C++11特性原子類型——多原子類型並行訪問的序列化測試

/*C++11 原子類型測試  問題：如果有多個原子需要操作，如何保障並行的序列化*/  #include <thread>#include <stdlib.h>#include <atomic> #include <iostream>#include <time.h>#include <vector>#include "mytimer.h"using namespace std;//用原子資料類型作為共用資源的資料類型atomic_long total(0);atomic_int i(0);//long total = 0;TimeVal a, b;void click(){//由於有兩個原子類型，必須保證更新的順序性    while (i++ < 10000000)  //原子操作，保障了順序操作        total += 1;        ///*錯誤while(i < 1000000)   //一次讀取{i++;    //上次讀取i到這裡未防止再次讀取i的值total++;}*/}//如果用單線程計算的話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]);//主線程測試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;//單線程測試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;    // 計時開始    get_current_time( a );    // 建立N個線程類比點擊統計    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();    // 計時結束    get_current_time( b );    // 輸出結果    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檔案

#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

編譯：

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

 

運行結果：

運行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.

 

運行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.

 

運行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.

 

運行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.

 

從最後結果看，每個線程由於任務少了，所用時間短了，但總的系統用於線程調度的時間多了。