windows 多线程问题 用windows api提供的函数创建4个线程 并用WaitForMultipleObjects 但结果有问题

Posted 2023-03-15

用windows api提供的函数创建4个线程 并用WaitForMultipleObjects 但结果有问题
这里附上代码和执行结果 求高手解答

#include <windows.h>
#include <process.h>
#include <iostream>

DWORD WINAPI thread_a(LPVOID hParameter)

printf("this is thread 1\n");
return 0;

DWORD WINAPI thread_b(LPVOID hParameter)

printf("this is thread 2\n");
return 0;

DWORD WINAPI thread_c(LPVOID hParameter)

printf("this is thread 3\n");
return 0;

DWORD WINAPI thread_d(LPVOID hParameter)

printf("this is thread 4\n");
return 0;


int main(int avg,char* arg)

HANDLE h[4];
DWORD dw;
h[0] = ::CreateThread(NULL,NULL,thread_a,NULL,0,&dw);
h[1] = ::CreateThread(NULL,NULL,thread_b,NULL,0,&dw);
h[2] = ::CreateThread(NULL,NULL,thread_c,NULL,0,&dw);
h[3] = ::CreateThread(NULL,NULL,thread_d,NULL,0,&dw);
::WaitForMultipleObjects(4,h,TRUE,INFINITE);
return 0;


4个线程 5个输出。。。 有个线程会被执行2次 求解决方案。。。

#include<windows.h>
#include<fstream.h>
#include<stdio.h>
#include<string>
#include<conio.h>

//定义一些常量；
//本程序允许的最大临界区数；
#define MAX_BUFFER_NUM 10
//秒到微秒的乘法因子；
#define INTE_PER_SEC 1000
//本程序允许的生产和消费线程的总数；
#define MAX_THREAD_NUM 64

//定义一个结构，记录在测试文件中指定的每一个线程的参数
struct ThreadInfo

int serial; //线程序列号
char entity; //是P还是C
double delay; //线程延迟
int thread_request[MAX_THREAD_NUM]; //线程请求队列
int n_request; //请求个数
;

//全局变量的定义

//临界区对象的声明,用于管理缓冲区的互斥访问；
CRITICAL_SECTION PC_Critical[MAX_BUFFER_NUM];
int Buffer_Critical[MAX_BUFFER_NUM]; //缓冲区声明，用于存放产品；
HANDLE h_Thread[MAX_THREAD_NUM]; //用于存储每个线程句柄的数组；
ThreadInfo Thread_Info[MAX_THREAD_NUM]; //线程信息数组；
HANDLE empty_semaphore; //一个信号量；
HANDLE h_mutex; //一个互斥量；
DWORD n_Thread = 0; //实际的线程的数目；
DWORD n_Buffer_or_Critical; //实际的缓冲区或者临界区的数目；
HANDLE h_Semaphore[MAX_THREAD_NUM]; //生产者允许消费者开始消费的信号量；

//生产消费及辅助函数的声明
void Produce(void *p);
void Consume(void *p);
bool IfInOtherRequest(int);
int FindProducePositon();
int FindBufferPosition(int);

int main(void)

//声明所需变量；
DWORD wait_for_all;
ifstream inFile;

//初始化缓冲区；
for(int i=0;i< MAX_BUFFER_NUM;i++)
Buffer_Critical[i] = -1;
//初始化每个线程的请求队列；
for(int j=0;j<MAX_THREAD_NUM;j++)
for(int k=0;k<MAX_THREAD_NUM;k++)
Thread_Info[j].thread_request[k] = -1;
Thread_Info[j].n_request = 0;

//初始化临界区；
for(i =0;i< MAX_BUFFER_NUM;i++)
InitializeCriticalSection(&PC_Critical[i]);

//打开输入文件，按照规定的格式提取线程等信息；
inFile.open("test.txt");
//从文件中获得实际的缓冲区的数目；
inFile >> n_Buffer_or_Critical;
inFile.get();
printf("输入文件是:\n");
//回显获得的缓冲区的数目信息；
printf("%d \n",(int) n_Buffer_or_Critical);
//提取每个线程的信息到相应数据结构中；
while(inFile)
inFile >> Thread_Info[n_Thread].serial;
inFile >> Thread_Info[n_Thread].entity;
inFile >> Thread_Info[n_Thread].delay;
char c;
inFile.get(c);
while(c!='\n'&& !inFile.eof())
inFile>> Thread_Info[n_Thread].thread_request[Thread_Info[n_Thread].n_request++];
inFile.get(c);

n_Thread++;


//回显获得的线程信息，便于确认正确性；
for(j=0;j<(int) n_Thread;j++)
int Temp_serial = Thread_Info[j].serial;
char Temp_entity = Thread_Info[j].entity;
double Temp_delay = Thread_Info[j].delay;
printf(" \n thread%2d %c %f ",Temp_serial,Temp_entity,Temp_delay);
int Temp_request = Thread_Info[j].n_request;
for(int k=0;k<Temp_request;k++)
printf(" %d ", Thread_Info[j].thread_request[k]);
cout<<endl;

printf("\n\n");

//创建在模拟过程中几个必要的信号量
empty_semaphore=CreateSemaphore(NULL,n_Buffer_or_Critical,n_Buffer_or_Critical,
"semaphore_for_empty");
h_mutex = CreateMutex(NULL,FALSE,"mutex_for_update");

//下面这个循环用线程的ID号来为相应生产线程的产品读写时所
//使用的同步信号量命名；
for(j=0;j<(int)n_Thread;j++)
std::string lp ="semaphore_for_produce_";
int temp =j;
while(temp)
char c = (char)(temp%10);
lp+=c;
temp/=10;

h_Semaphore[j+1]=CreateSemaphore(NULL,0,n_Thread,lp.c_str());


//创建生产者和消费者线程；
for(i =0;i< (int) n_Thread;i++)
if(Thread_Info[i].entity =='P')
h_Thread[i]= CreateThread(NULL,0,(LPTHREAD_START_ROUTINE)(Produce),
&(Thread_Info[i]),0,NULL);
else
h_Thread[i]=CreateThread(NULL,0,(LPTHREAD_START_ROUTINE)(Consume),
&(Thread_Info[i]),0,NULL);


//主程序等待各个线程的动作结束；
wait_for_all = WaitForMultipleObjects(n_Thread,h_Thread,TRUE,-1);
printf(" \n \nALL Producer and consumer have finished their work. \n");
printf("Press any key to quit!\n");
_getch();
return 0;


//确认是否还有对同一产品的消费请求未执行；
bool IfInOtherRequest(int req)

for(int i=0;i<n_Thread;i++)
for(int j=0;j<Thread_Info[i].n_request;j++)
if(Thread_Info[i].thread_request[j] == req)
return TRUE;

return FALSE;


//找出当前可以进行产品生产的空缓冲区位置；
int FindProducePosition()

int EmptyPosition;
for (int i =0;i<n_Buffer_or_Critical;i++)
if(Buffer_Critical[i] == -1)
EmptyPosition = i;
//用下面这个特殊值表示本缓冲区正处于被写状态；
Buffer_Critical[i] = -2;
break;

return EmptyPosition;


//找出当前所需生产者生产的产品的位置；
int FindBufferPosition(int ProPos)

int TempPos;
for (int i =0 ;i<n_Buffer_or_Critical;i++)
if(Buffer_Critical[i]==ProPos)
TempPos = i;
break;

return TempPos;


//生产者进程
void Produce(void *p)

//局部变量声明；
DWORD wait_for_semaphore,wait_for_mutex,m_delay;
int m_serial;

//获得本线程的信息；
m_serial = ((ThreadInfo*)(p))->serial;
m_delay = (DWORD)(((ThreadInfo*)(p))->delay *INTE_PER_SEC);

Sleep(m_delay);
//开始请求生产
printf("Producer %2d sends the produce require.\n",m_serial);

//确认有空缓冲区可供生产，同时将空位置数empty减1；用于生产者和消费者的同步；
wait_for_semaphore = WaitForSingleObject(empty_semaphore,-1);

//互斥访问下一个可用于生产的空临界区，实现写写互斥；
wait_for_mutex = WaitForSingleObject(h_mutex,-1);
int ProducePos = FindProducePosition();
ReleaseMutex(h_mutex);

//生产者在获得自己的空位置并做上标记后，以下的写操作在生产者之间可以并发；
//核心生产步骤中，程序将生产者的ID作为产品编号放入，方便消费者识别;
printf("Producer %2d begin to produce at position %2d.\n",m_serial,ProducePos);
Buffer_Critical[ProducePos] = m_serial;
printf("Producer %2d finish producing :\n ",m_serial);
printf(" position[ %2d ]:%3d \n" ,ProducePos,Buffer_Critical[ProducePos]);
//使生产者写的缓冲区可以被多个消费者使用，实现读写同步；
ReleaseSemaphore(h_Semaphore[m_serial],n_Thread,NULL);


//消费者进程
void Consume(void * p)

//局部变量声明；
DWORD wait_for_semaphore,m_delay;
int m_serial,m_requestNum; //消费者的序列号和请求的数目；
int m_thread_request[MAX_THREAD_NUM];//本消费线程的请求队列；

//提取本线程的信息到本地；
m_serial = ((ThreadInfo*)(p))->serial;
m_delay = (DWORD)(((ThreadInfo*)(p))->delay *INTE_PER_SEC);
m_requestNum = ((ThreadInfo *)(p))->n_request;
for (int i = 0;i<m_requestNum;i++)
m_thread_request[i] = ((ThreadInfo*)(p))->thread_request[i];

Sleep(m_delay);
//循环进行所需产品的消费
for(i =0;i<m_requestNum;i++)

//请求消费下一个产品
printf("Consumer %2d request to consume %2d product\n",m_serial,m_thread_request[i]);
//如果对应生产者没有生产，则等待；如果生产了,允许的消费者数目-1；实现了读写同步；
wait_for_semaphore=WaitForSingleObject(h_Semaphore[m_thread_request[i]],-1);

//查询所需产品放到缓冲区的号
int BufferPos=FindBufferPosition(m_thread_request[i]);

//开始进行具体缓冲区的消费处理，读和读在该缓冲区上仍然是互斥的；
//进入临界区后执行消费动作；并在完成此次请求后，通知另外的消费者本处请求已
//经满足；同时如果对应的产品使用完毕，就做相应处理；并给出相应动作的界面提
//示；该相应处理指将相应缓冲区清空，并增加代表空缓冲区的信号量；
EnterCriticalSection(&PC_Critical[BufferPos]);

printf("Consumer%2d begin to consume %2d product \n",m_serial,m_thread_request[i]);
((ThreadInfo*)(p))->thread_request[i] =-1;

if(!IfInOtherRequest(m_thread_request[i]))
Buffer_Critical[BufferPos] = -1;//标记缓冲区为空；
printf("Consumer%2d finish consuming %2d:\n ",m_serial,m_thread_request[i]);
printf(" position[ %2d ]:%3d \n" ,BufferPos,Buffer_Critical[BufferPos]);
ReleaseSemaphore(empty_semaphore,1,NULL);

else
printf("Consumer %2d finish consuming product %2d\n ",m_serial,m_thread_request[i]);

//离开临界区
LeaveCriticalSection(&PC_Critical[BufferPos]);




看函数就行
参考技术A 控制台是个共享资源，要进行互斥访问，修改如下：
#include <windows.h>
#include <process.h>
#include <iostream>
HANDLE mute=::CreateMutex(0,0,0);
DWORD WINAPI thread_a(LPVOID hParameter)
::WaitForSingleObject(mute,-1);
printf("this is thread 1\n");
::ReleaseMutex(mute);
return 0;

DWORD WINAPI thread_b(LPVOID hParameter)

::WaitForSingleObject(mute,-1);
printf("this is thread 2\n");
::ReleaseMutex(mute);
return 0;

DWORD WINAPI thread_c(LPVOID hParameter)
::WaitForSingleObject(mute,-1);
printf("this is thread 3\n");
::ReleaseMutex(mute);
return 0;

DWORD WINAPI thread_d(LPVOID hParameter)
::WaitForSingleObject(mute,-1);
printf("this is thread 4\n");
::ReleaseMutex(mute);
return 0;


int main(int avg,char* arg)

HANDLE h[4];
DWORD dw;
h[0] = ::CreateThread(NULL,NULL,thread_a,NULL,0,&dw);
h[1] = ::CreateThread(NULL,NULL,thread_b,NULL,0,&dw);
h[2] = ::CreateThread(NULL,NULL,thread_c,NULL,0,&dw);
h[3] = ::CreateThread(NULL,NULL,thread_d,NULL,0,&dw);
::WaitForMultipleObjects(4,h,TRUE,INFINITE);
return 0;
本回答被提问者采纳