Win32API、MFC、.NET并行求和运算（星星笔记）

1.运用Win32API实现并行求和运算，程序如下：

// Win32APIadd.cpp : 定义控制台应用程序的入口点。//#include "stdafx.h"#include <Windows.h>#include "time.h"HANDLE finish[2];HANDLE finish2;long long sum[2];DWORD WINAPI ThreadOne(LPVOID param){for (long i =1;i<= 500000000;i++){sum[0] +=i;}SetEvent(finish[0]);return 0;}DWORD WINAPI ThreadTwo(LPVOID param){for (long i =500000001;i<= 1000000000;i++){sum[1] +=i;}SetEvent(finish[1]);return 0;}DWORD WINAPI ThreadThree(LPVOID param){long long sumserial = 0;for (long i =1;i<= 1000000000;i++){sumserial +=i;}printf("sumserial=%lld\n",sumserial);SetEvent(finish2);return 0;}int _tmain(int argc, _TCHAR* argv[]){clock_t start = clock();long long sumand = 0;finish[0] = CreateEvent(NULL ,false,false,NULL);finish[1] = CreateEvent(NULL ,false,false,NULL);finish2 = CreateEvent(NULL ,false,false,NULL);HANDLE thread1 = CreateThread(NULL,0,ThreadOne,NULL,0,NULL);HANDLE thread2 = CreateThread(NULL,0,ThreadTwo,NULL,0,NULL);WaitForMultipleObjects(2,finish,true,INFINITE);for (int i=0;i<2;i++){sumand +=sum[i];}clock_t end = clock();printf("sumand=%lld\n",sumand);printf("andtime=%d\n",end - start);clock_t start2 = clock();HANDLE thread3 = CreateThread(NULL,0,ThreadThree,NULL,0,NULL);WaitForSingleObject(finish2,INFINITE);clock_t end2 = clock();printf("serialtime=%d\n",end2 - start2);system("pause");return 0;}

运行结果如下：

2.运用MFC实现并行求和，程序如下：

// MFC_A_ADD.cpp : 定义控制台应用程序的入口点。//#include "stdafx.h"#include <afxmt.h>#include <iostream>#include <afxwin.h>#include "time.h"using namespace std;long long sum[2] = {0,0} ;CEvent faxEvent1(false); CEvent faxEvent2(false);CEvent faxEvent(false);UINT threadProc4(LPVOID param){long long temp = 0;for(long i=1;i<=500000000;i++){temp  += i;}sum[0] = temp;SetEvent(faxEvent1);return 0;}UINT threadProc5(LPVOID param){long long temp = 0;for(long i=500000001;i<=1000000000;i++){temp += i;}sum[1] = temp;SetEvent(faxEvent2);return 0;}UINT threadProc6(LPVOID param){long long sum = 0;for(long i=1;i<=1000000000;i++){sum +=i;}printf("sumserial=%lld\n",sum);SetEvent(faxEvent);return 0;}int _tmain(int argc, _TCHAR* argv[]){long long sumand = 0;clock_t start = clock();AfxBeginThread(threadProc4,NULL);AfxBeginThread(threadProc5,NULL);WaitForSingleObject(faxEvent1,INFINITE);WaitForSingleObject(faxEvent2,INFINITE);for (int i=0;i<2;i++){sumand+=sum[i];}clock_t end = clock();printf("sumand=%lld\n",sumand);printf("paralleltime=%d\n",end - start);clock_t start2 = clock();AfxBeginThread(threadProc6,NULL);WaitForSingleObject(faxEvent,INFINITE);clock_t end2 = clock();printf("serialtime=%d\n",end2 - start2);system("pause");return 0;}

运行结果如下：

3.运用.NET实现并行求和，程序如下：

using System;using System.Collections.Generic;using System.Linq;using System.Text;using System.Threading;using System.Diagnostics;namespace NETadd{    class Program    {        static void Main(string[] args)        {            Stopwatch stopwatch = new Stopwatch();            Work work1 = new Work(1);            ThreadStart thread1 = new ThreadStart(work1.pSumto);            Thread newthread1 = new Thread(thread1);            Work work2 = new Work(2);            ThreadStart thread2 = new ThreadStart(work2.pSumto);            Thread newthread2 = new Thread(thread2);            stopwatch.Start();            newthread1.Start();            newthread2.Start();            newthread1.Join();            newthread2.Join();            stopwatch.Stop();            TimeSpan timeSpan = stopwatch.Elapsed;            double milliseconds = timeSpan.TotalMilliseconds;            Console.Write("parallel sum = {0}\n",(work1.getSum()+work2.getSum()));            Console.Write("parallel time =");            Console.Write(milliseconds);            stopwatch.Start();            Console.Write("\nserial sum = {0}\n", new Work(1).sumto());            stopwatch.Stop();            TimeSpan timeSpan2 = stopwatch.Elapsed;            double milliseconds2 = timeSpan2.TotalMilliseconds;            Console.Write("serial time =");            Console.Write(milliseconds2);            Console.Read();        }    }    class Work    {        private long sum = 0;        private long start;        public Work(long i)        {            this.start = i;        }        public void pSumto()        {            for (long i = start; i <= 1000000000; i += 2)            {                sum += i;            }        }        public long sumto()        {            long sumto = 0;            for (long i = start; i <= 1000000000; i++)            {                sumto += i;            }            return sumto;        }        public long getSum()        {            return sum;        }    }}

运行结果如下：