Calculate the Fibonacci number (recursive approach) in compile time (constexpr) in C++11

Mantosh Kumar picture Mantosh Kumar · Mar 25, 2014 · Viewed 7.2k times · Source

I wrote the program Fibonacci number calculation in compile time (constexpr) problem using the template metaprogramming techniques supported in C++11. The purpose of this is to calculate the difference in the run-time between the template metaprogramming approach and the old conventional approach.

// Template Metaprograming Approach
template<int  N>
constexpr int fibonacci() {return fibonacci<N-1>() + fibonacci<N-2>(); }
template<>
constexpr int fibonacci<1>() { return 1; }
template<>
constexpr int fibonacci<0>() { return 0; }



// Conventional Approach
 int fibonacci(int N) {
   if ( N == 0 ) return 0;
   else if ( N == 1 ) return 1;
   else
      return (fibonacci(N-1) + fibonacci(N-2));
} 

I ran both programs for N = 40 on my GNU/Linux system and measured the time and found that that conventional solution (1.15 second) is around two times slower than the template-based solution (0.55 second). This is a significant improvement as both approaches are based on the recursion.

To understand it more I compiled the program (-fdump-tree-all flag) in g++ and found that compiler actually generated the 40 different functions (like fibonacci<40>, fibonacci<39>...fibonacci<0>).

constexpr int fibonacci() [with int N = 40] () {
  int D.29948, D.29949, D.29950;
  D.29949 = fibonacci<39> ();
  D.29950 = fibonacci<38> ();
  D.29948 = D.29949 + D.29950;
  return D.29948;
}

constexpr int fibonacci() [with int N = 39] () {
  int D.29952, D.29953, D.29954;
  D.29953 = fibonacci<38> ();
  D.29954 = fibonacci<37> ();
  D.29952 = D.29953 + D.29954;
  return D.29952;
}
...
...
...
constexpr int fibonacci() [with int N = 0] () {
  int D.29962;
  D.29962 = 0;
  return D.29962;
}

I also debugged the program in GDB and found that all the above functions are executed an equal number of times as with the conventional recursive approach. If both versions of the program are executing the function an equal number of times (recursive), then how is this achieved by template metaprogramming techniques? I would also like to know your opinion about how and why a template metaprogramming based approach is taking half time compared to the other version? Can this program be made faster than the current one?

Basically my intention here is to understand what's going on internally as much as possible.

My machine is GNU/Linux with GCC 4.8.1, and I used the optimization -o3 for both programs.

Answer

iavr picture iavr · Mar 25, 2014

Try this:

template<size_t N>
struct fibonacci : integral_constant<size_t, fibonacci<N-1>{} + fibonacci<N-2>{}> {};

template<> struct fibonacci<1> : integral_constant<size_t,1> {};
template<> struct fibonacci<0> : integral_constant<size_t,0> {};

With clang and -Os, this compiles in roughly 0.5s and runs in zero time for N=40. Your "conventional" approach compiles in roughly 0.4s and runs in 0.8s. Just for checking, the result is 102334155 right?

When I tried your own constexpr solution the compiler run for a couple of minutes and then I stopped it because apparently memory was full (computer started freezing). The compiler was trying to compute the final result and your implementation is extremely inefficient to be used at compile time.

With this solution, template instantiations at N-2, N-1 are re-used when instantiating N. So fibonacci<40> is actually known at compile time as a value, and there is nothing to do at run-time. This is a dynamic programming approach and of course you can do the same at run time if you store all values at 0 through N-1 before computing at N.

With your solution, the compiler can evaluate fibonacci<N>() at compile time but is not required to. In your case, all or part of computation is left for run time. In my case, all computation is attempted at compile time, hence never ending.