哲学家用餐问题求助急!!!

发布网友发布时间：2023-07-14 08:35

共1个回答

热心网友时间：2024-11-25 13:16

#include <pthread.h>
#include <stdio.h>

#define N 5
#define LEFT (i+N-1)%N
#define RIGHT (i+1)%N
#define THINK_TIME 3
#define EAT_TIME 2

enum { THINKING, HUNGRY, EATING } state[N];

pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER, s[N];

void test(int i)
{
if (state[i] == HUNGRY
&& state[LEFT] != EATING
&& state[RIGHT] != EATING)
{
state[i] = EATING;
pthread_mutex_unlock(&s[i]);
}
}

void take_forks(int i)
{
pthread_mutex_lock(&mutex);
state[i] = HUNGRY;
test(i);
pthread_mutex_unlock(&mutex);
pthread_mutex_lock(&s[i]);
}

void put_forks(int i)
{
pthread_mutex_lock(&mutex);
state[i] = THINKING;
test(LEFT);
test(RIGHT);
pthread_mutex_unlock(&mutex);
}

void think(int i)
{
printf("philosopher %d is thinking...\n", i);
sleep(THINK_TIME);
}

void eat(int i)
{
printf("philosopher %d is eating...\n", i);
sleep(EAT_TIME);
}

void* phi(void* vargp)
{
int i = *(int*)vargp;
while (1)
{
think(i);
take_forks(i);
eat(i);
put_forks(i);
}
return NULL;
}

int main()
{
int i;
pthread_t tid[N];
for (i = 0; i < N; i++)
pthread_create(&tid[i], NULL, phi, (void*)(&i));
for (i = 0; i < N; i++)
pthread_join(tid[i], NULL);
return 0;
}追问能说明一下一些重要的变量定义吗
就注释一下
谢谢了

追答其实哲学家就餐就是一个这样的逻辑，只有当某个哲学家左边和右边的筷子都空闲就可以就餐，一旦遇到有一个或者两个都不空闲，就继续等待，

哲学家用餐问题 求助 急!!!

哲学家用餐问题求助急!!!