构造函数采用 std::array 时无法推断 class 中的类型

Question

I am trying to create a templated class containing enum-string pairs and unable to make type deduction work. With a following code i have two problems:

namespace {
template<typename T, size_t S>
using EnumStringArray = std::array<std::pair<T, const char*>, S>;
}

template<typename T, size_t S>
class EnumToString {
public:
    constexpr EnumToString(const EnumStringArray<T, S>& array) :
            _array(array)
    {}
private:
    EnumStringArray<T, S> _array;
};

template<typename T, size_t S>
EnumToString(const EnumStringArray<T, S>&) -> EnumToString<T, S>;

enum MyEnum {
    One, 
    Two
};

constexpr EnumToString enumStrings = {{{    //<---- does not compile without explicit types
        {One, "One"},
        {Two, "Two"}
}}};

1. Why compiler can't deduce parameters of EnumToString by himself within constructor?
2. Why user deduction guide does not help?

Answer 1

{..} has not type and cannot be deduced (except as std::initializer_list<T> or T(&)[N] ).

so regular constructor, or deduction guide doesn't help with CTAD here.

Answer 2

The reason, as has mentioned by @Jarod42, is that in C++ brace initializers do not have any types and cannot be deduced by the compiler in the context of templates. The reason of this has been summarized here .

To fix it, you can use an auto variable to deduce the type and then pass it to your template method.

auto x = {{One, "One"}, {Two, "Two"}};
// Now x has a type, you can pass it to a template function.