Menu
Consider this output:
int foo (int, char) {std::cout << "foo\n"; return 0;}
double bar (bool, double, long ) {std::cout << "bar\n"; return 3.5;}
bool baz (char, short, float) {std::cout << "baz\n"; return true;}
int main() {
const auto tuple = std::make_tuple(5, 'a', true, 3.5, 1000, 't', 2, 5.8);
multiFunction<2,3,3> (tuple, foo, bar, baz); // foo bar baz
}
So multiFunction<2,3,3> takes the first 2 elements of tuple and passes them to foo, the next 3 elements of tuple and passes them to bar, etc... I got this working (except when the functions have overloads, which is a separate problem). But the return values of each function called are lost. I want those return values stored somewhere, something like
std::tuple<int, double, bool> result = multiFunction<2,3,3> (tuple, foo, bar, baz);
But I don't know how to implement that. For those who want to help get this done, here is my (updated) working code so far, which stores the outputs into a stringstream only. Not easy to get all the values back, especially if the objects saved in the stream are complex classes.
#include <iostream>
#include <tuple>
#include <utility>
#include <sstream>
template <std::size_t N, typename Tuple>
struct TupleHead {
static auto get (const Tuple& tuple) { // The subtuple from the first N components of tuple.
return std::tuple_cat (TupleHead<N-1, Tuple>::get(tuple), std::make_tuple(std::get<N-1>(tuple)));
}
};
template <typename Tuple>
struct TupleHead<0, Tuple> {
static auto get (const Tuple&) { return std::tuple<>{}; }
};
template <std::size_t N, typename Tuple>
struct TupleTail {
static auto get (const Tuple& tuple) { // The subtuple from the last N components of tuple.
return std::tuple_cat (std::make_tuple(std::get<std::tuple_size<Tuple>::value - N>(tuple)), TupleTail<N-1, Tuple>::get(tuple));
}
};
template <typename Tuple>
struct TupleTail<0, Tuple> {
static auto get (const Tuple&) { return std::tuple<>{}; }
};
template <typename Tuple, typename F, std::size_t... Is>
auto functionOnTupleHelper (const Tuple& tuple, F f, const std::index_sequence<Is...>&) {
return f(std::get<Is>(tuple)...);
}
template <typename Tuple, typename F>
auto functionOnTuple (const Tuple& tuple, F f) {
return functionOnTupleHelper (tuple, f, std::make_index_sequence<std::tuple_size<Tuple>::value>{});
}
template <typename Tuple, typename... Functions> struct MultiFunction;
template <typename Tuple, typename F, typename... Fs>
struct MultiFunction<Tuple, F, Fs...> {
template <std::size_t I, std::size_t... Is>
static inline auto execute (const Tuple& tuple, std::ostringstream& oss, const std::index_sequence<I, Is...>&, F f, Fs... fs) {
const auto headTuple = TupleHead<I, Tuple>::get(tuple);
const auto tailTuple = TupleTail<std::tuple_size<Tuple>::value - I, Tuple>::get(tuple);
// functionOnTuple (headTuple, f); // Always works, though return type is lost.
oss << std::boolalpha << functionOnTuple (headTuple, f) << '\n'; // What about return types that are void???
return MultiFunction<std::remove_const_t<decltype(tailTuple)>, Fs...>::execute (tailTuple, oss, std::index_sequence<Is...>{}, fs...);
}
};
template <>
struct MultiFunction<std::tuple<>> {
static auto execute (const std::tuple<>&, std::ostringstream& oss, std::index_sequence<>) { // End of recursion.
std::cout << std::boolalpha << oss.str();
// Convert 'oss' into the desired tuple? But how?
return std::tuple<int, double, bool>(); // This line is just to make the test compile.
}
};
template <std::size_t... Is, typename Tuple, typename... Fs>
auto multiFunction (const Tuple& tuple, Fs... fs) {
std::ostringstream oss;
return MultiFunction<Tuple, Fs...>::execute (tuple, oss, std::index_sequence<Is...>{}, fs...);
}
// Testing
template <typename T> int foo (int, char) {std::cout << "foo<T>\n"; return 0;}
double bar (bool, double, long ) {std::cout << "bar\n"; return 3.5;}
template <int...> bool baz (char, short, float) {std::cout << "baz<int...>\n"; return true;}
int main() {
const auto tuple = std::make_tuple(5, 'a', true, 3.5, 1000, 't', 2, 5.8);
std::tuple<int, double, bool> result = multiFunction<2,3,3> (tuple, foo<bool>, bar, baz<2,5,1>); // foo<T> bar baz<int...>
}
700
tuples are immutable. You can create a tuple. But you cannot store elements to an already created tuple. To create (or convert) your list of elements to a tuple.
Tuple. Tuples are used to store multiple items in a single variable. Tuple is one of 4 built-in data types in Python used to store collections of data, the other 3 are List, Set, and Dictionary, all with different qualities and usage. A tuple is a collection which is ordered and unchangeable.
Example: Pass a Tuple as an Argument using *args Syntax Unpacking in Python uses *args syntax. As functions can take an arbitrary number of arguments, we use the unpacking operator * to unpack the single argument into multiple arguments. This is a special way of receiving parameters to a function as a tuple.
Here's an approach where the number of arguments is deduced greedily:
#include <tuple>
namespace detail {
using namespace std;
template <size_t, size_t... Is, typename Arg>
constexpr auto call(index_sequence<Is...>, Arg&&) {return tuple<>{};}
template <size_t offset, size_t... Is, typename ArgT, typename... Fs>
constexpr auto call(index_sequence<Is...>, ArgT&&, Fs&&...);
template <size_t offset, size_t... Is,
typename ArgT, typename F, typename... Fs,
typename=decltype(declval<F>()(get<offset+Is>(declval<ArgT>())...))>
constexpr auto call(index_sequence<Is...>, ArgT&& argt, F&& f, Fs&&... fs) {
return tuple_cat(make_tuple(f(get<offset+I>(forward<ArgT>(argt))...)),
call<offset+sizeof...(Is)>(index_sequence<>{},
forward<ArgT>(argt),
forward<Fs>(fs)...));}
template <size_t offset, size_t... Is, typename ArgT, typename... Fs>
constexpr auto call(index_sequence<Is...>, ArgT&& argt, Fs&&... fs) {
return call<offset>(index_sequence<Is..., sizeof...(Is)>{},
forward<ArgT>(argt), forward<Fs>(fs)...);}
}
template <typename ArgT, typename... Fs>
constexpr auto multifunction(ArgT&& argt, Fs&&... fs) {
return detail::call<0>(std::index_sequence<>{},
std::forward<ArgT>(argt), std::forward<Fs>(fs)...);}
Demo. However, the above has quadratic time complexity in the number of return values, because tuple_cat is called recursively. Instead, we can use a slightly modified version of call to obtain the indices for each call - the actual tuple is then obtained directly:
#include <tuple>
namespace detail {
using namespace std;
template <size_t, size_t... Is, typename Arg>
constexpr auto indices(index_sequence<Is...>, Arg&&) {return tuple<>{};}
template <size_t offset, size_t... Is, typename ArgT, typename... Fs>
constexpr auto indices(index_sequence<Is...>, ArgT&&, Fs&&...);
template <size_t offset, size_t... Is, typename ArgT, typename F, class... Fs,
typename=decltype(declval<F>()(get<offset+Is>(declval<ArgT>())...))>
constexpr auto indices(index_sequence<Is...>, ArgT&& argt, F&& f, Fs&&... fs){
return tuple_cat(make_tuple(index_sequence<offset+Is...>{}),
indices<offset+sizeof...(Is)>(index_sequence<>{},
forward<ArgT>(argt),
forward<Fs>(fs)...));}
template <size_t offset, size_t... Is, typename ArgT, typename... Fs>
constexpr auto indices(index_sequence<Is...>, ArgT&& argt, Fs&&... fs) {
return indices<offset>(index_sequence<Is..., sizeof...(Is)>{},
forward<ArgT>(argt), forward<Fs>(fs)...);}
template <typename Arg, typename F, size_t... Is>
constexpr auto apply(Arg&& a, F&& f, index_sequence<Is...>) {
return f(get<Is>(a)...);}
template <typename ITuple, typename Args, size_t... Is, typename... Fs>
constexpr auto apply_all(Args&& args, index_sequence<Is...>, Fs&&... fs) {
return make_tuple(apply(forward<Args>(args), forward<Fs>(fs),
tuple_element_t<Is, ITuple>{})...);
}
}
template <typename ArgT, typename... Fs>
constexpr auto multifunction(ArgT&& argt, Fs&&... fs) {
return detail::apply_all<decltype(detail::indices<0>(std::index_sequence<>{},
std::forward<ArgT>(argt),
std::forward<Fs>(fs)...))>
(std::forward<ArgT>(argt), std::index_sequence_for<Fs...>{},
std::forward<Fs>(fs)...);}
Demo 2.
170
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
