Question

如果我有一个元组类型不同的元组，如

std::tuple<T0, T1, T2, ...>

如何获取元素类型的索引？

template<class T, class Tuple>
struct Index
{
    enum {value = ?;}
};

感谢。

Answer 1

template <class T, class Tuple>
struct Index;

template <class T, class... Types>
struct Index<T, std::tuple<T, Types...>> {
    static const std::size_t value = 0;
};

template <class T, class U, class... Types>
struct Index<T, std::tuple<U, Types...>> {
    static const std::size_t value = 1 + Index<T, std::tuple<Types...>>::value;
};

See it live at Coliru

此实现返回给定类型第一次出现的索引。请求一个不在元组中的类型的索引会导致编译错误（并且会产生相当丑陋的错误）。

Answer 2

template< size_t I, typename T, typename Tuple_t>
constexpr size_t index_in_tuple_fn(){
    static_assert(I < std::tuple_size<Tuple_t>::value,"The element is not in the tuple");

    typedef typename std::tuple_element<I,Tuple_t>::type el;
    if constexpr(std::is_same<T,el>::value ){
        return I;
    }else{
        return index_in_tuple_fn<I+1,T,Tuple_t>();
    }
}

template<typename T, typename Tuple_t>
struct index_in_tuple{
    static constexpr size_t value = index_in_tuple_fn<0,T,Tuple_t>();
};

上面的示例避免了生成大量的子元组，这在您为大型元组调用index_in_tuple时使编译失败（内存不足）

Answer 3

另一个使用折叠表达式。找不到时，还会将值设置为-1。

template <class X, class Tuple>
class Idx;

template <class X, class... T>
class Idx<X, std::tuple<T...>> {
    template <std::size_t... idx>
    static constexpr ssize_t find_idx(std::index_sequence<idx...>) {
        return -1 + ((std::is_same<X, T>::value ? idx + 1 : 0) + ...);
    }
public:
    static constexpr ssize_t value = find_idx(std::index_sequence_for<T...>{});
};

实时：https://onlinegdb.com/SJE8kOYdv

获取元组元素类型的索引？

3 个答案: