User defined literal arguments are not constexpr?

23
votes

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, however it doesn't let me do it with templates as the compiler complains that the arguments are not and cannot be constexpr.

I'm confused by this - aren't literals constant expressions? The 5 in 5_fac is always a literal that can be evaluated during compile time, so why can't I use it as such?

First method:

constexpr int factorial_function(int x) {
  return (x > 0) ? x * factorial_function(x - 1) : 1;
}

constexpr int operator "" _fac(unsigned long long x) {
  return factorial_function(x); // this works
}

Second method:

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

constexpr int operator "" _fac(unsigned long long x) {
  return factorial_template<x>::value; // doesn't work - x is not a constexpr
}
5
c++c++11constexpruser-defined-literals
I don't think any compiler implements those user-defined literals yet. How are you testing this? – Kerrek SB
Nice! I didn't know that was supported yet. I have one of those installed on a machine somewhere, now I'm tempted to work out those variadic templates :-) – Kerrek SB

5 Answers

11
votes

I don't know if there's a better way in C++11 to do this than the current accepted answer, but with relaxed constexpr in C++14, you can just write "normal" code:

constexpr unsigned long long int operator "" _fac(unsigned long long int x) {
    unsigned long long int result = 1;
    for (; x >= 2; --x) {
        result *= x;
    }
    return result;
}

static_assert(5_fac == 120, "!");
5
votes

This is how I ended up doing it:

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

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

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

template <char ...cv>
constexpr unsigned int operator "" _fac()
{
  return factorial<literal<cv...>::to_int>::value;
}

Huge thanks to KerrekSB!

4
votes

I may be wrong, but I think constexpr functions can also be called with non-constant arguments (in which case they don't give a constant expression and are evaluated at runtime). Which wouldn't work well with non-type template arguments.

3
votes

In order to make use of constexpr with user defined literals, you apparently have to use a variadic template. Take a look at the second listing in the wikipedia article for an example.

1
votes

@Pubby. The easy way to digest the char non-type parameter pack is to cature it into an initializer list for a string. Then you can use atoi, atof, etc:

#include <iostream>

template<char... Chars>
  int
  operator "" _suffix()
  {
    const char str[]{Chars..., '\0'};
    return atoi(str);
  }

int
main()
{
  std::cout << 12345_suffix << std::endl;
}

Remember to tack on a null character for the C-style functions.