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标题:急~可不可以帮我看看这个程序能不能运行~
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lily_0406
Rank: 1
等 级:新手上路
帖 子:6
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注 册:2007-4-9
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 问题点数:0 回复次数:2 
急~可不可以帮我看看这个程序能不能运行~
#include <windows.h>
#include <iostream>

const unsigned short SIZE_OF_BUFFER = 10; //缓冲区长度
unsigned short ProductID = 0; //产品号
unsigned short ConsumeID = 0; //将被消耗的产品号
unsigned short in = 0; //产品进缓冲区时的缓冲区下标
unsigned short out = 0; //产品出缓冲区时的缓冲区下标

int g_buffer[SIZE_OF_BUFFER]; //缓冲区是个循环队列
bool g_continue = true; //控制程序结束
HANDLE g_hMutex; //用于线程间的互斥
HANDLE g_hFullSemaphore; //当缓冲区满时迫使生产者等待
HANDLE g_hEmptySemaphore; //当缓冲区空时迫使消费者等待

DWORD WINAPI Producer(LPVOID); //生产者线程
DWORD WINAPI Consumer(LPVOID); //消费者线程

int main()
{
//创建各个互斥信号
g_hMutex = CreateMutex(NULL,FALSE,NULL);
g_hFullSemaphore = CreateSemaphore(NULL,SIZE_OF_BUFFER-1,SIZE_OF_BUFFER-1,NULL);
g_hEmptySemaphore = CreateSemaphore(NULL,0,SIZE_OF_BUFFER-1,NULL);

//调整下面的数值,可以发现,当生产者个数多于消费者个数时,
//生产速度快,生产者经常等待消费者;反之,消费者经常等待
const unsigned short PRODUCERS_COUNT = 3; //生产者的个数
const unsigned short CONSUMERS_COUNT = 1; //消费者的个数

//总的线程数
const unsigned short THREADS_COUNT = PRODUCERS_COUNT+CONSUMERS_COUNT;

HANDLE hThreads[PRODUCERS_COUNT]; //各线程的handle
DWORD producerID[CONSUMERS_COUNT]; //生产者线程的标识符
DWORD consumerID[THREADS_COUNT]; //消费者线程的标识符

//创建生产者线程
for (int i=0;i<PRODUCERS_COUNT;++i){
hThreads[i]=CreateThread(NULL,0,Producer,NULL,0,&producerID[i]);
if (hThreads[i]==NULL) return -1;
}
//创建消费者线程
for (int i=0;i<CONSUMERS_COUNT;++i){
hThreads[PRODUCERS_COUNT+i]=CreateThread(NULL,0,Consumer,NULL,0,&consumerID[i]);
if (hThreads[i]==NULL) return -1;
}

while(g_continue){
if(getchar()){ //按回车后终止程序运行
g_continue = false;
}
}

return 0;
}

//生产一个产品。简单模拟了一下,仅输出新产品的ID号
void Produce()
{
std::cerr << "Producing " << ++ProductID << " ... ";
std::cerr << "Succeed" << std::endl;
}

//把新生产的产品放入缓冲区
void Append()
{
std::cerr << "Appending a product ... ";
g_buffer[in] = ProductID;
in = (in+1)%SIZE_OF_BUFFER;
std::cerr << "Succeed" << std::endl;

//输出缓冲区当前的状态
for (int i=0;i<SIZE_OF_BUFFER;++i){
std::cout << i <<": " << g_buffer[i];
if (i==in) std::cout << " <-- 生产";
if (i==out) std::cout << " <-- 消费";
std::cout << std::endl;
}
}

//从缓冲区中取出一个产品
void Take()
{
std::cerr << "Taking a product ... ";
ConsumeID = g_buffer[out];
out = (out+1)%SIZE_OF_BUFFER;
std::cerr << "Succeed" << std::endl;

//输出缓冲区当前的状态
for (int i=0;i<SIZE_OF_BUFFER;++i){
std::cout << i <<": " << g_buffer[i];
if (i==in) std::cout << " <-- 生产";
if (i==out) std::cout << " <-- 消费";
std::cout << std::endl;
}
}

//消耗一个产品
void Consume()
{
std::cerr << "Consuming " << ConsumeID << " ... ";
std::cerr << "Succeed" << std::endl;
}

//生产者
DWORD WINAPI Producer(LPVOID lpPara)
{
while(g_continue){
WaitForSingleObject(g_hFullSemaphore,INFINITE);
WaitForSingleObject(g_hMutex,INFINITE);
Produce();
Append();
Sleep(1500);
ReleaseMutex(g_hMutex);
ReleaseSemaphore(g_hEmptySemaphore,1,NULL);
}
return 0;
}

//消费者
DWORD WINAPI Consumer(LPVOID lpPara)
{
while(g_continue){
WaitForSingleObject(g_hEmptySemaphore,INFINITE);
WaitForSingleObject(g_hMutex,INFINITE);
Take();
Consume();
Sleep(1500);
ReleaseMutex(g_hMutex);
ReleaseSemaphore(g_hFullSemaphore,1,NULL);
}
return 0;
}
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2007-04-10 20:36
布拉莫斯
Rank: 1
来 自:中国太平洋舰队
等 级:新手上路
帖 子:169
专家分:0
注 册:2007-3-31
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得分:0 
回复:(lily_0406)急~可不可以帮我看看这个程序能不...

你的程序我改了一下 运行是能运行了 但结束的条件我改了 ,违背了你的意愿!

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


const unsigned short SIZE_OF_BUFFER = 10; //缓冲区长度
unsigned short ProductID = 0; //产品号
unsigned short ConsumeID = 0; //将被消耗的产品号
unsigned short in = 0; //产品进缓冲区时的缓冲区下标
unsigned short out = 0; //产品出缓冲区时的缓冲区下标
char g;
int g_buffer[SIZE_OF_BUFFER]; //缓冲区是个循环队列
bool g_continue = true; //控制程序结束
HANDLE g_hMutex; //用于线程间的互斥
HANDLE g_hFullSemaphore; //当缓冲区满时迫使生产者等待
HANDLE g_hEmptySemaphore; //当缓冲区空时迫使消费者等待

DWORD WINAPI Producer(LPVOID); //生产者线程
DWORD WINAPI Consumer(LPVOID); //消费者线程

int main()
{
//创建各个互斥信号
g_hMutex = CreateMutex(NULL,FALSE,NULL);
g_hFullSemaphore = CreateSemaphore(NULL,SIZE_OF_BUFFER-1,SIZE_OF_BUFFER-1,NULL);
g_hEmptySemaphore = CreateSemaphore(NULL,0,SIZE_OF_BUFFER-1,NULL);

//调整下面的数值,可以发现,当生产者个数多于消费者个数时,
//生产速度快,生产者经常等待消费者;反之,消费者经常等待
const unsigned short PRODUCERS_COUNT = 3; //生产者的个数
const unsigned short CONSUMERS_COUNT = 1; //消费者的个数

//总的线程数
const unsigned short THREADS_COUNT = PRODUCERS_COUNT+CONSUMERS_COUNT;

HANDLE hThreads[PRODUCERS_COUNT]; //各线程的handle
DWORD producerID[CONSUMERS_COUNT]; //生产者线程的标识符
DWORD consumerID[THREADS_COUNT]; //消费者线程的标识符

//创建生产者线程
for (int i=0;i<PRODUCERS_COUNT;++i){
hThreads[i]=CreateThread(NULL,0,Producer,NULL,0,&producerID[i]);
if (hThreads[i]==NULL) return -1;
}
//创建消费者线程
for ( i=0;i<CONSUMERS_COUNT;++i){
hThreads[PRODUCERS_COUNT+i]=CreateThread(NULL,0,Consumer,NULL,0,&consumerID[i]);
if (hThreads[i]==NULL) return -1;
}

while(g_continue){  //这里在VC环境下运行 只要銉入任意一个字符 再按回车就可以退出了!
cin>>g;
if(g){ //按回车后终止程序运行
g_continue = false;
}
}

return 0;
}

//生产一个产品。简单模拟了一下,仅输出新产品的ID号
void Produce()
{
/*std::cerr*/cout<<"Producing"<<++ProductID<<" ... ";
/*std::cerr*/cout<< "Succeed" <</*std::*/endl;
}

//把新生产的产品放入缓冲区
void Append()
{
cout<< "Appending a product ... ";
g_buffer[in] = ProductID;
in = (in+1)%SIZE_OF_BUFFER;
cout<< "Succeed" <<endl;

//输出缓冲区当前的状态
for (int i=0;i<SIZE_OF_BUFFER;++i){
cout << i <<": " << g_buffer[i];
if (i==in) cout << " <-- 生产";
if (i==out) cout << " <-- 消费";
cout << endl;
}
}

//从缓冲区中取出一个产品
void Take()
{
cout << "Taking a product ... ";
ConsumeID = g_buffer[out];
out = (out+1)%SIZE_OF_BUFFER;
cout << "Succeed" << endl;

//输出缓冲区当前的状态
for (int i=0;i<SIZE_OF_BUFFER;++i){
cout << i <<": " << g_buffer[i];
if (i==in) cout << " <-- 生产";
if (i==out) cout << " <-- 消费";
cout << endl;
}
}

//消耗一个产品
void Consume()
{
cout << "Consuming " << ConsumeID << " ... ";
cout << "Succeed" << endl;
}

//生产者
DWORD WINAPI Producer(LPVOID lpPara)
{
while(g_continue){
WaitForSingleObject(g_hFullSemaphore,INFINITE);
WaitForSingleObject(g_hMutex,INFINITE);
Produce();
Append();
Sleep(1500);
ReleaseMutex(g_hMutex);
ReleaseSemaphore(g_hEmptySemaphore,1,NULL);
}
return 0;
}

//消费者
DWORD WINAPI Consumer(LPVOID lpPara)
{
while(g_continue){
WaitForSingleObject(g_hEmptySemaphore,INFINITE);
WaitForSingleObject(g_hMutex,INFINITE);
Take();
Consume();
Sleep(1500);
ReleaseMutex(g_hMutex);
ReleaseSemaphore(g_hFullSemaphore,1,NULL);
}
return 0;
}


真理往往掌握在少数人手中,可现实却是少数服从多数!
2007-04-11 18:52
yushui
Rank: 3Rank: 3
等 级:论坛游民
威 望:7
帖 子:1355
专家分:22
注 册:2006-7-19
收藏
得分:0 
^_^有和我一样学进程管理的啊 ,不过 怎么不用 PV操作来实现呢,那样就很少了啊 代码 ,不过我还没真正写全过 呵呵

fighting!from now on!
2007-04-11 20:23
快速回复:急~可不可以帮我看看这个程序能不能运行~
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