//<pre name="code" class="cpp">//这是2个线程模拟卖火车票的小程序
#include <windows.h>
#include <iostream>
using namespace std;
DWORD WINAPI Fun1Proc(LPVOID lpParameter);//thread data
DWORD WINAPI Fun2Proc(LPVOID lpParameter);//thread data
int index=0;
int tickets=10;
HANDLE hMutex;
int main()
{
HANDLE hThread1;
HANDLE hThread2;
HANDLE m_hMutex;
//创建线程
hThread1=CreateThread(NULL,0,Fun1Proc,NULL,0,NULL);
hThread2=CreateThread(NULL,0,Fun2Proc,NULL,0,NULL);
CloseHandle(hThread1);
CloseHandle(hThread2);
//创建互斥对象
hMutex=CreateMutex(NULL,TRUE,"tickets");
if (hMutex)
{
if (ERROR_ALREADY_EXISTS==GetLastError())//得到错误:已经存在
{
CloseHandle(m_hMutex);
m_hMutex = NULL;
cout<<"only one instance can run!"<<endl;
return 0;
}
}
WaitForSingleObject(hMutex,INFINITE);
ReleaseMutex(hMutex);//申请了两次就要施放两次(没搞懂在哪里申请了两次?难道建立一次,wait也算一次?)
ReleaseMutex(hMutex);//谁申请谁施放
Sleep(4000);//让主线程睡4秒,让其他线程有时间执行完他们的代码,如果不睡就会出现其他线程执行不完或出错的情况,但时如果不知道线程需要多少时间执行,那应该写多少时间?
}
//线程1的入口函数
DWORD WINAPI Fun1Proc(LPVOID lpParameter)//thread data
{
while (true)//无限循环线程
{
WaitForSingleObject(hMutex,INFINITE);//得到互斥体才能执行
if (tickets>0)
{
Sleep(1);
cout<<"thread1 sell ticket :"<<tickets--<<endl;
}
else
break;
ReleaseMutex(hMutex);//施放互斥体
}
return 0;
}
//线程2的入口函数
DWORD WINAPI Fun2Proc(LPVOID lpParameter)//thread data
{
while (true)
{
WaitForSingleObject(hMutex,INFINITE);
if (tickets>0)
{
Sleep(1);
cout<<"thread2 sell ticket :"<<tickets--<<endl;
}
else
break;
ReleaseMutex(hMutex);
}
return 0;
}
食用须知:从网上找的代码,有错,已调试成功。
https://me.csdn.net/mnhdxhcky