Forward string literal to consteval function

Question

My api would require calling runtime(constFoo(str)) to get the final result, I wanted to shorten this because the real function names are kind of verbose. To do this I need to somehow wrap the consteval function, but I can't do that since the argument cannot be constexpr, it's also a string literal which requires the its size to be known at compile time.

I'm aware of Pass const expression to consteval function through non-consteval functions but it doesn't really solve my issue because I need the size to be known at compile time.

int runtime(int val) {
  std::string foo = "foo"; // some run time operation
  return val + 2;
}

template <std::size_t N> consteval int constFoo(const char (&data)[N]) {
  // ...
  const auto d = data[0];
  return N;
}

template <std::size_t N> int outer(const char (&data)[N]) {
  int val = constFoo(data); // Call to consteval function 'constFoo<2ULL>' is not a constant expression
  return runtime(val);
}

int main() {
  auto d = outer("d");
}

Also if this would be somehow possible I'd rather avoid a solution where the literal is passed as a template argument, but as far as I'm aware this is impossible.

Answer 1

Since C++20, non-type template parameter allows some literal class types, so turn your string literal into one of those classes:

template <std::size_t N>
struct LiteralString
{
    consteval LiteralString(const char (&s)[N]) { std::copy(s, s + N, &data[0]); }

    static constexpr std::size_t size = N;
    char data[N]{};
};

// [..]

template <LiteralString data>
int outer()
{
  int val = constFoo(data.data);
  return runtime(val);
}

int main()
{
  auto d = outer<LiteralString("d")>();
  // [..]
}

Demo

Answer 2

Make the parameter of the outer (non-consteval) function a class with a consteval constructor taking a string.