并行使用 std::thread？

Question

我是 std::thread 的新手，我尝试编写 parallel_for。 我编写了以下代码：

// parallel_for.cpp
// compilation: g++ -O3 -std=c++0x parallel_for.cpp -o parallel_for -lpthread
// execution: time ./parallel_for 100 50000000 
// (100: number of threads, 50000000: vector size)
#include <iostream>
#include <iomanip>
#include <cstdlib>
#include <vector>
#include <thread>
#include <cmath>
#include <algorithm>
#include <numeric>
#include <utility>

// Parallel for
template<typename Iterator, class Function>
void parallel_for(const Iterator& first, const Iterator& last, Function&& f, const int nthreads = 1, const int threshold = 1000)
{
    const unsigned int group = std::max(std::max(1, std::abs(threshold)), (last-first)/std::abs(nthreads));
    std::vector<std::thread> threads;
    for (Iterator it = first; it < last; it += group) {
        threads.push_back(std::thread([=](){std::for_each(it, std::min(it+group, last), f);}));
    }
    std::for_each(threads.begin(), threads.end(), [=](std::thread& x){x.join();});
}

// Function to apply
template<typename Type>
void f1(Type& x)
{
    x = std::sin(x)+std::exp(std::cos(x))/std::exp(std::sin(x)); 
}

// Main
int main(int argc, char* argv[]) {

    const unsigned int nthreads = (argc > 1) ? std::atol(argv[1]) : (1);
    const unsigned int n = (argc > 2) ? std::atol(argv[2]) : (100000000);
    double x = 0;
    std::vector<double> v(n);
    std::iota(v.begin(), v.end(), 0);

    parallel_for(v.begin(), v.end(), f1<double>, nthreads);

    for (unsigned int i = 0; i < n; ++i) x += v[i];
    std::cout<<std::setprecision(15)<<x<<std::endl;
    return 0;
}

但这不起作用：（来自 g++ 4.6 的错误代码）

parallel_for.cpp: In instantiation of ‘parallel_for(const Iterator&, const Iterator&, Function&&, int, int) [with Iterator = __gnu_cxx::__normal_iterator<double*, std::vector<double> >, Function = void (&)(double&)]::<lambda()>’:
parallel_for.cpp:22:9:   instantiated from ‘void parallel_for(const Iterator&, const Iterator&, Function&&, int, int) [with Iterator = __gnu_cxx::__normal_iterator<double*, std::vector<double> >, Function = void (&)(double&)]’
parallel_for.cpp:43:58:   instantiated from here
parallel_for.cpp:22:89: erreur: field ‘parallel_for(const Iterator&, const Iterator&, Function&&, int, int) [with Iterator = __gnu_cxx::__normal_iterator<double*, std::vector<double> >, Function = void (&)(double&)]::<lambda()>::__f’ invalidly declared function type

如何解决这个问题？

编辑：这个新版本编译但没有给出好的结果：

// parallel_for.cpp
// compilation: g++ -O3 -std=c++0x parallel_for.cpp -o parallel_for -lpthread
// execution: time ./parallel_for 100 50000000 
// (100: number of threads, 50000000: vector size)
#include <iostream>
#include <iomanip>
#include <cstdlib>
#include <vector>
#include <thread>
#include <cmath>
#include <algorithm>
#include <numeric>
#include <utility>

// Parallel for
template<typename Iterator, class Function>
void parallel_for(const Iterator& first, const Iterator& last, Function&& f, const int nthreads = 1, const int threshold = 1000)
{
    const unsigned int group = std::max(std::max(1, std::abs(threshold)), (last-first)/std::abs(nthreads));
    std::vector<std::thread> threads;
    for (Iterator it = first; it < last; it += group) {
        threads.push_back(std::thread([=, &f](){std::for_each(it, std::min(it+group, last), f);}));
    }
    std::for_each(threads.begin(), threads.end(), [](std::thread& x){x.join();});
}

// Function to apply
template<typename Type>
void f(Type& x)
{
    x = std::sin(x)+std::exp(std::cos(x))/std::exp(std::sin(x)); 
}

// Main
int main(int argc, char* argv[]) {

    const unsigned int nthreads = (argc > 1) ? std::atol(argv[1]) : (1);
    const unsigned int n = (argc > 2) ? std::atol(argv[2]) : (100000000);
    double x = 0;
    double y = 0;
    std::vector<double> v(n);

    std::iota(v.begin(), v.end(), 0);
    std::for_each(v.begin(), v.end(), f<double>);
    for (unsigned int i = 0; i < n; ++i) x += v[i];

    std::iota(v.begin(), v.end(), 0);
    parallel_for(v.begin(), v.end(), f<double>, nthreads);
    for (unsigned int i = 0; i < n; ++i) y += v[i];

    std::cout<<std::setprecision(15)<<x<<" "<<y<<std::endl;
    return 0;
}

结果是：

./parallel_for 1 100
155.524339894552 4950

并行版本返回 4950，而顺序版本返回 155..... 问题出在哪里？

Answer 1

您需要在 (last-first) 进行强制转换或类型转换。原因是在模板参数推导期间从不进行类型转换。

这很好用（也解决了 DeadMG 和 Ben Voigt 发现的问题）。 两个版本都给出 156608294.151782，n=100000000。

template<typename Iterator, class Function>
void parallel_for(const Iterator& first, const Iterator& last, Function&& f, const int nthreads = 1, const int threshold = 1000)
{
    const unsigned int group = std::max(std::max(ptrdiff_t(1), ptrdiff_t(std::abs(threshold))), ((last-first))/std::abs(nthreads));
    std::vector<std::thread> threads;
    threads.reserve(nthreads);
    Iterator it = first;
    for (; it < last-group; it += group) {
        threads.push_back(std::thread([=,&f](){std::for_each(it, std::min(it+group, last), f);}));
    }
    std::for_each(it, last, f); // last steps while we wait for other threads

    std::for_each(threads.begin(), threads.end(), [](std::thread& x){x.join();});
}

由于步长for_each(it, last, f)比其他步小，我们不妨在等待其他结果的同时使用调用线程完成。

Answer 2

• 您必须通过引用来捕获函数。

  [=, &f] () { /* your code */ };

• 查看代码。

  #include <iostream>
  
  template <class T>
  void foo(const T& t)
  {
      const int a = t;
      [&]
      {
          std::cout << a << std::endl;
      }();
  }
  
  
  int main()
  {
      foo(42);
      return 0;
  }
  

  clang 给出输出 42，但 g++ 抛出警告：‘a’ is used uninitialized in this function，并打印 0。看起来像一个错误。

  解决方法：使用const auto（用于代码中的变量group）。

  UPD：我想，就是这样。 http://gcc.gnu.org/bugzilla/show_bug.cgi?id=52026

Answer 3

一个问题是it += group 可以合法地成为last，但在最后创建一个值是未定义的行为。仅仅检查it < last 为时已晚，无法解决这个问题。

您需要在 it 仍然有效时测试 last - it。 （it + group 和 last - group 都不一定是安全的，尽管后者应该是由于 group 的计算方式造成的。）

例如：

template<typename Iterator, class Function>
void parallel_for(const Iterator& first, const Iterator& last, Function f, const int nthreads = 1, const int threshold = 100)
{
    const unsigned int group = std::max(std::max(1, std::abs(threshold)), (last-first)/std::abs(nthreads));
    std::vector<std::thread> threads;
    threads.reserve(nthreads);
    Iterator it = first;
    for (; last - it > group; it += group) {
        threads.push_back(std::thread([=, &f](){std::for_each(it, it+group, last), f);}));
    }
    threads.push_back(std::thread([=, &f](){std::for_each(it, last, f);}));

    std::for_each(threads.begin(), threads.end(), [](std::thread& x){x.join();});
}

Answer 4

您将std::min(it+group, last) 赋予std::for_each，但始终在末尾添加group。这意味着如果last 不是it 上的group 的倍数，您将把it 移到last 后面，即UB。

Answer 5

您需要通过引用进行捕获，并且需要在（后到先）进行强制转换或类型转换。 原因是在模板参数推导期间从不进行类型转换。

另外，修复 DeadMG 发现的问题，最终得到以下代码。

效果很好，两个版本都给出 156608294.151782，n=100000000。

template<typename Iterator, class Function>
void parallel_for(const Iterator& first, const Iterator& last, Function&& f, const int nthreads = 1, const int threshold = 1000)
{
    const unsigned int group = std::max(std::max(ptrdiff_t(1), ptrdiff_t(std::abs(threshold))), ((last-first))/std::abs(nthreads));
    std::vector<std::thread> threads;
    Iterator it = first;
    for (; it < last-group; it += group) {
        threads.push_back(std::thread([=,&f](){std::for_each(it, std::min(it+group, last), f);}));
    }
    std::for_each(it, last, f); // use calling thread while we wait for the others
    std::for_each(threads.begin(), threads.end(), [](std::thread& x){x.join();});
}

Answer 6

vc11 解决方案，如果它不适用于 gcc，请告诉我。

template<typename Iterator, class Function>
void parallel_for( const Iterator& first, const Iterator& last, Function&& f, const size_t nthreads = std::thread::hardware_concurrency(), const size_t threshold = 1 )
{
    const size_t portion = std::max( threshold, (last-first) / nthreads );
    std::vector<std::thread> threads;
    for ( Iterator it = first; it < last; it += portion )
    {
        Iterator begin = it;
        Iterator end = it + portion;
        if ( end > last )
            end = last;

        threads.push_back( std::thread( [=,&f]() {
            for ( Iterator i = begin; i != end; ++i )
                f(i);
        }));
    }
    std::for_each(threads.begin(), threads.end(), [](std::thread& x){x.join();});
}