死锁之五个哲学家就餐问题 C语言实现

#include<stdio.h>
#include<process.h>
#include<windows.h>
#include<stdlib.h>
#include<math.h>
HANDLE Mutext[5];
int chopstick = 0;
int *Physical;
unsigned int _stdcall ThreadFun(void *p)
{
int* current = (int*)p;
while (1)
{
if (abs(*current - chopstick)<4)
{
WaitForSingleObject(Mutext[*current], INFINITE);//加锁
WaitForSingleObject(Mutext[(*current + 1) % 5], INFINITE);
printf("哲学家%d正在就餐...\n", *current);
printf("哲学家放下左边%d筷子\n", chopstick);
chopstick = (chopstick + 1) % 5;
printf("哲学家放下右边%d筷子\n\n", chopstick);
Sleep(2000);
ReleaseMutex(Mutext[(*current + 1) % 5]);//解锁
ReleaseMutex(Mutext[(*current) % 5]);
}
}
return 0;
}
int main()
{
//初始化互斥变量,初始化要传递给线程的数组
//HANDLE类型可以为(Event,Mutex,Process,Thread,Semaphore)数组
int Physical[5];
int i;
for (i = 0; i<5; i++)
{
Mutext[i] = CreateMutexA(NULL, false, NULL);//创建了一个一个匿名的互斥锁创建后
Physical[i] = i;
}
//然后开始进行创建线程
HANDLE *hThread = NULL;
hThread = new HANDLE[5];
for (i = 0; i<5; i++)
{
hThread[i] = (HANDLE)_beginthreadex(NULL, 0, ThreadFun, &Physical[i], 0, NULL);
}
WaitForMultipleObjects(5, hThread, true, INFINITE);//第二个参数为指定对象句柄组合中的第一个元素
for (i = 0; i<5; i++)
{
CloseHandle(hThread[i]);//销毁创建的进程
CloseHandle(Mutext[i]);//销毁锁
}
return 0;
}

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转载自blog.csdn.net/qq_40394168/article/details/80452794