哲学家就餐问题是多线程同步中的经典问题，本文在 windows 上用 C/C++ 实现该问题。

哲学家就餐问题是多线程同步中的经典问题，本文在 windows 上用 C/C++ 实现该问题。

之前在网上看到一个该问题的实现：http://dev.csdn.net/article/77/77383.shtm，是用临界区作的，但是感觉不是太好，因为当一个哲学家在试图取叉子用餐的时候，其他哲学家必须等待，而实际上，当 1 号哲学家

取叉子时，3号哲学家也可以取叉子，他们是不冲突的。

假设有 5 个哲学家，有 5 把叉子，每个哲学家有思考，饥饿，和吃饭三个状态。我们用 5 个信号量分别表示 5 把叉子。在打印哲学家状态时，使用一个互斥体 mutex ，否则打印出来是乱的。

源代码如下：

// PhilosopherDining.cpp : Defines the entry point for the console application.

//

#include "stdafx.h"

#include "windows.h"

#include

#include

#include

#include

#include

using namespace std;

const unsigned int PHILOSOPHER_NUM=5;

const char THINKING=1;

const char HUNGRY=2;

const char DINING=3;

// each fork has a semaphore

HANDLE semph[PHILOSOPHER_NUM];

// Mutex for printing

HANDLE mutex;

void philosopherProc(void* param);

int main(int argc, char* argv[])

{

int i;

srand(time(0));

mutex = CreateMutex(NULL, false, NULL);

for (i=0; i

{

semph[i] = CreateSemaphore(NULL, 1, 1, NULL);

_beginthread(philosopherProc, 0, (void*)&i);

Sleep(10);

}

Sleep(2000);

return 0;

}

void philosopherProc(void* param)

{

int myid;

char idStr[128];

char stateStr[128];

char mystate;

int ret;

unsigned int leftFork;

unsigned int rightFork;

myid = *((int*)(param));

itoa(myid, idStr, 10);

//cout << "philosopher " << myid << " begin......" << endl;

Sleep(10);

// initial state is THINKING

mystate = THINKING;

leftFork = (myid) % PHILOSOPHER_NUM;

rightFork = (myid + 1) % PHILOSOPHER_NUM;

while (true)

{

switch(mystate)

{

case THINKING:

// changing my state

mystate = HUNGRY;

strcpy(stateStr, "HUNGRY");

break;

case HUNGRY:

strcpy(stateStr, "HUNGRY");

// first test the left fork ...

ret = WaitForSingleObject(semph[leftFork], 0);

if (ret == WAIT_OBJECT_0)

{

// left fork is ok, take it up !

// then test the right fork ...

ret = WaitForSingleObject(semph[rightFork], 0);

if (ret == WAIT_OBJECT_0)

{

// right fork is also ok !

// changing my state

mystate = DINING;

strcpy(stateStr, "DINING");

}

else

{

// right fork is being used by others, so I must put down

// the left fork.

ReleaseSemaphore(semph[leftFork], 1, NULL);

}

}

break;

case DINING:

// put down both the left and right fork

ReleaseSemaphore(semph[leftFork], 1, NULL);

ReleaseSemaphore(semph[rightFork], 1, NULL);

// changing my state

mystate = THINKING;

strcpy(stateStr, "THINKING");

break;

}

// print my state

WaitForSingleObject(mutex, INFINITE);

cout << "philosopher " << myid << " is : " << stateStr << endl;

ReleaseMutex(mutex);

// sleep a random time : between 1 - 5 s

int sleepTime;

sleepTime = 1 + (int)(5.0*rand()/(RAND_MAX+1.0));

Sleep(sleepTime*10);

}

}