C++11特性原子类型——多原子类型并行访问的串行化测试

/*C++11 原子类型测试
  问题：如果有多个原子需要操作，如何保障并行的序列化
*/  

#include <thread>
#include <stdlib.h>
#include <atomic> 
#include <iostream>
#include <time.h>
#include <vector>
#include "mytimer.h"

using namespace std;

//用原子数据类型作为共享资源的数据类型
atomic_long total(0);
atomic_int i(0);
//long total = 0;

TimeVal a, b;

void click()
{
	//由于有两个原子类型，必须保证更新的顺序性
    while (i++ < 10000000)  //原子操作，保障了顺序操作
        total += 1;        //
	/*错误
		while(i < 1000000)   //一次读取
		{
			i++;    //上次读取i到这里未防止再次读取i的值
			total++;
		}
	*/
}

//如果用单线程计算的话
long commonclick()
{	
	long total = 0;
	for(int i = 0; i < 	10000000; i++)
		total++;
	return total;
}

int main(int argc, char* argv[])
{
	int N = 16;
	double time0, time1;
	if (argc == 2)
		N = atoi(argv[1]);
	
	//主线程测试
	get_current_time( a );
	long val = commonclick();
	get_current_time( b );
	cout << "Main thread" << endl;
	cout<<"result:"<<val<<endl;
	time0 = getDeltaTime(b, a) * 1e3;
    cout<<"duration:"<< time0 <<"ms"<<endl;	
	
	cout << "Single thread" << endl;
	//单线程测试
	get_current_time( a );
	thread t1(commonclick);
	t1.join();
	get_current_time( b );
    cout<<"duration:"<< getDeltaTime(b, a) * 1e3 <<"ms"<<endl;	
	
	cout << "Multi thread " << N << endl;
    // 计时开始
    get_current_time( a );
    // 创建N个线程模拟点击统计
    vector<thread> threads(N);
    for(int i=0; i<N;++i) 
    {
        threads[i] = thread(click);
    }
    for(int i=0; i<N;++i) 
		threads[i].join();
    // 计时结束
    get_current_time( b );
    // 输出结果
    cout<<"result:"<<total<<endl;
	time1 = getDeltaTime(b, a) * 1e3;
    cout<<"duration:"<< time1 <<"ms"<<endl;
	
	cout << "Delay for each thread: " << time1 / N - time0 << "ms, about " << time1/N / time0 << " time of main thread." << endl;

    return 0;
}

mytimer.h文件

#ifndef __MYGLIB_H__
#define __MYGLIB_H__

#include <windows.h>

typedef LARGE_INTEGER TimeVal;
LARGE_INTEGER freq;

void get_current_time(TimeVal& s)
{
	QueryPerformanceFrequency(&freq);
	QueryPerformanceCounter(&s);    
}

double getDeltaTime(TimeVal t2, TimeVal t1)
{
	double time = (double)(t2.QuadPart-t1.QuadPart)/(double)freq.QuadPart;  
	return time;
}

#endif

编译：

g++ -std=c++11 atomic_par.cpp

运行结果：

运行a 2

Main thread
result:10000000
duration:25.7743ms
Single thread
duration:29.4485ms
Multi thread 2
result:10000000
duration:372.861ms
Delay for each thread: 160.656ms, about 7.23319 time of main thread.

运行a 4

Main thread
result:10000000
duration:22.2332ms
Single thread
duration:23.3911ms
Multi thread 4
result:10000000
duration:472.065ms
Delay for each thread: 95.783ms, about 5.3081 time of main thread.

运行a 8

Main thread
result:10000000
duration:22.2252ms
Single thread
duration:22.7899ms
Multi thread 8
result:10000000
duration:468.485ms
Delay for each thread: 36.3355ms, about 2.63488 time of main thread.

运行a 16

Main thread
result:10000000
duration:22.399ms
Single thread
duration:23.7603ms
Multi thread 16
result:10000000
duration:489.68ms
Delay for each thread: 8.20599ms, about 1.36636 time of main thread.

从最后结果看，每个线程由于任务少了，所用时间短了，但总的系统用于线程调度的时间多了。