为什么我的程序在1个线程上比在8个线程上运行更快？

Question

请看下面的代码：

#include <iostream>
#include <thread>
#include <numeric>
#include <algorithm>
#include <vector>
#include <chrono>

template<typename Iterator, typename T>
struct accumulate_block
{
    void operator()(Iterator begin, Iterator end, T& result)
    {
        result = std::accumulate(begin, end, result);
    }    
};

template<typename Iterator, typename T>
int accumulate_all(Iterator begin, Iterator end, T& init)
{
    auto numOfThreads = std::thread::hardware_concurrency();
    std::vector<std::thread> threads(numOfThreads);
    auto step = std::distance(begin, end) / numOfThreads;
    std::vector<int> results(numOfThreads,0);
    for(int i=0; i<numOfThreads-1; ++i)
    {
        auto block_end = begin;
        std::advance(block_end, step);
        threads[i] = std::thread(accumulate_block<Iterator, T>(), begin, block_end, std::ref(results[i]));
        begin = block_end;
    }
    threads[numOfThreads-1] = std::thread(accumulate_block<Iterator, T>(), begin, end, std::ref(results[numOfThreads-1]));
    for_each(threads.begin(), threads.end(), std::mem_fn(&std::thread::join));
    return accumulate(results.begin(), results.end(), 0);
}

int main()
{ 
   int x=0;
   std::vector<int> V(20000000,1);
   auto t1 = std::chrono::high_resolution_clock::now();
   //std::accumulate(std::begin(V), std::end(V), x); singe threaded option
   std::cout<<accumulate_all(std::begin(V), std::end(V), x);
   auto t2 = std::chrono::high_resolution_clock::now();
   std::cout << "process took: "
    << std::chrono::duration_cast<std::chrono::nanoseconds>(t2 - t1).count()
    << " nanoseconds\n";
    return 0;
}

当我在并发版本上运行时（基本上在8个线程上运行，因为我的std::thread::hardware_concurrency();返回8）
输出为：处理花费： 8895404 nanoseconds 。

但是单线程选项的输出为： process took: 124 nanoseconds

谁能解释这个奇怪的行为？

Answer 1

编译器删除对std::accumulate的调用，因为它没有副作用，并且不使用结果。

固定：

auto sum = std::accumulate(std::begin(V), std::end(V), x); // singe threaded option

// At the very end.
std::cout << sum << '\n';