我想以下列方式从另一个可变参数函数调用一个可变参数函数:
template <typename ...Ts>
void f(Ts const & ...) { /* ... */ }
template <typename ...Args>
void g(Args const & ...args)
{
// shall call f(arg0.a(), arg0.b(), arg1.a(), arg1.b(), ...)
}
我是通过以下方式完成的:
struct sentinel_t { };
template <typename Arg, typename ...Args>
void g_impl(Arg const & arg, Args const & ...args)
{
g_impl(args..., arg.a(), arg.b());
}
template <typename ...Ts>
void g_impl(sentinel_t , Ts const & ...ts)
{
f(ts...);
}
template <typename ...Args>
void g(Args const & ...args)
{
g_impl(args..., sentinel_t{});
}
有没有其他/更好的方法来实现这种模式?
答案 0 :(得分:5)
template <class... Args> void g(Args const &... args) {
impl::apply([](auto const &... p) { f(p...); },
std::tuple_cat(std::forward_as_tuple(args.a(), args.b())...));
}
直到std::apply
为standardized,您可以使用c ++ 14(取自referenced paper)使自己非常简单:
namespace impl {
template <typename F, typename Tuple, size_t... I>
decltype(auto) apply_impl(F &&f, Tuple &&t, std::index_sequence<I...>) {
return std::forward<F>(f)(std::get<I>(std::forward<Tuple>(t))...);
}
template <typename F, typename Tuple> decltype(auto) apply(F &&f, Tuple &&t) {
using Indices =
std::make_index_sequence<std::tuple_size<std::decay_t<Tuple>>::value>;
return apply_impl(std::forward<F>(f), std::forward<Tuple>(t), Indices{});
}
}
答案 1 :(得分:4)
你可以做
namespace detail
{
// dispatcher
template <typename T>
decltype(auto) call_a_b(std::integral_constant<std::size_t, 0u>, const T& arg) {return arg.a();}
// dispatcher
template <typename T>
decltype(auto) call_a_b(std::integral_constant<std::size_t, 1u>, const T& arg) {return arg.b();}
template <std::size_t... Is, typename Tuple>
void call_f_a_b(std::index_sequence<Is...>, const Tuple& tuple)
{
f(call_a_b(std::integral_constant<std::size_t, Is % 2>{}, std::get<Is / 2>(tuple))...);
}
}
template <typename...Ts>
void g(const Ts&... args)
{
return detail::call_f_a_b(std::make_index_sequence<2 * sizeof...(Ts)>{}, std::tie(args...));
// call f(arg0.a(), arg0.b(), arg1.a(), arg1.b(), ...)
}