Uniform initialization by tuple - c++

Question

Today, I arrived at a situation, where I have a vector of tuples, where the tuples might contain several entries. Now I wanted to convert my vector of tuples to a vector of objects, such that the entries of the tuples will exactly match the uniform initialization of my object.
The following code does the job for me, but it is a bit clumsy. I'm asking myself if it might be possible to derive a generic solution that can construct the Objects if the tuples matches exactly the uniform initialization order of the objects.
This might be a very desirable functionality, when the number of parameters to pass grows.
#include <vector>
#include <tuple>
#include <string>
#include <algorithm>
struct Object
{
std::string s;
int i;
double d;
};
int main() {
std::vector<std::tuple<std::string, int, double>> values = { {"A",0,0.},{"B",1,1.} };
std::vector<Object> objs;
std::transform(values.begin(), values.end(), std::back_inserter(objs), [](auto v)->Object
{
// This might get tedious to type, if the tuple grows
return { std::get<0>(v), std::get<1>(v), std::get<2>(v) };
// This is my desired behavior, but I don't know what magic_wrapper might be
// return magic_wrapper(v);
});
return EXIT_SUCCESS;
}

Answer 1

Provide Object an std::tuple constructor. You can use std::tie to assign your members:
template<typename ...Args>
Object(std::tuple<Args...> t) {
std::tie(s, i, d) = t;
}
Now it gets automatically constructed:
std::transform(values.begin(), values.end(), std::back_inserter(objs),
[](auto v) -> Object {
return { v };
});
To reduce the amount of copying you might want to replace auto v with const auto& v and make the constructor accept a const std::tuple<Args...>& t.
Also, it's good practise to access the source container via const iterator:
std::transform(values.cbegin(), values.cend(), std::back_inserter(objs), ...

Answer 2

Here is a non-intrusive version (i.e. not touching Object) that extracts the number of specified data members. Note that this relies on aggregate initialization.
template <class T, class Src, std::size_t... Is>
constexpr auto createAggregateImpl(const Src& src, std::index_sequence<Is...>) {
return T{std::get<Is>(src)...};
}
template <class T, std::size_t n, class Src>
constexpr auto createAggregate(const Src& src) {
return createAggregateImpl<T>(src, std::make_index_sequence<n>{});
}
You invoke it like this:
std::transform(values.cbegin(), values.cend(), std::back_inserter(objs),
[](const auto& v)->Object { return createAggregate<Object, 3>(v); });
Or, without the wrapping lambda:
std::transform(values.cbegin(), values.cend(), std::back_inserter(objs),
createAggregate<Object, 3, decltype(values)::value_type>);
As #Deduplicator pointed out, the above helper templates implement parts of std::apply, which can be used instead.
template <class T>
auto aggregateInit()
{
return [](auto&&... args) { return Object{std::forward<decltype(args)>(args)...}; };
}
std::transform(values.cbegin(), values.cend(), std::back_inserter(objs),
[](const auto& v)->Object { return std::apply(aggregateInit<Object>(), v); });

Answer 3

Since C++17, you might use std::make_from_tuple:
std::transform(values.begin(),
values.end(),
std::back_inserter(objs),
[](const auto& t)
{
return std::make_from_tuple<Object>(t);
});
Note: requires appropriated constructor for Object.