? WindowsAPI里的多線程簡單例子，先記下來，以后再學(xué)學(xué)
#include?<windows.h>
#include?<iostream>
using?namespace?std;
//線程處理函數(shù)
DWORD?WINAPI?Fun1(LPVOID?lpParam);
DWORD?WINAPI?Fun2(LPVOID?lpParam);
int?tickets?=?20;
void?main()
{
????HANDLE?hThread1,?hThread2;
????hThread1?=?CreateThread(NULL,?0,?Fun1,?NULL,?0,?NULL);
????hThread2?=?CreateThread(NULL,?0,?Fun2,?NULL,?0,?NULL);
????CloseHandle(hThread1);
????CloseHandle(hThread2);
????cin.get();
}
DWORD?WINAPI?Fun1(LPVOID?lpParam)
{
????while(true)
????{
????????if(tickets?>?0)
????????????cout<<"Thread1?sell?ticket?:?"<<tickets--<<endl;
????????else?break;
????}
????return?0;
}
DWORD?WINAPI?Fun2(LPVOID?lpParam)
{
????while(true)
????{
????????if(tickets?>?0)
????????????cout<<"Thread2?sell?ticket?:?"<<tickets--<<endl;
????????else?break;
????}
????return?0;
}
這里在main里用兩個(gè)函數(shù)Fun1和Fun2建了兩個(gè)線程，有一個(gè)全局變量ticket＝20表示20張票，分由兩個(gè)線程進(jìn)行出售。
但是這個(gè)程序運(yùn)行起來結(jié)果會(huì)出現(xiàn)混亂和不確定：
原因很簡單，在操作系統(tǒng)里都學(xué)過關(guān)于同步的問題，這里的兩個(gè)線程之間就沒有對(duì)臨界資源ticket進(jìn)行同步。類似于操作系統(tǒng)里的PV操作，可以將程序改為：
#include?<windows.h>
#include?<iostream>
using?namespace?std;
//線程處理函數(shù)
DWORD?WINAPI?Fun1(LPVOID?lpParam);
DWORD?WINAPI?Fun2(LPVOID?lpParam);
int?tickets?=?20;
HANDLE?hMutex;
void?main()
{
????hMutex?=?CreateMutex(NULL,?false,?NULL);
????HANDLE?hThread1,?hThread2;
????hThread1?=?CreateThread(NULL,?0,?Fun1,?NULL,?0,?NULL);
????hThread2?=?CreateThread(NULL,?0,?Fun2,?NULL,?0,?NULL);
????CloseHandle(hThread1);
????CloseHandle(hThread2);
????cin.get();
}
DWORD?WINAPI?Fun1(LPVOID?lpParam)
{
????while(true)
????{
????????WaitForSingleObject(hMutex,?INFINITE);
????????if(tickets?>?0)
????????????cout<<"Thread1?sell?ticket?:?"<<tickets--<<endl;
????????else?break;
????????ReleaseMutex(hMutex);
????}
????return?0;
}
DWORD?WINAPI?Fun2(LPVOID?lpParam)
{
????while(true)
????{
????????WaitForSingleObject(hMutex,?INFINITE);
????????if(tickets?>?0)
????????????cout<<"Thread2?sell?ticket?:?"<<tickets--<<endl;
????????else?break;
????????ReleaseMutex(hMutex);
????}
????return?0;
}
這里用了一個(gè)信息量Mutex來對(duì)ticket進(jìn)行同步操作，進(jìn)程要操作ticket之前必須先獲得信號(hào)量，使用完之后釋放以使整個(gè)過程能繼續(xù)下去。
最后得到正常的結(jié)果為：
