一道经典多线程题

来源：互联网发布：周扬青开的淘宝店铺编辑：程序博客网时间：2024/05/02 16:16

问题的描述

启动3个线程打印递增的数字, 线程1先打印1,2,3,4,5, 然后是线程2打印6,7,8,9,10, 然后是线程3打印11,12,13,14,15. 接着再由线程1打印16,17,18,19,20....以此类推, 直到打印到75. 程序的输出结果应该为:

线程1: 1

线程1: 2

线程1: 3

线程1: 4

线程1: 5

线程2: 6

线程2: 7

线程2: 8

线程2: 9

线程2: 10

...

线程3: 71

线程3: 72

线程3: 73

线程3: 74

线程3: 75

java实现见http://coolxing.iteye.com/blog/1236696

这里我用c++实现。大致思路跟http://blog.csdn.net/jlncu/article/details/8891978里的差不多，代码结构一样。

线程顺次触发，全局变量每次加1，每个线程输出5个数后触发下个线程。

#include <windows.h>#include <iostream>using namespace std;const int THREAD_NUM = 3;//线程个数  HANDLE g_hThreadEvent[THREAD_NUM];  //互斥事件  int LOOP = 1;DWORD WINAPI fun1(LPVOID lpParam);DWORD WINAPI fun2(LPVOID lpParam);DWORD WINAPI fun3(LPVOID lpParam);void main(){int i = 0;HANDLE hdl[THREAD_NUM];//线程句柄//事件开启的时候是没有触发的  for (int i = 0; i < THREAD_NUM; i++)   g_hThreadEvent[i] = CreateEvent(NULL,FALSE,FALSE,NULL);  //线程句柄hdl[0] = ::CreateThread(NULL, 0, fun1, NULL, 0, NULL); hdl[1] = ::CreateThread(NULL, 0, fun2, NULL, 0, NULL); hdl[2] = ::CreateThread(NULL, 0, fun3, NULL, 0, NULL);       //触发第一个线程，线程函数会按顺序自己触发下一个要执行的线程      SetEvent(g_hThreadEvent[0]);        WaitForMultipleObjects(THREAD_NUM , hdl, TRUE, INFINITE);        //清理      for (i = 0; i < THREAD_NUM; i++)      {          CloseHandle(hdl[i]);          CloseHandle(g_hThreadEvent[i]);      }  getchar();}DWORD WINAPI fun1(LPVOID lpParam){while(true){WaitForSingleObject(g_hThreadEvent[0],INFINITE);if(LOOP < 76){int i = 0;while(i < 5){cout<<"线程1："<<LOOP++<<endl;i++;}::SetEvent(g_hThreadEvent[1]);}else{break;}}return 0;}DWORD WINAPI fun2(LPVOID lpParam){while(true){WaitForSingleObject(g_hThreadEvent[1],INFINITE); if(LOOP < 76){int i = 0;while(i < 5){cout<<"线程2："<<LOOP++<<endl;i++;}::SetEvent(g_hThreadEvent[2]);}else{break;}}return 0;}DWORD WINAPI fun3(LPVOID lpParam){while(true){WaitForSingleObject(g_hThreadEvent[2],INFINITE); if(LOOP < 76){int i = 0;while(i < 5){cout<<"线程3："<<LOOP++<<endl;i++;}::SetEvent(g_hThreadEvent[0]);}else{break;}}return 0;