Transpose a container of structs - c++

Question

Let we have a struct Record{uint8_t x, y;};, a container Container<Record> of structs and a struct Transposed{Container<uint8_t> x,y};. Container c is a template which first arg is a type of value, all of the rest args have default values. For example it can be a std::vector (the rest args are types) or std::span (the rest arg is a value). The template should work with all of them. Also we may like to pass tne rest of template arguments to underlying template.
How we can get the Transposed from the container using templates?
I have tried variadic templates,
#include <iostream>
#include <vector>
template <typename value_type=uint8_t> struct Record{
value_type x, y;
};
template<typename value_type, typename value_type2=uint8_t> class ContainerA: public std::vector<value_type>{
value_type2 b=1u;
};
template<typename value_type, uint8_t int_value=1u> class ContainerB: public std::vector<value_type>{};
template<typename value_type, template <typename ...> typename container_type> class Transposed{
container_type<value_type> x, y;
public:
Transposed(container_type<Record<value_type>> & recs){
x.reserve(recs.size());
y.reserve(recs.size());
x.resize(recs.size());
y.resize(recs.size());
size_t i=0;
for(auto &rec :recs){
x[i] = rec.x;
y[i] = rec.y;
++i;
}
}
};
int main(){
std::vector<Record<uint8_t>> recsV{
{1, 2},
{3, 4}
};
Transposed trV{recsV};
std::cout<<"vec"<<std::endl;
ContainerA<Record<uint8_t>> recsA{
{1, 2},
{3, 4}
};
Transposed trA{recsA};
std::cout<<"A"<<std::endl;
/*ContainerB<Record<uint8_t>> recsB{
{1, 2},
{3, 4}
};
Transposed trB{recsB};
std::cout<<"B"<<std::endl;*/
return 0;
}
but it seems they cannot match both types and values. Usage of more than 1 variadic template argument is not allowed. Is it a flaw in C++ language or a deliberate design choice and do we need something like any_template_arg keyword or should just specifying 2 variadic arguments of different types be allowed to allow this use case?

Answer 1

As far I know, there isn't a way to match types and values (and template-template) template arguments together. And I've searched it for a long time.
So I don't see a way to make what do you want in a simple and elegant way.
Trying to respond to your question
How we can get the Transposed from the container using templates?
the best I can imagine is declare (isn't necessary define them taking in count are used only inside a decltype()) a couple of trivial functions as follows
template <typename VT,
template <typename...> typename CT>
CT<VT> extract_func (CT<Record<VT>>);
template <typename VT,
template <typename, auto...> typename CT>
CT<VT> extract_func (CT<Record<VT>>);
The first one is to remove the Record part when the CT container accepts a variadic list of types (std::vector and ContainerA cases); the second one is for CT containers accepting (after the type parameter) one or more values (ContainerB case).
Obviously this doen't cover all possible cases, but it's trivial declare other extract_func() functions to cover other cases.
Now you can declare Tranposed as follows
template <typename T,
typename CT = decltype(extract_func(std::declval<T>()))>
class Transposed
Observe that, now, Transposed accept a generic type T but is SFINAE enabled only when the T type matches (as argument) an extract_func() declaration.
In the Transposed body you can use CT to declare x and y and T for the argument of the constructor.
The following is a full compiling example
#include <iostream>
#include <vector>
template <typename value_type=std::uint8_t>
struct Record
{ value_type x, y; };
template <typename value_type, typename value_type2=std::uint8_t>
class ContainerA : public std::vector<value_type>
{ value_type2 b=1u; };
template <typename value_type, std::uint8_t int_value=1u>
class ContainerB : public std::vector<value_type>
{ };
template <typename VT,
template <typename...> typename CT>
CT<VT> extract_func (CT<Record<VT>>);
template <typename VT,
template <typename, auto...> typename CT>
CT<VT> extract_func (CT<Record<VT>>);
template <typename T,
typename CT = decltype(extract_func(std::declval<T>()))>
class Transposed
{
private:
CT x, y;
public:
Transposed (T & recs)
{
x.reserve(recs.size());
y.reserve(recs.size());
x.resize(recs.size());
y.resize(recs.size());
std::size_t i=0u;
for(auto &rec :recs){
x[i] = rec.x;
y[i] = rec.y;
++i;
}
}
};
int main ()
{
std::vector<Record<std::uint8_t>> recsV { {1, 2}, {3, 4} };
Transposed trV{recsV};
std::cout<<"vec"<<std::endl;
ContainerA<Record<std::uint8_t>> recsA { };
Transposed trA{recsA};
std::cout<<"A"<<std::endl;
ContainerB<Record<std::uint8_t>> recsB { };
Transposed trB{recsB};
std::cout<<"B"<<std::endl;
}

Answer 2

One working example for you:
template<typename value_type=uint8_t>
struct Record{
value_type x, y;
};
template<class T>
std::vector<T> rebind_container(std::vector<Record<T>> const&);
// Span transposes into vector.
template<class T>
std::vector<T> rebind_container(std::span<Record<T>> const&);
template<class T>
std::list<T> rebind_container(std::list<Record<T>> const&);
inline void reserve(...) {}
template<class... Args>
inline void reserve(std::vector<Args...>* v, size_t n) {
v->reserve(n);
}
template<class container_type>
struct Transposed {
using container_type2 = decltype(rebind_container(std::declval<container_type>()));
container_type2 x, y;
Transposed(container_type const& recs) {
auto const n = recs.size();
reserve(&x, n);
reserve(&y, n);
for(auto& rec : recs) {
x.push_back(rec.x);
y.push_back(rec.y);
}
}
};
int main(){
std::vector<Record<uint8_t>> recsV{
{1, 2},
{3, 4}
};
Transposed<decltype(recsV)> trV{recsV};
std::cout << trV.x.size() << std::endl;
std::list<Record<uint8_t>> recsV2{
{1, 2},
{3, 4}
};
Transposed<decltype(recsV2)> trV2{recsV2};
std::cout << trV2.x.size() << std::endl;
}