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可能重复：
并行使用 std::thread？

考虑以下实现 a 的代码parallel_for：

// 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 = 100)
{
    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]) : (7);
    const unsigned int n = (argc > 2) ? std::atol(argv[2]) : (1000);
    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;
}

在 ideone（http://ideone.com/GrYJsQ）上它给出：

1567.22616587821 1567.22616587821

而在我使用 g++ 4.6 的计算机上，它给出 (g++ -O3 -std=c++0x parallel_for.cpp -o parallel_for -lpthread和./parallel_for) ：

1567.22616587821 499500

问题出在哪里 ？

4

0 回答 0