Array size template deduction

Question

I have a class that stores an std::array.

The size of the array is evaluated in compile time, this is because the application runs on an embedded device so no dynamic allocations :(. The code looks something like this:

template<uint8_t size>
class A
{
    //some constructor
    A(...);
    std::array<int, size> what;
}
//Wanted use cases
A instance1({1,2,3});
//Unwanted use case
A<3> instance2({1,2,3});

I don't know how to construct the constructor that I want. I've tried for a week now dozens of designs and none got what I wanted. Here are the names of thing that I have tried:

0. Template deduction guides - also templated versions, which I'm not sure if they are legal...
1. std::initializer_list - the size of the list cannot be put in the template argument. At least not in a non-constexpr context.
2. std::array
3. plain old array
4. using keyword - also templated.

Tl;Dr:

How to deduce a template parameter value signifying a size of an array type from the given array in the signature of the constructor function?

Answer 1

A small example using deduction guides:

template<uint8_t size>
class A
{
public:
    template <typename... Args>
    constexpr A(Args&&... args)
    : what { std::forward<decltype(args)>(args)... }
    {}
    constexpr size_t getSize() const { return what.size(); }
private:
    std::array<int, size> what;
};

//deduction guide
template <typename... Args> A(Args... args) -> A<sizeof...(args)>;

int main()
{
    //Wanted use cases
    A instance1 (1,2,3);
    static_assert (instance1.getSize() == 3);
    
    //or
    //A instance1 {1,2,3};
    return 0;
}

Answer 2

#include <array>

template<class... U>
class A
{
    std::array<int,sizeof...(U)> what;
public:
    constexpr A(U&&... u) : what{std::forward<U>(u)...} { }
    constexpr std::size_t size() const { 
        return what.size();
    }
};

int main() {
    A a(1,2,3,4,5,6);
    static_assert(a.size() == 6);

    A b(1);
    static_assert(b.size() == 1);
    
    return 0;
}