用信号量解决读者学者问题c语言
时间: 2023-08-21 07:05:45 浏览: 73
读者学者问题是一个典型的同步问题,可以使用信号量来解决。在这个问题中,有多个读者和多个学者,他们需要在一个资源(比如图书馆)中读书或写作。为了避免冲突和死锁,可以使用以下程序来解决读者学者问题:
```c
#include <stdio.h>
#include <pthread.h>
#include <semaphore.h>
#define MAX_READERS 5
#define MAX_WRITERS 5
int read_count = 0;
sem_t mutex, r_mutex, w_mutex;
void *reader(void *arg)
{
int id = *(int *)arg;
printf("Reader %d is trying to read.\n", id);
sem_wait(&mutex);
read_count++;
if (read_count == 1)
sem_wait(&w_mutex);
sem_post(&mutex);
sem_wait(&r_mutex);
printf("Reader %d is now reading.\n", id);
sem_post(&r_mutex);
sem_wait(&mutex);
read_count--;
if (read_count == 0)
sem_post(&w_mutex);
sem_post(&mutex);
printf("Reader %d finished reading.\n", id);
pthread_exit(NULL);
}
void *writer(void *arg)
{
int id = *(int *)arg;
printf("Writer %d is trying to write.\n", id);
sem_wait(&w_mutex);
printf("Writer %d is now writing.\n", id);
sem_post(&w_mutex);
printf("Writer %d finished writing.\n", id);
pthread_exit(NULL);
}
int main()
{
pthread_t readers[MAX_READERS], writers[MAX_WRITERS];
sem_init(&mutex, 0, 1);
sem_init(&r_mutex, 0, 1);
sem_init(&w_mutex, 0, 1);
int reader_ids[MAX_READERS], writer_ids[MAX_WRITERS];
for (int i = 0; i < MAX_READERS; i++) {
reader_ids[i] = i;
pthread_create(&readers[i], NULL, reader, &reader_ids[i]);
}
for (int i = 0; i < MAX_WRITERS; i++) {
writer_ids[i] = i;
pthread_create(&writers[i], NULL, writer, &writer_ids[i]);
}
for (int i = 0; i < MAX_READERS; i++)
pthread_join(readers[i], NULL);
for (int i = 0; i < MAX_WRITERS; i++)
pthread_join(writers[i], NULL);
sem_destroy(&mutex);
sem_destroy(&r_mutex);
sem_destroy(&w_mutex);
return 0;
}
```
在这个程序中,我们使用了三个信号量:mutex、r_mutex 和 w_mutex。mutex 用于对读者数量进行互斥访问,r_mutex 用于对读者进行互斥访问,w_mutex 用于对写者进行互斥访问。当读者进入读区时,会首先尝试获取 mutex 信号量,以保证只有一个读者能够修改读者数量。如果是第一个读者进入读区,它会尝试获取 w_mutex 信号量,从而阻止写者进入写区。然后它会获取 r_mutex 信号量,表示它正在读取资源。当读者离开读区时,会先释放 r_mutex 信号量,再修改读者数量,最后如果是最后一个读者离开读区,它会释放 w_mutex 信号量,允许写者进入写区。
当写者进入写区时,它会尝试获取 w_mutex 信号量,表示它正在写入资源。当它离开写区时,会释放 w_mutex 信号量。
这个程序可以避免读者和写者同时访问资源,从而保证了资源的正确性。