Can I separate creation and usage locations of compile-time strategies?

Question

#include 
#include 
#include 
#include 

using namespace std;

struct SubAlgorithm1 { void operator ()          


        

Answer 1

Yes. I call the technique the magic switch.

You create a std::tuple of your algorithms. You ceate a template function that will be passed one of those algorithms.

You can add other arguments via perfect variardic forwarding if you want.

template<size_t Max, typename...Ts, typename Func>
bool magic_switch( int n, Func&& f,  std::tuple<Ts...> const & pick ) {
  if( n==Max-1 ) {
    f(std::get<Max-1>(pick));
    return true;
  } else {
    return magic_switch<Max-1>( n, std::forward<Func>(f), pick );
  }
}

In pseudo code. Specialize Max==0 to just return false, and you might have to make it a functor so you can partial specialize.

The passed in functor is annoying to write, as a downside.

Another variation is to use a meta-factory (well, a meta programming type factory? Maybe it is a meta-map. Well, whatever.)

#include <iostream>
#include <tuple>
#include <vector>
#include <utility>
#include <cstddef>
#include <functional>
#include <array>
#include <iostream>

// metaprogramming boilerplate:
template<template<typename>class Factory, typename SourceTuple>
struct tuple_map;
template<template<typename>class Factory, template<typename...>class L, typename... SourceTypes>
struct tuple_map<Factory, L<SourceTypes...>> {
  typedef L< Factory<SourceTypes>... > type;
};
template<template<typename>class Factory, typename SourceTuple>
using MapTuple = typename tuple_map<Factory, SourceTuple>::type;
template<std::size_t...> struct seq {};
template<std::size_t max, std::size_t... s>
struct make_seq: make_seq<max-1, max-1, s...> {};
template<std::size_t... s>
struct make_seq<0, s...> {
  typedef seq<s...> type;
};
template<std::size_t max>
using MakeSeq = typename make_seq<max>::type;

// neat little class that lets you type-erase the contents of a tuple,
// and turn it into a uniform array:
template<typename SourceTuple, typename DestType>
struct TupleToArray;
template<template<typename...>class L, typename... Ts, typename DestType>
struct TupleToArray<L<Ts...>, DestType> {
  template<std::size_t... Index>
  std::array< DestType, sizeof...(Ts) > operator()( L<Ts...> const& src, seq<Index...> ) const {
    std::array< DestType, sizeof...(Ts) > retval{ DestType( std::get<Index>(src) )... };
    return retval;
  }

  std::array< DestType, sizeof...(Ts) > operator()( L<Ts...> const& src ) const {
    return (*this)( src, MakeSeq<sizeof...(Ts)>() );
  }
};
template< typename DestType, typename SourceTuple >
auto DoTupleToArray( SourceTuple const& src )
  -> decltype( TupleToArray<SourceTuple, DestType>()( src ) )
{
  return TupleToArray<SourceTuple, DestType>()( src );
}

// Code from here on is actually specific to this problem:
struct SubAlgo { int operator()(int x) const { return x; } };
struct SubAlgo2 { int operator()(int x) const { return x+1; } };

template<typename Sub>
struct FullAlgo {
  void operator()( std::vector<int>& v ) const {
    for( auto& x:v )
      x = Sub()( x );
  }
};

// a bit messy, but I think I could clean it up:
typedef std::tuple< SubAlgo, SubAlgo2 > subAlgos;
MapTuple< FullAlgo, subAlgos > fullAlgos;
typedef std::function< void(std::vector<int>&) > funcType;
std::array< funcType, 2 > fullAlgoArray =
  DoTupleToArray< funcType >( fullAlgos );

int main() {
  std::vector<int> test{1,2,3};
  fullAlgoArray[0]( test );
  for (auto&& x: test)
    std::cout << x;
  std::cout << "\n";
  fullAlgoArray[1]( test );
  for (auto&& x: test)
    std::cout << x;
  std::cout << "\n";
}

which is lots of boilerplate, but what I've just done is allowed you to take your stateless sub algorithm and plug it into your full algorithm one element at a time, then type-erase the resulting full algorithm and store it in a std::function array.

There is a virtual call overhead, but it occurs at the top level.

Answer 2

You should use an interface, with both SubAlgorithm1 and SubAlgorithm2 (you'll need better names than that) implementing the interface. The create an object of either class depending on runtime_flag.