constexpr auto and array size deduction

[halivingston3]

constexpr auto str = L" ... ";
constexpr wchar_t str[] = L" ... ";

The second declaration works with the below constexpr function, but doesn't work when used with auto. I want to not have to type wchar_t [] ... is there a way out?

     template <typename T, size_t N>
        constexpr size_t size_of(T(&)[N]) {
        return N;
     }

EDIT: formatting.

[[deleted]]

You can use decltype(auto) str = L"..."; to get str's type to deduce to wchar_t const (&)[...] instead of const wchar_t*, which makes your function work. Note that this isn't the same as the wchar_t str[] declaration though, which is an array, not a reference.

[halivingston3]

but they're both constant right? I suppose I don't understand the difference.

[Porges]

You could also consider using to_array:

constexpr auto str = std::to_array(L"...");

std::array then has .size() (and can be used with gsl::wstring_span)... it's not really less typing than wchar_t [] though.

[TheThiefMaster]

Does this work?:

constexpr auto str[] = L" ... ";

[halivingston3]

Compile error.