操作系統課程設計實驗報告之多線程模擬哲學家就餐問題

多線程模擬哲學家就餐問題 
-----Amoon 2005/09/23 

1）問題描述 

學操作系統的進程同步都要涉及到三個經典問題：生產者-消費者問題、讀者-寫者問題和哲學家就餐問題。下面來介紹一下哲學家就餐問題： 哲學家就餐問題中，一組哲學家圍坐在一個圓桌旁，每個哲學家的左邊都只有一只筷子（當然他的右邊也有一只筷子，但是這是他右邊哲學家的左邊的筷子），他們吃完了就思考，思考了一會就會餓，餓了就想吃，然而，為了吃飯，他們必須獲得左邊和右邊的筷子。當每個哲學家只拿有一只筷子的時候，會坐者等另一只筷子，在每個哲學家都只拿一個筷子的時候，就會發生死鎖。 

2）關于源碼的一點說明： 

應用CRITICAL_SECTION 變量解決數據共享問題。 

3）vc6下源碼如下( 經作者在vc6下測試通過運行 ）： 

#include <windows.h> 
#include <process.h> /* _beginthread, _endthread */ 
#include <time.h> 
#include <fstream> 
#include <string> 
#include <iostream> 
#include <assert.h> 
#include <time.h> 

using namespace std; 


ofstream out("out.txt"); 
BOOL b[6];//共享數據,用于表示 5 雙筷子的使用情況,FALSE代表被占用 
CRITICAL_SECTION cs; 

class philosopher 
{ 
private: 
int m_num; //編號 
int m_index; //標示裝態,0-------waiting 1-----eating 2-----thinking 

BOOL m_lhand; //true -----筷子空閑,false----筷子被占用 
BOOL m_rhand; 

public: 
philosopher(int num=0) { m_num=num; m_index=2; m_lhand=TRUE; m_rhand = FALSE ;} 

int GetNum() const { return m_num; } 

int GetStatus() const { return m_index ; } 

void ChangeStatus() ; 
}; 

void philosopher::ChangeStatus() 
{ 
EnterCriticalSection (&cs) ; 

if(m_index==1) //eating----->thinking 
{ 
assert(!b[m_num]); 
b[m_num]=TRUE; 

if( m_num==1) 
{ 
assert(!b[5]); 
b[5]= TRUE; 
} 
else 
b[m_num-1]=TRUE; 

m_index=2; // change to thinking 
srand(time(NULL)*10000); 
Sleep(rand()%10); 
} 

else if( m_index==2) //thinking 
{ 
m_index=0; //change to waiting 

} 
else if( m_index==0) // waiting 
{ 

if(m_num==1) 
{ 
if(b[1]&&b[5]) 
{ 
b[1]=FALSE; 
b[5]=FALSE; 
m_index=1; 
srand(time(NULL)*10000); 
Sleep(rand()%10); 
} 
} 
else 
{ 

if(b[m_num]&&b[m_num-1]) 
{ 
b[m_num]= FALSE; 
b[m_num-1]= FALSE; 
m_index=1; 
srand(time(NULL)*10000); 
Sleep(rand()%10); 
} 
} 


} 
LeaveCriticalSection (&cs) ; 

} 

//ust to out the philosopher info 

string PrintStatus(philosopher *pPh) 
{ 
pPh->ChangeStatus(); 
int i=pPh->GetStatus(); 
string str; 
if(i==0) 
str="Waiting"; 
else if(i==1) 
str="eating"; 
else str="thinking"; 
return str; 
} 

void ThreadFunc1(PVOID param) 
{ 
while(1) 

{ 
philosopher *pPh; 
pPh=( philosopher *) param; 
pPh->ChangeStatus(); 


} 

} 
void ThreadFunc2(PVOID param) 
{ 
while(1) 
{ 
philosopher *pPh; 
pPh=( philosopher *) param; 
pPh->ChangeStatus(); 
} 
} 
void ThreadFunc3(PVOID param) 
{ 
while(1) 
{ 
philosopher *pPh; 
pPh=( philosopher *) param; 
pPh->ChangeStatus(); 
} 
} 
void ThreadFunc4(PVOID param) 
{ 
while(1) 
{ 
philosopher *pPh; 
pPh=( philosopher *) param; 
pPh->ChangeStatus(); 
} 
} 
void ThreadFunc5(PVOID param) 
{ 
while(1) 
{ 
philosopher *pPh; 
pPh=( philosopher *) param; 
pPh->ChangeStatus(); 
} 
} 


int main() 
{ 

for(int j=1;j<=5;j++) b[j]=TRUE; 
out<<b[1]<<b[2]<<b[3]<<b[4]<<b[5]; 
philosopher ph1(1),ph2(2),ph3(3),ph4(4),ph5(5); 
InitializeCriticalSection (&cs) ; 

_beginthread(ThreadFunc1,0,&ph1); 
_beginthread(ThreadFunc2,0,&ph2); 
_beginthread(ThreadFunc3,0,&ph3); 
_beginthread(ThreadFunc4,0,&ph4); 
_beginthread(ThreadFunc5,0,&ph5); 
cout<<"press ctrl+C to exit"; 
while(1) 
{ 
out<<endl<<b[1]<<b[2]<<b[3]<<b[4]<<b[5]<<endl; // use to test 
out<<"The status at :"<<clock()<<" ms "<<endl; 
out<<"哲學家"<<ph1.GetNum() <<"所處狀態:"<<ph1.GetNum()<<"--"<<PrintStatus(&ph1)<<endl; 
out<<"哲學家"<<ph2.GetNum() <<"所處狀態:"<<ph2.GetNum()<<"--"<<PrintStatus(&ph2)<<endl; 
out<<"哲學家"<<ph3.GetNum() <<"所處狀態:"<<ph3.GetNum()<<"--"<<PrintStatus(&ph3)<<endl; 
out<<"哲學家"<<ph4.GetNum() <<"所處狀態:"<<ph4.GetNum()<<"--"<<PrintStatus(&ph4)<<endl; 
out<<"哲學家"<<ph5.GetNum() <<"所處狀態:"<<ph5.GetNum()<<"--"<<PrintStatus(&ph5)<<endl; 

Sleep(20); 
} 

DeleteCriticalSection (&cs) ; 

return 0; 
}