HPX transform_reduce

2019-08-05 11:37发布

问题:

I've tried to use transform_reduce from hpx as given in the Answer https://stackoverflow.com/a/54481320/11008404 but I can't compile it. My code so far:

#include <hpx/hpx_main.hpp>
#include <hpx/hpx.hpp>
#include <hpx/include/parallel_transform_reduce.hpp>
#include <hpx/include/iostreams.hpp>

#include <vector>

class A {
public:
  double residual() {
    // Calculate actual local residual
    double i = 1.0;
    return i;
  }
};

int main() {

  std::vector<A> vec(300);
  double res = hpx::parallel::transform_reduce(hpx::parallel::execution::par,        
                      vec.begin(), vec.end(),                         // (1)
                      [](A& a_ref){ return a_ref.residual(); },       // (2)
                      0, [](double a, double b){ return a + b; });    // (3)

  hpx::cout << "residual: " << res << hpx::endl;

  return 0;
}

The compiler throws this errors:

hpx.cpp:23:65: error: no matching function for call to ‘transform_reduce(const hpx::parallel::execution::parallel_policy&, std::vector<A>::iterator, std::vector<A>::iterator, main()::<lambda(A&)>, int, main()::<lambda(double, double)>)’
                   0, [](double a, double b){ return a + b; });    // (3)

.../include/hpx/parallel/algorithms/transform_reduce.hpp:255:22: error: no type named ‘type’ in ‘struct hpx::util::invoke_result<main()::<lambda(double, double)>, A>’

Has someone a suggestion what is wrong or another solution for the question asked in Parallel reduce (e.g. sum) a vector of hpx::futures<double>?

回答1:

If I change hkaiser's answer to

#include <hpx/hpx_main.hpp>
#include <hpx/hpx.hpp>
#include <hpx/include/parallel_transform_reduce.hpp>
#include <hpx/include/iostreams.hpp>

#include <vector>

class A {
public:
  double residual() const {
    // Calculate actual local residual
    double i = 1.0;
    return i;
  }
};

int main() {

  std::vector<A> vec(300);
  double res = hpx::parallel::transform_reduce(hpx::parallel::execution::par,        
                      vec.begin(), vec.end(),
                      0.,
                      [](double a, double b){ return a + b; },
                      [](const A& a_ref){ return a_ref.residual(); }); // note: const!

  hpx::cout << "residual: " << res << hpx::endl;

  return 0;
}

the code compiles. It also compiles if you pass A by value or as a pointer.

I don't know if this behaviour is intended so i opened an issue on HPX's github (https://github.com/STEllAR-GROUP/hpx/issues/3651)



回答2:

The signature of transform_reduce has changed several times during the process of its standardization (see here for what got actually standardized: https://en.cppreference.com/w/cpp/algorithm/transform_reduce). I think that in order to compile you just have to get the sequence of arguments right:

#include <hpx/hpx_main.hpp>
#include <hpx/hpx.hpp>
#include <hpx/include/parallel_transform_reduce.hpp>
#include <hpx/include/iostreams.hpp>

#include <vector>

class A {
public:
  double residual() {
    // Calculate actual local residual
    double i = 1.0;
    return i;
  }
};

int main() {

  std::vector<A> vec(300);
  double res = hpx::parallel::transform_reduce(hpx::parallel::execution::par,        
                      vec.begin(), vec.end(),
                      0.,
                      [](double a, double b){ return a + b; },
                      [](A& a_ref){ return a_ref.residual(); });

  hpx::cout << "residual: " << res << hpx::endl;

  return 0;
}


标签: c++ hpx