Why std::get does not work with variables?

Question

I am running into a trouble of understanding how functions, especially template functions and local variables behave during the compilation time.

So this code works well with std::get:

enum class UserInfoFields{name, email, address};

using UserInfo = std::tuple<std::string, std::string, std::string>;

int main()
{
    UserInfo s{"Edmund", "[email protected]", "Denver street 19"};
    std::cout << std::get<static_cast<size_t>(UserInfoFields::name)>(s) << std::endl;
    return 0;
}

This from my understanding is because std::get is a template function and it requires a template argument to be known during the compilation. That makes sense as static_cast<... gives us the value during the compilation time.

What I do not understand, if I change the main() code to this:

 int main()
{
    UserInfo s{"Edmund", "[email protected]", "Denver street 19"};
    auto a = static_cast<size_t>(UserInfoFields::name);
    std::cout << std::get<a>(s) << std::endl;
    return 0;
}

This is not allowed. I know that I have to use constexpr but I want to know, why exactly the second code does not work?

Answer 1

You wrote yourself that

std::get is a template function and it requires a template argument to be known during the compilation

The value of a local variable is not (in the general case) known during compilation; a local variable's value is a runtime property. As such, a local variable cannot be used as a template argument.

If you want to use it as one, you have to make it a compile-time value. This is achieved by making it constexpr (as you've also stated in the question).