问题
currently I'm working on a dynamic container structure, which represents one pod value or has vector of pointers with same container type. The container has an interface optional<T> expect_value<T>() 。 For pod types the implemention is simple. For the non pod value, I would call expect_value<tuple<args...>>(), the args would be tuple as well. But when implement this function, I come across a trouble: how to redirect a.expect_value<tuple<args...>>() to a.expect_value_tuple<args...>>(). For example, the call to a.expect_value<tuple<int,int>() would return the result of a.expect_value_tuple<int, int>(). Because the argument is empty, I cant use the type deduce of unpacked arguments. Then the whole project just cant progress any more. Any ideas? Below is the minimal example for my intention.
#include <tuple>
#include <vector>
#include <optional>
#include <functional>
using namespace std;
template<typename T>
struct is_tuple_impl : std::false_type {};
template<typename... Ts>
struct is_tuple_impl<std::tuple<Ts...>> : std::true_type {};
template<typename T>
struct is_tuple : is_tuple_impl<std::decay_t<T>> {};
class my_container;
template<typename... args, size_t... arg_idx>
optional<tuple<args>...> get_tuple_value_from_vector(const vector<my_container*>& v_list, std::index_sequence<arg_idx...>)
{
auto temp_result = make_tuple((*v_list[arg_idx]).expect_value<arg>()...);
if(!(get<arg_idx>(temp_result) &&...))
{
return nullopt;
}
return make_tuple(get<arg_idx>(temp_result).value()...);
}
class my_container
{
public:
int value_type; // 1 for v_int 2 for v_list 0 empty
union
{
int v_int;
};
vector<my_container*> v_list;
template<typename T>
optional<T> expect_simple_value();
template<typename... args>
optional<tuple<args...>> expect_tuple_value();
template<typename T>
optional<T> expect_value();
};
template <typename T>
optional<T> my_container::expect_simple_value()
{
return nullopt;
}
template <>
optional<int> my_container::expect_simple_value()
{
if(value_type == 1)
{
return v_int;
}
return nullopt;
}
template<typename... args>
optional<tuple<args...>> my_container::expect_tuple_value()
{
if(v_list.size() == 0)
{
return nullopt;
}
for(const auto i: v_list)
{
if(!i)
{
return nullopt;
}
}
auto the_tuple_size = sizeof...(args);
if(v_list.size() != the_tuple_size)
{
return nullopt;
}
return get_tuple_value_from_vector<args...>(v_list, index_sequence_for<args...>{});
}
template <typename T>
optional<T> my_container::expect_value()
{
if(is_tuple<T>::value)
{
return expect_tuple_value<T>();
}
else
{
return expect_simple_value<T>();
}
}
int main()
{
my_container test_value;
test_value.value_type = 1;
test_value.v_int = 1;
auto result = test_value.expect_value<tuple<int, int>>();
if(result)
{
return 0;
}
else
{
return 1;
}
}
the heart of the problem is the line return expect_tuple_value<T>(); When logic goes there, the T should be tuple<args...>, but what I want is return return expect_tuple_value<args...>().
回答1:
What about using template argument deduction and overload resolution through partial ordering of function template:
class my_container
{
public:
template<class T> optional<T> expect_value_simple();
template<class...Args> optional<tuple<Args...>> expect_value_tuple();
private:
template<class T> struct deduce_type{};
template<typename T>
auto expect_value_dispatching(deduce_type<T>){
return expect_value_simple<T>();
}
template<typename...Args>
auto expect_value_dispatching(deduce_type<tuple<Args...>>){
return expect_value_tuple<Args...>();
}
public:
template<typename T>
auto expect_value(){
return expect_value_dispatching(deduce_type<T>{});
}
};
(Demo)
回答2:
The if before the line in question should be a constexpr if.
Unpacking of types is annoying to do without using a class helper. I can do it with some fancy c++14 lambda action tho.
template<class T>
struct tag_t{using type=T;};
template<class Tag>
using type=typename Tag::type;
template<class Tuple>
struct unpack_tuple;
template<class...Ts>
struct unpack_tuple<std::tuple<Ts...>> {
template<class F>
decltype(auto) operator()(F&& f)const {
return std::forward<F>(f)( tag_t<Ts>{}... );
}
};
#define TYPE_FROM(...) \
type< std::decay_t<decltype(__VA_ARGS__)> >
then we get
if constexpr(is_tuple<T>::value)
{
return unpack_tuple<T>{}([&](auto...tags){
return expect_tuple_value<TYPE_FROM(tags)...>();
});
}
else
{
return expect_simple_value<T>();
}
and done.
回答3:
The core issue here is that you need to do argument deduction at least once to go back from a type to its variadic parameters. To do that, you must pass some instance of such a variadically-templated type to a function - but it does not have to be the original one.
Yakk's solution does this via a variadic lambda that is passed instances of tag-types (one per tuple type). The advantage here is that you can use a lambda instead of an explicit intermediary function every time.
Oliv's solution uses a monostate type which we can instantiate and pass to a function for type deduction. It's much cleaner but requires such an intermediary function for every use case.
Here is a (more or less theoretical) version combining both, using templated variadic lambdas (C++20, and they apparently don't even have clang support as of now):
template<class... Args>
struct MonostateTuple
{};
template<class... Args>
auto tupleToMonostate(std::tuple<Args...>)
{
return MonostateTuple<Args...>{};
}
template<class T, class F>
auto unpack_tuple(F&& f)
{
using MT = decltype(tupleToMonostate(std::declval<T>()));
return std::forward<F>(f)(MT{});
}
/// User code
template<class Tuple>
auto foo()
{
return unpack_tuple<Tuple>([&] <typename... Args> (MonostateTuple<Args...>) {
return expect_tuple_value<Args...>();
});
}
It's a bit more ugly in the lambda signature (not to mention the lack of compiler support again) but theoretically combines both advantages.
来源：https://stackoverflow.com/questions/53744497/how-to-forward-the-types-of-tuple-to-specialize-other-template