如何在编译时使用C ++模板存储递归函数的中间结果？

Question

I asked How do I capture the results of a recursive function at compile-time? , but I think my approach was wrong.

I have a program like so:

#include <iostream>
#include <list>

std::list<unsigned int> recursive_case(std::list<unsigned int>& result, unsigned int& i) {
    result.push_front(1 + (i % 10));
    i /= 10;
    return i != 0 ? recursive_case(result, i) : result;
}

std::list<unsigned int> initial_case(unsigned int i) {
    std::list<unsigned int> result;
    result.push_back(i % 10);
    i /= 10;
    return i != 0 ? recursive_case(result, i) : result;
}

int main() {
    auto list = initial_case(123);
    bool first = true;
    for (auto i: list) {
        if (first) {
            first = false;
        } else {
            std::cout << ", ";
        }
        std::cout << i;
    }
    std::cout << std::endl;
}

The output is 2, 3, 3 .

I want to perform the above computation and get the same output but in compile-time (the loop iteration and output-printing would be at runtime ie everything starting from the for loop). Templates seem like a possibility (that's why I tagged this ask as such), but I am open to anything that gets the job done in compile-time.

Answer 1

You can use constexpr to calculate the list at compile time. I converted the recursion to iteration and used the indices trick to call calculate as often as necessary.

#include <iostream>
#include <array>
#include <iterator>
#include <utility>

constexpr std::size_t count_digits(std::size_t N, std::size_t Count = 0)
{
  return (N > 0) ? count_digits(N/10, Count+1) : Count; 
}

constexpr std::size_t ipow(std::size_t N, std::size_t Base)
{
  return (N > 0) ? Base*ipow(N-1,Base) : 1; 
}

constexpr std::size_t calculate(std::size_t n, std::size_t i)
{
    std::size_t p = ipow(i,10);
    std::size_t t = (n/p) % 10; 
    return i > 0 ? (t+1) : t;
}

template<std::size_t Num, std::size_t C, std::size_t... Is>
constexpr std::array<std::size_t, C> build_list(std::index_sequence<Is...>)
{
  return {{ calculate(Num, C-Is-1)... }};
}

template <std::size_t Num, std::size_t C = count_digits(Num)>
constexpr auto build_list()
{
  return build_list<Num, C>(std::make_index_sequence<C>{});
}


int main()
{
    constexpr auto list = build_list<123>();

    for(auto e : list)
    {
        std::cout << e << " ";
    }

    return 0;
}

output:

2 3 3 

live example

Answer 2

Here's one solution.

#include <iostream>

// Print one digit.
template <unsigned int N, bool Initial> struct AtomicPrinter
{
   static void print()
   {
      std::cout << N%10;
   }
};

template <unsigned int N> struct AtomicPrinter<N, false>
{
   static void print()
   {
      std::cout << 1 + N%10 << ", ";
   }
};

// Recursive printer for a number
template <unsigned int N, bool Initial> struct Printer
{
   static void print()
   {
      Printer<N/10, false>::print();
      AtomicPrinter<N, Initial>::print();
   }
};

// Specialization to end recursion.
template <bool TF> struct Printer<0, TF>
{
   static void print()
   {
   }
};

void printList()
{
   Printer<123, true>::print();
   std::cout << std::endl;
}

int main() {
   printList();
}

If there is a need to separate printing of the digits from constructing the list of digits, you can use:

#include <iostream>
#include <list>

template <unsigned int N, bool Initial> struct Digit
{
   static void get(std::list<int>& l)
   {
      l.push_back(N%10);
   }
};

template <unsigned int N> struct Digit<N, false>
{
   static void get(std::list<int>& l)
   {
      l.push_back(1 + N%10);
   }
};

template <unsigned int N, bool Initial> struct Digits
{
   static void get(std::list<int>& l)
   {
      Digits<N/10, false>::get(l);
      Digit<N, Initial>::get(l);
   }
};

template <bool TF> struct Digits<0, TF>
{
   static void get(std::list<int>& l)
   {
   }
};

void printList()
{
   std::list<int> l;
   Digits<123, true>::get(l);
   bool first = true;
   for (auto i: l) {
      if (first) {
         first = false;
      } else {
         std::cout << ", ";
      }
      std::cout << i;
   }
   std::cout << std::endl;
}

int main() {
   printList();
}

Answer 3

You may use something like the following to split number at compile time:

#include <utility>
#include <iostream>

template <char... Cs>
std::integer_sequence<char, Cs...> operator "" _seq() { return {}; }

template <char...Cs>
void print(std::integer_sequence<char, Cs...>)
{
    const char* sep = "";
    for (const auto& c : {Cs...}) {
        std::cout << sep << c;
        sep = ", ";
    }
}

int main() {
    auto seq = 123_seq;
    print(seq);
}

Demo