在没有互斥体的情况下在C++ 11中实现共享整数计数器的最简单方法:

Question

假设我们有以下代码来计算出现事件的次数:

int i=0;
void f() {
   // do stuff  . . .
   if(something_happens) ++i;
}

int main() {
    std::vector<std::thread> threads;
    for(int j = 0; j< std::thread::hardware_concurrency(); ++j) {
        threads.push_back(std::thread(f));
    }

    std::for_each(threads.begin(), threads.end(), std::mem_fn(&std::thread_join));
    std::cout << "i = " << i << '\n';
}

目前看来我的竞争条件很明显.使用C++ 11,什么是(1)消除这种竞争条件的最简单方法,以及(2)最快的方法？,最好不使用互斥锁.谢谢.

更新:使用注释使用原子,我得到了一个工作程序,编译在英特尔编译器,版本13:

#include <iostream>
#include <thread>
#include <vector>
#include <atomic>
#include <algorithm>

std::atomic<unsigned long long> i = 0;

void f(int j) {
    if(j%2==0) {
        ++i;
    }  
}

int main() {
    std::cout << "Atomic i = " << i << "\n";
    int numThreads = 8; //std::thread::hardware_concurrency() not yet implemented by Intel
    std::vector<std::thread> threads;
    for(int k=0; k< numThreads; ++k) {
        threads.push_back(std::thread(f, k));
    }

    std::for_each(threads.begin(), threads.end(), std::mem_fn(&std::thread::join));
        std::cout << "Atomic i = " << i << "\n";
    }

Answer 1

您可能想要查看原子类型.您可以在不需要锁/互斥锁的情况下访问它们.