c++ - 并行使用 std::thread？

Question

我是 std::thread 的新手，我尝试编写一个parallel_for. 我编码了以下内容：

// 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

如何解决这个问题呢？

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？

score 5 · Accepted Answer

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。

这工作得很好（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();});
}

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

score 1 · Accepted Answer

您必须通过引用来捕获函数。

[=, &f] () { /* your code */ };
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 output 42，但 g++ 抛出 warning:‘a’ is used uninitialized in this function和 prints 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

score 1 · Accepted Answer

一个问题是这it += group可能是last合法的，但最终创造价值是未定义的行为。仅仅检查it < last为时已晚，无法解决这个问题。

您需要改为测试last - itwhileit仍然有效。（既不it + group也不last - group一定是安全的，虽然后者应该是由于group计算方式所致。）

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();});
}

score 0 · Accepted Answer

你给std::min(it+group, last)，std::for_each但总是group加到最后。这意味着如果last不是groupon from的倍数it，您将移动it过去last，即 UB。

score 0 · Accepted Answer

您需要通过引用捕获，并且您需要在（最后一个）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。

此外，修复 DeadMG 发现的问题，最终得到以下代码。

它工作得很好，两个版本都给出了 n=100000000 的 156608294.151782。

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();});
}

score 0 · Accepted Answer

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();});
}

c++ - 并行使用 std::thread？

6 回答 6

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