c++ - 安全类型-std::tuple的all_of/any_of/none_of

因此，我开始实现一些模仿STL算法行为的算法，但是使用了一个异构容器，也就是std::tuple。

template<typename UnaryPredicate, typename Tuple>
bool all_of(UnaryPredicate&& p, Tuple&& t) noexcept
{
    return std::apply([&p](auto&& ...xs){ return (p(std::forward<decltype(xs)>(xs)) && ...); }, std::forward<Tuple>(t));
}

template<typename UnaryPredicate, typename Tuple>
bool any_of(UnaryPredicate&& p, Tuple&& t) noexcept
{
    return std::apply([&p](auto&& ...xs){ return (p(std::forward<decltype(xs)>(xs)) || ...); }, std::forward<Tuple>(t));
}

template<typename UnaryPredicate, typename Tuple>
bool none_of(UnaryPredicate&& p, Tuple&& t) noexcept
{
    return std::apply([&p](auto&& ...xs){ return !(p(std::forward<decltype(xs)>(xs)) || ...); }, std::forward<Tuple>(t));
}

如果您使用返回 bool(boolean) 值的UnaryPredicate，则所有这些方法都可以正常工作。但是，如果没有呢？用元组的每个元素调用时，如何确保UnaryPredicate返回 bool(boolean) 值？另外，我该如何检查UnaryPredicate实际上没有引发任何异常。

我知道有类似“is_nothrow_invocable”和“invoke_result”的类型特征，但是所有这些都需要元组包含的元素类型。我真的必须使用“algorithm_impl”模式吗？

namespace impl
{
    template<typename UnaryPredicate, typename Tuple, auto ...Is>
    bool all_of_impl(UnaryPredicate&& p, Tuple&& t, std::index_sequence<Is...>) noexcept
    {
        return std::apply([&p](auto&& ...xs){ return (p(std::forward<decltype(xs)>(xs)) && ...); }, std::forward<Tuple>(t));
    }
}

template<typename UnaryPredicate, typename Tuple>
bool all_of(UnaryPredicate&& p, Tuple&& t) noexcept
{
    return impl::all_of_impl(std::forward<UnaryPredicate>(p), std::forward<Tuple>(t), std::make_index_sequence<std::tuple_size_v<std::decay_t<Tuple>>>{});
}

现在我可以做这样的事情:

std::enable_if_t<std::conjunction_v<std::is_same<std::invoke_result_t<std::decay_t<UnaryPredicate>, std::tuple_element_t<Is, std::decay_t<Tuple>>>, bool>...>, bool>

但这真的是要走的路吗？

编辑:

好吧，一如既往，我把事情复杂化了。我想我找到了可以接受的解决方案:

template<typename UnaryPredicate, typename Tuple>
struct helper;

template<typename UnaryPredicate, typename Tuple>
struct helper2;

template<typename UnaryPredicate, typename ...Ts>
struct helper<UnaryPredicate, std::tuple<Ts...>>
    : std::bool_constant<std::conjunction_v<std::is_same<bool, std::invoke_result_t<std::decay_t<UnaryPredicate>, std::decay_t<Ts>>>...>>
{};

template<typename UnaryPredicate, typename ...Ts>
struct helper2<UnaryPredicate, std::tuple<Ts...>>
    : std::bool_constant<std::conjunction_v<std::is_nothrow_invocable<std::decay_t<UnaryPredicate>, std::decay_t<Ts>>...>>
{};

template<typename UnaryPredicate, typename Tuple>
inline constexpr auto helper_v{ helper<UnaryPredicate, Tuple>::value };

template<typename UnaryPredicate, typename Tuple>
inline constexpr auto helper2_v{ helper2<UnaryPredicate, Tuple>::value };


template<typename UnaryPredicate, typename Tuple>
std::enable_if_t<helper_v<UnaryPredicate, Tuple>, bool> all_of(UnaryPredicate&& p, Tuple&& t) noexcept(helper2_v<UnaryPredicate, Tuple>)
{
    return std::apply([&p](auto&& ...xs){ return (p(std::forward<decltype(xs)>(xs)) && ...); }, std::forward<Tuple>(t));
}

template<typename UnaryPredicate, typename Tuple>
std::enable_if_t<helper_v<UnaryPredicate, Tuple>, bool> any_of(UnaryPredicate&& p, Tuple&& t) noexcept(helper2_v<UnaryPredicate, Tuple>)
{
    return std::apply([&p](auto&& ...xs){ return (p(std::forward<decltype(xs)>(xs)) || ...); }, std::forward<Tuple>(t));
}

template<typename UnaryPredicate, typename Tuple>
std::enable_if_t<helper_v<UnaryPredicate, Tuple>, bool> none_of(UnaryPredicate&& p, Tuple&& t) noexcept(helper2_v<UnaryPredicate, Tuple>)
{
    return std::apply([&p](auto&& ...xs){ return !(p(std::forward<decltype(xs)>(xs)) || ...); }, std::forward<Tuple>(t));
}

最佳答案

#define RETURNS(...) \
  noexcept(noexcept(__VA_ARGS__)) \
  -> decltype(__VA_ARGS__) \
  { return __VA_ARGS__; }

template<class UnaryPredicate, class Tuple>
auto all_of(UnaryPredicate&& p, Tuple&& t)
RETURNS(
  std::apply(
    [&p](auto&& ...xs){
      return (p(std::forward<decltype(xs)>(xs)) && ...);
    },
    std::forward<Tuple>(t)
  )
)

等等