C++ template meta-programming member function loop unrolling - c++

Question

I have just started to use template meta-programming in my code. I have a class which has as a member which is a vector of a multi-dimensional Cartesian points. Here is a basic setup of the class:
template<size_t N>
class TGrid{
public:
void round_points_3(){
for(std::size_t i = 0; i < Xp.size();i++){
Xp[i][0] = min[0] + (std::floor((Xp[i][0] - min[0]) * nbins[0] / (max[0] - min[0])) * bin_w[0]) + bin_w[0]/2.0;
Xp[i][1] = min[1] + (std::floor((Xp[i][1] - min[1]) * nbins[1] / (max[1] - min[1])) * bin_w[1]) + bin_w[1]/2.0;
Xp[i][2] = min[2] + (std::floor((Xp[i][2] - min[2]) * nbins[2] / (max[2] - min[2])) * bin_w[2]) + bin_w[2]/2.0;
}
}
void round_points_2(){
for(std::size_t i = 0; i < Xp.size();i++){
Xp[i][0] = min[0] + (std::floor((Xp[i][0] - min[0]) * nbins[0] / (max[0] - min[0])) * bin_w[0]) + bin_w[0]/2.0;
Xp[i][1] = min[1] + (std::floor((Xp[i][1] - min[1]) * nbins[1] / (max[1] - min[1])) * bin_w[1]) + bin_w[1]/2.0;
}
}
void round_points_1(){
for(std::size_t i = 0; i < Xp.size();i++){
Xp[i][0] = min[0] + (std::floor((Xp[i][0] - min[0]) * nbins[0] / (max[0] - min[0])) * bin_w[0]) + bin_w[0]/2.0;
}
}
public:
std::vector<std::array<double,N> > Xp;
std::vector<double> min, max, nbins, bin_w;
};
This class represented a multidimensional Grid. The dimension is specified by the template value N. I will be having many operations which can be made more efficient by having template specific member functions tailored to the specific dimensions, such as loop unrolling.
In the class TGrid, I have 3 functions specific for dimensions D=1,D=2 and D=3. This is indicated by the subscript _1,_2 and _3 of the functions.
I am looking for a template meta-programming oriented approach to write
these three functions more compactly.
I have seen examples of loop unrolling but all of these examples don't consider member functions of a template class.

Answer 1

Putting to one side the question of whether or not this is an appropriate optimisation, or if other optimisations should be regarded first, this is how I would do it. (But I do agree, sometimes it is demonstrably better to explicitly unroll loops — the compiler isn't always the best judge.)
One can't partially specialize a member function, and one can't specialize a nested struct without specializing the outer struct, so the only solution is to use a separate templated struct for the unrolling mechanism. Feel free to put this in some other namespace :)
The unrolling implementation:
template <int N>
struct sequence {
template <typename F,typename... Args>
static void run(F&& f,Args&&... args) {
sequence<N-1>::run(std::forward<F>(f),std::forward<Args>(args)...);
f(args...,N-1);
}
};
template <>
struct sequence<0> {
template <typename F,typename... Args>
static void run(F&& f,Args&&... args) {}
};
This takes an arbitrary functional object and a list of arguments, and then calls the object with the arguments and an additional final argument N times, where the final argument ranges from 0 to N-1. The universal references and variadic templates are not necessary; the same idea can be employed in C++98 with less generality.
round_points<K> then calls sequence::run<K> with a helper static member function:
template <size_t N>
class TGrid {
public:
template <size_t K>
void round_points(){
for (std::size_t i = 0; i < Xp.size();i++) {
sequence<K>::run(TGrid<N>::round_item,*this,i);
}
}
static void round_item(TGrid &G,int i,int j) {
G.Xp[i][j] = G.min[j] + (std::floor((G.Xp[i][j] - G.min[j]) * G.nbins[j] / (G.max[j] - G.min[j])) * G.bin_w[j]) + G.bin_w[j]/2.0;
}
// ...
};
Edit: Addendum
Doing the equivalent with a pointer-to-member function appears to be hard for compilers to inline. As an alternative, to avoid the use of a static round_item, you can use a lambda, e.g.:
template <size_t N>
class TGrid {
public:
template <size_t K>
void round_points(){
for (std::size_t i = 0; i < Xp.size();i++) {
sequence<K>::run([&](int j) {round_item(i,j);});
}
}
void round_item(int i,int j) {
Xp[i][j] = min[j] + (std::floor((Xp[i][j] - min[j]) * nbins[j] / (max[j] - min[j])) * bin_w[j]) + bin_w[j]/2.0;
}
// ...
};