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什么更快 - 使用元素选择运算符访问多数组的元素,或使用迭代器遍历多数组?

就我而言,我每次都需要对多数组的所有元素进行一次完整的传递。

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1 回答 1

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访问 a 的每个元素的最快方法boost::multi_array是 viadata()num_elements()

随着data()您获得对底层原始存储(包含数组数据的连续块)的访问权限,因此不需要多个索引计算(还考虑multi_array可以从不同于 0 的基数索引数组,这是一个更复杂的问题)。

一个简单的测试给出:

g++ -O3 -fomit-frame-pointer -march=native   (GCC v4.8.2)
Writing (index): 9.70651
Writing (data):  2.22353
Reading (index): 4.5973 (found 1)
Reading (data):  3.53811 (found 1)

clang++ -O3 -fomit-frame-pointer -march=native   (CLANG v3.3)
Writing (index): 5.49858
Writing (data):  2.13678
Reading (index): 5.07324 (found 1)
Reading (data):  2.55109 (found 1)

默认情况下,boost 访问方法执行范围检查。如果提供的索引超出为数组定义的范围,则断言将中止程序。要禁用范围检查,您可以在包含在应用程序中BOOST_DISABLE_ASSERTS之前定义预处理器宏。multi_array.hpp

这将减少很多性能差异:

g++ -O3 -fomit-frame-pointer -march=native   (GCC v4.8.2)
Writing (index): 3.15244
Writing (data):  2.23002
Reading (index): 1.89553 (found 1)
Reading (data):  1.54427 (found 1)

clang++ -O3 -fomit-frame-pointer -march=native   (CLANG v3.3)
Writing (index): 2.24831
Writing (data):  2.12853
Reading (index): 2.59164 (found 1)
Reading (data):  2.52141 (found 1)

性能差异增加(即data()更快):

  • 具有更多的维度;
  • 使用更少的元素(对于大量元素,对元素的访问不会像将这些元素加载到 CPU 缓存中的缓存压力那样重要。预取器将坐在那里尝试加载这些元素,即会占用很大一部分时间)。

无论如何,这种优化不太可能在实际程序中产生可衡量的差异。除非您通过广泛的测试最终确定它是某种瓶颈的根源,否则您不应该担心这一点。

资源:

#include <chrono>
#include <iostream>

// #define BOOST_DISABLE_ASSERTS
#include <boost/multi_array.hpp>

int main()
{
  using array3 = boost::multi_array<unsigned, 3>;
  using index = array3::index;

  using clock = std::chrono::high_resolution_clock;
  using duration = std::chrono::duration<double>;

  constexpr unsigned d1(300), d2(400), d3(200), sup(100);

  array3 A(boost::extents[d1][d2][d3]);

  // Writing via index
  const auto t_begin1(clock::now());
  unsigned values1(0);
  for (unsigned n(0); n < sup; ++n)
    for (index i(0); i != d1; ++i)
      for (index j(0); j != d2; ++j)
        for (index k(0); k != d3; ++k)
          A[i][j][k] = ++values1;
  const auto t_end1(clock::now());

  // Writing directly
  const auto t_begin2(clock::now());
  unsigned values2(0);
  for (unsigned n(0); n < sup; ++n)
  {
    const auto sup(A.data() + A.num_elements());

    for (auto i(A.data()); i != sup; ++i)
      *i = ++values2;
  }
  const auto t_end2(clock::now());

  // Reading via index
  const auto t_begin3(clock::now());
  bool found1(false);
  for (unsigned n(0); n < sup; ++n)
    for (index i(0); i != d1; ++i)
      for (index j(0); j != d2; ++j)
        for (index k(0); k != d3; ++k)
          if (A[i][j][k] == values1)
            found1 = true;
  const auto t_end3(clock::now());

  // Reading directly
  const auto t_begin4(clock::now());
  bool found2(false);
  for (unsigned n(0); n < sup; ++n)
  {
    const auto sup(A.data() + A.num_elements());

    for (auto i(A.data()); i != sup; ++i)
      if (*i == values2)
        found2 = true;
  }
  const auto t_end4(clock::now());

  std::cout << "Writing (index): "
            << std::chrono::duration_cast<duration>(t_end1 - t_begin1).count()
            << std::endl
            << "Writing (data):  "
            << std::chrono::duration_cast<duration>(t_end2 - t_begin2).count()
            << std::endl
            << "Reading (index): "
            << std::chrono::duration_cast<duration>(t_end3 - t_begin3).count()
            << " (found " << found1 << ")" << std::endl
            << "Reading (data):  "
            << std::chrono::duration_cast<duration>(t_end4 - t_begin4).count()
            << " (found " << found2 << ")" << std::endl;

  return 0;
}
于 2014-05-14T08:22:53.297 回答