原生并行版std::accumulate
代码来自《c++并发编程实战》
#include <iostream>
#include <numeric>
#include <algorithm>
#include <thread>
#include <functional>
#include <vector>
#include <chrono>
typedef long long LL;
template<typename Iterator, typename T>
struct accumulate_block{
void operator()(Iterator first, Iterator last, T& result){
result = std::accumulate(first, last, result);
}
};
template<typename Iterator, typename T>
T parallel_accumulate(Iterator first, Iterator last, T init){
unsigned long const length = std::distance(first, last);
// 若输入范围为空 返回init的值
if(!length)
return init;
// 确定启动的线程数
unsigned long const min_per_thread = 25;
unsigned long const max_threads = (length + min_per_thread - 1) / min_per_thread;
// 能并发在一个程序的线程数量
unsigned long const hardware_threads = std::thread::hardware_concurrency();
// 确定最终的线程数量 若hardware_concurrency返回0则选择两个线程
unsigned long const num_threads = std::min(hardware_threads != 0 ? hardware_threads : 2, max_threads);
// 每个线程中处理的元素数量
unsigned long const block_size = length / num_threads;
std::vector<T> results(num_threads); // 存放中间结果
std::vector<std::thread> threads(num_threads-1); // 已有主线程 -1
Iterator block_start = first;
for(unsigned long i = 0; i < (num_threads - 1); i++){
Iterator block_end = block_start;
std::advance(block_end, block_size); // block_end指向当前块的末尾
// 启动一个新线程为当前块累加结果
threads[i] = std::thread(accumulate_block<Iterator, T>(), block_start, block_end, std::ref(results[i]));
block_start = block_end; // 当迭代器指向当前块末尾时启动一下一个块
}
// 处理最终块的结果
accumulate_block<Iterator, T>()(block_start, last, results[num_threads - 1]);
// 等待所有线程执行完毕
std::for_each(threads.begin(), threads.end(), std::mem_fn(&std::thread::join));
// 返回最终结果
return std::accumulate(results.begin(), results.end(), init);
}
int main(){
std::vector<LL> v;
LL sum = 0;
for(LL i = 0;i <= 10000000; i++) v.push_back(i);
auto beforeTime = std::chrono::steady_clock::now();
//std::cout << parallel_accumulate(v.begin(), v.end(), sum) << std::endl;
//std::cout << std::accumulate(v.begin(), v.end(), sum) << std::endl;
auto afterTime = std::chrono::steady_clock::now();
double duration_millsecond = std::chrono::duration<double, std::milli>(afterTime - beforeTime).count();
std::cout << duration_millsecond << std::endl;
return 0;
}
0~10000000相加时测试如下。
