#include <windows.h> //包含頭文件#include <stdio.h>DWORD WINAPI myfun1( //聲明線程函數LPVOID lpParameter );const int stop=1; //stop 為一const int go_on=2; //開始為2int record=2; static HANDLE hmutex;int a=0; //定義全局變量aint main(){ HANDLE h1; //定義線程句柄h1=::CreateThread(NULL,0,myfun1,NULL,0,NULL);printf("haha\n");//創建線程1::CloseHandle(h1); Sleep(1000);//關閉線程句柄對象 return 0;}DWORD WINAPI myfun1(LPVOID lpParameter) //線程函數1{ while(1){::WaitForSingleObject(hmutex,INFINITE); //請求事件對象//變量自加a++; //線程睡眠1秒printf("qiubai\n");::ReleaseMutex(hmutex); if(a>0)break;//釋放互斥對象句柄}return 0; //線程}
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慕容708150
TA貢獻1831條經驗 獲得超4個贊
你應該建立兩個進程, 這樣才可以觀察, 還有你的 haha 是在 main 函數里面, 只執行一次, 不會輸出多次, qiubai, 在進程中, 但是 if (a>0) 也就是執行一次之后這個進程雖然沒有關閉但是, 已經不可能進入輸出的程序了.
#include <windows.h>#include <stdio.h>DWORD WINAPI myfun1(LPVOID lpParameter);DWORD WINAPI myfun2(LPVOID lpParameter);static HANDLE hmutex, hmutex2;int main(){ HANDLE h1; h1=::CreateThread(NULL,0,myfun1,NULL,0,NULL); HANDLE h2; h2=::CreateThread(NULL,0,myfun2,NULL,0,NULL); printf("Start:\n"); Sleep(5000); ::CloseHandle(h1); ::CloseHandle(h2); return 0;}DWORD WINAPI myfun1(LPVOID lpParameter){ int a = 0; while (1) { ::WaitForSingleObject(hmutex,INFINITE); a++; printf("qiubai\n"); ::ReleaseMutex(hmutex); Sleep(200); if(a > 15) break; } return 0;}DWORD WINAPI myfun2(LPVOID lpParameter){ int b = 0; while (1) { ::WaitForSingleObject(hmutex2,INFINITE); b++; printf("haha\n"); ::ReleaseMutex(hmutex2); Sleep(300); if(b > 10) break; } return 0;} |
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