请看这段代码:
#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(基本上是在 8 个线程上,因为我的 std::thread::hardware_concurrency();
返回 8)
输出是:进程耗时:8895404 纳秒
。
但是单线程选项输出是:process toked: 124 nanoseconds
谁能解释这种奇怪的行为??
最佳答案
编译器移除了对 std::accumulate
的调用,因为它没有副作用并且不使用结果。
修复:
auto sum = std::accumulate(std::begin(V), std::end(V), x); // singe threaded option
// At the very end.
std::cout << sum << '\n';
关于c++ - 为什么我的程序在 1 个线程上比在 8.C++ 上运行得更快,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/49978209/