How to make_from_tuple on the heap? - c++

Question

So in C++ there is now make_from_tuple as:
T obj = std::make_from_tuple<T>( { Args... args } ); // args represents a tuple
but how would one do:
T* obj = std::make_new_from_tuple<T*>( { Args... args } );
There is make_shared and make_unique but neither of those takes a tuple (and I'm not sure how one would extract the arguments from the tuple if that is the direction to go, as you can always make_unique then release if you want the raw pointer).
Very simple example 1:
struct A
{
int i_; double d_; std::string s_;
A( int i, double d, const std::string& s ) : i_(i), d_(d), s_(s) {}
};
auto aTuple = std::make_tuple( 1, 1.5, std::string("Hello") );
For a more complex example, if the tuple contains a unique_ptr you want to forward, I will want that to work too.

Answer 1

You can basically copy the implementation from cppreference's Possible implementation, chuck in a make_unique and it just works:
#include <string>
#include <memory>
namespace detail {
template<class T, class Tuple, std::size_t... I>
constexpr std::unique_ptr<T> make_new_from_tuple_impl(Tuple&& t, std::index_sequence<I...>)
{
static_assert(std::is_constructible_v<T,
decltype(std::get<I>(std::declval<Tuple>()))...>);
#if __cpp_lib_reference_from_temporary >= 202202L
if constexpr (std::tuple_size_v<std::remove_reference_t<Tuple>> == 1) {
using tuple_first_t = decltype(std::get<0>(std::declval<Tuple>()));
static_assert(!std::reference_constructs_from_temporary_v<T, tuple_first_t>);
}
#endif
return std::make_unique<T>(std::get<I>(std::forward<Tuple>(t))...);
}
} // namespace detail
template<class T, class Tuple>
constexpr std::unique_ptr<T> make_new_from_tuple(Tuple&& t)
{
return detail::make_new_from_tuple_impl<T>(std::forward<Tuple>(t),
std::make_index_sequence<std::tuple_size_v<std::remove_reference_t<Tuple>>>{});
}
struct A
{
int i_; double d_; std::string s_;
A( int i, double d, const std::string& s ) : i_(i), d_(d), s_(s) {}
};
int main()
{
auto aTuple = std::make_tuple( 1, 1.5, std::string("Hello") );
auto a = make_new_from_tuple<A>(std::move(aTuple));
}
Or with raw pointers:
#include <string>
#include <memory>
namespace detail {
template<class T, class Tuple, std::size_t... I>
constexpr T* make_new_from_tuple_impl(Tuple&& t, std::index_sequence<I...>)
{
static_assert(std::is_constructible_v<T,
decltype(std::get<I>(std::declval<Tuple>()))...>);
#if __cpp_lib_reference_from_temporary >= 202202L
if constexpr (std::tuple_size_v<std::remove_reference_t<Tuple>> == 1) {
using tuple_first_t = decltype(std::get<0>(std::declval<Tuple>()));
static_assert(!std::reference_constructs_from_temporary_v<T, tuple_first_t>);
}
#endif
return new T(std::get<I>(std::forward<Tuple>(t))...);
}
} // namespace detail
template<class T, class Tuple>
constexpr T* make_new_from_tuple(Tuple&& t)
{
return detail::make_new_from_tuple_impl<T>(std::forward<Tuple>(t),
std::make_index_sequence<std::tuple_size_v<std::remove_reference_t<Tuple>>>{});
}
struct A
{
int i_; double d_; std::string s_;
A( int i, double d, const std::string& s ) : i_(i), d_(d), s_(s) {}
};
int main()
{
auto aTuple = std::make_tuple( 1, 1.5, std::string("Hello") );
auto a = make_new_from_tuple<A>(std::move(aTuple));
}

Answer 2

You can use std::apply. It unpacks arguments from a tuple and passes them to a callable. You just need to wrap the constructor.
#include <tuple>
#include <string>
struct A
{
int i_; double d_; std::string s_;
A( int i, double d, const std::string& s ) : i_(i), d_(d), s_(s) {}
};
int main() {
auto aTuple = std::make_tuple( 1, 1.5, std::string("Hello") );
A* a = std::apply( [](int i,double d,const std::string& s) { return new A(i,d,s);},aTuple);
}
Perfect forwarding omitted for the sake of brevity.

Answer 3

Simply call new with it:
T* obj = new auto(std::make_from_tuple<T>( { Args... args } ));
// unique ptr
std::unique_ptr obj{new auto(std::make_from_tuple<T>({ Args... args }))};
// shared ptr
std::shared_ptr obj{new auto(std::make_from_tuple<T>({ Args... args }))};
// doesn't work for make_shared unfortunately
auto obj = std::apply([](auto&&... args) {
return std::make_shared<T>(std::forward<decltype(args)>(args)...);
}, { Args... args });
Because make_from_tuple returns a prvalue, this will construct the object directly in the heap with no copies or move.

Answer 4

Explicit use of new and delete is considered a bad practice since C++11/C++14.
So use of std::make_unique (C++14) and std::make_shared (C++11) is recommended.
template<typename T, typename...Args>
std::shared_ptr<T> make_shared_from_tuple(const std::tuple<Args...>& t)
{
return std::apply([](auto...args){ return std::make_shared<T>(args...); }, t);
}
template<typename T, typename...Args>
std::unique_ptr<T> make_unique_from_tuple(const std::tuple<Args...>& t)
{
return std::apply([](auto...args){ return std::make_unique<T>(args...); }, t);
}
https://godbolt.org/z/6zjTqK8f1
Probably it would be nice to provide also some overloads to handle move semantics.

Answer 5

The simplest I can write is:
template<typename obj, typename tup>
auto unique_from_tuple(tup&& t) {
return std::apply(
[] (auto&& ... args) {
return std::make_unique<obj>
(std::forward<decltype(args)>(args)...);
},
std::forward<tup>(t)
);
};
// Create a unique pointer of type obj:
auto x =unique_from_tuple<obj>(y,z,w);