以下情况，请问我该怎么修改呢？有了解的大佬么？

Question

#include <windows.h> //包含头文件#include <stdio.h>DWORD WINAPI myfun1( //声明线程函数LPVOID lpParameter  );const int stop=1; //stop 为一const int go_on=2; //开始为2int record=2;  static HANDLE hmutex;int a=0; //定义全局变量aint main(){  HANDLE h1;  //定义线程句柄h1=::CreateThread(NULL,0,myfun1,NULL,0,NULL);printf("haha\n");//创建线程1::CloseHandle(h1);  Sleep(1000);//关闭线程句柄对象  return 0;}DWORD WINAPI myfun1(LPVOID lpParameter) //线程函数1{ while(1){::WaitForSingleObject(hmutex,INFINITE); //请求事件对象//变量自加a++; //线程睡眠1秒printf("qiubai\n");::ReleaseMutex(hmutex);  if(a>0)break;//释放互斥对象句柄}return 0; //线程}

Answer 1

你应该建立两个进程, 这样才可以观察, 还有你的 haha 是在 main 函数里面, 只执行一次, 不会输出多次, qiubai, 在进程中, 但是 if (a>0) 也就是执行一次之后这个进程虽然没有关闭但是, 已经不可能进入输出的程序了.

#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;}