.i, src.b, src.d);
}

MyStruct to_struct(std::tuple<int, bool, double> const& src)
{
  MyStruct s;
  std::tie(s.i, s.b, s.d) = src;
  return s;
}

I know that trivial memory copying can do the trick, but it is alignment and implementation dependent?

You mention the "trivial memory copy" would work - only for copying the individual members. So basically, a memcpy of the entire structure to the tuple and vice-versa is not going to always behave as you expect (if ever); the memory layout of a tuple is not standardised. If it does work, it is highly dependent on the implementation.

Answer 4

Tuple to struct conversion is trivial, but backward I think is impossible at current C++ level in general.

#include <type_traits>
#include <utility>

#include <tuple>

namespace details
{

template< typename result_type, typename ...types, std::size_t ...indices >
result_type
make_struct(std::tuple< types... > t, std::index_sequence< indices... >) // &, &&, const && etc.
{
    return {std::get< indices >(t)...};
}

}

template< typename result_type, typename ...types >
result_type
make_struct(std::tuple< types... > t) // &, &&, const && etc.
{
    return details::make_struct< result_type, types... >(t, std::index_sequence_for< types... >{}); // if there is repeated types, then the change for using std::index_sequence_for is trivial
}

#include <cassert>
#include <cstdlib>

int main()
{
    using S = struct { int a; char b; double c; };
    auto s = make_struct< S >(std::make_tuple(1, '2', 3.0));
    assert(s.a == 1);
    assert(s.b == '2');
    assert(s.c == 3.0);
    return EXIT_SUCCESS;
}

Live example .