User defined literal arguments are not constexpr?

Question

I\'m testing out user defined literals. I want to make _fac return the factorial of the number.

Having it call a constexpr function works,

Answer 1

This is how I ended up doing it:

template 
constexpr t pow(t base, int exp) {
  return (exp > 0) ? base * pow(base, exp-1) : 1;
};

template  struct literal;
template <> struct literal<> {
  static const unsigned int to_int = 0;
};
template  struct literal {
  static const unsigned int to_int = (c - '0') * pow(10, sizeof...(cv)) + literal::to_int;
};

template  struct factorial {
  static const unsigned int value = N * factorial::value;
};
template <> struct factorial<0> {
  static const unsigned int value = 1;
};

template 
constexpr unsigned int operator "" _fac()
{
  return factorial::to_int>::value;
}

Huge thanks to KerrekSB!