函数模板如何推断出 initializer_list 的嵌套次数？

Question

我有一个函数模板，它采用任意嵌套列表和 returns 数组：

#include <array>
#include <initializer_list>

template<size_t N, typename List>
std::array<size_t,N> some_function (const List& list)
{
    // N is the number of times the list is nested.
    std::array<size_t,N> arr;
    return arr;
}

当我将此函数用于某些嵌套 std::initializer_list 时，如下所示：

int main () {
    using List = std::initializer_list<std::initializer_list<double>>;
    List list = {{1.,2.,3.},{4.,5.,6.}};

    std::array<size_t,2> arr;
    arr = some_function (list);
    return 0;
}

我收到无法推导类型 N 的错误

couldn't deduce template parameter ‘N’

问题

• 如何改进我的函数模板来推断列表的嵌套次数？
• 对于这种情况，是否有比 std::initializer_list 更好的选择？

Answer 1

你可以写两个重载constexprfunction templates to calculate the nested times, with the help of std::enable_if and SFINAE.

// types have not member type value_type
template <typename T, typename = void>
struct has_value_type: std::false_type {};
// types have member type value_type
template <typename T>
struct has_value_type<T, std::void_t<typename T::value_type>> : std::true_type {};

// return nested times as 0 for types without member type value_type
template<typename T>
constexpr std::enable_if_t<!has_value_type<T>::value, size_t> get_nested_times() {
    return 0;
}
// return nested times as 1 plus times got on the nested type recursively
template<typename T>
constexpr std::enable_if_t<has_value_type<T>::value, size_t> get_nested_times() {
    return 1 + get_nested_times<typename T::value_type>();
}

那么你可以在编译时得到嵌套时间

template<typename List>
auto some_function (const List& list)
{
    // N is the number of times the list is nested.
    constexpr auto N = get_nested_times<List>();
    std::array<size_t, N> arr;
    return arr;
}

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