生产者消费者问题c实现

生产者消费者问题是计算机科学中经典的同步问题，也是操作系统课程中一个重要的概念。在该问题中，有一组生产者线程和一组消费者线程共享一个缓冲区。生产者线程向缓冲区中放入产品，而消费者线程则从缓冲区中取走产品。为了避免竞态条件（多个线程同时访问同一共享资源）和死锁问题，需要通过某种机制来同步这些线程的操作。下面是一种基于信号量的C语言实现方式。

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

#define BUFFER_SIZE 5
#define PRODUCER_NUM 3
#define CONSUMER_NUM 2

int buffer[BUFFER_SIZE];    // 缓冲区
int in = 0;    // 缓冲区插入位置
int out = 0;    // 缓冲区读取位置

sem_t empty;    // 空槽数量
sem_t full;    // 满槽数量
pthread_mutex_t mutex;    // 互斥锁

void *producer(void *arg) {
    int id = *(int*) arg;
    while (1) {
        int item = rand() % 100;    // 生产随机整数
        sem_wait(&empty);    // 等待空槽
        pthread_mutex_lock(&mutex);    // 获取互斥锁
        buffer[in] = item;    // 插入数据
        printf("Producer %d produced item %d at position %d.\n", id, item, in);
        in = (in + 1) % BUFFER_SIZE;    // 更新插入位置
        pthread_mutex_unlock(&mutex);    // 释放互斥锁
        sem_post(&full);    // 增加满槽数量
        sleep(rand() % 3);    // 随机休眠一段时间
    }
}

void *consumer(void *arg) {
    int id = *(int*) arg;
    while (1) {
        sem_wait(&full);    // 等待满槽
        pthread_mutex_lock(&mutex);    // 获取互斥锁
        int item = buffer[out];    // 取出数据
        printf("Consumer %d consumed item %d from position %d.\n", id, item, out);
        out = (out + 1) % BUFFER_SIZE;    // 更新读取位置
        pthread_mutex_unlock(&mutex);    // 释放互斥锁
        sem_post(&empty);    // 增加空槽数量
        sleep(rand() % 3);    // 随机休眠一段时间
    }
}

int main() {
    pthread_t producer_threads[PRODUCER_NUM];
    pthread_t consumer_threads[CONSUMER_NUM];
    int producer_ids[PRODUCER_NUM];
    int consumer_ids[CONSUMER_NUM];

    sem_init(&empty, 0, BUFFER_SIZE);   // 初始化空槽数量为缓冲区大小
    sem_init(&full, 0, 0);   // 初始化满槽数量为0
    pthread_mutex_init(&mutex, NULL);   // 初始化互斥锁

    for (int i = 0; i < PRODUCER_NUM; i++) {
        producer_ids[i] = i + 1;
        pthread_create(&producer_threads[i], NULL, producer, &producer_ids[i]);
    }
    for (int i = 0; i < CONSUMER_NUM; i++) {
        consumer_ids[i] = i + 1;
        pthread_create(&consumer_threads[i], NULL, consumer, &consumer_ids[i]);
    }

    for (int i = 0; i < PRODUCER_NUM; i++) {
        pthread_join(producer_threads[i], NULL);
    }
    for (int i = 0; i < CONSUMER_NUM; i++) {
        pthread_join(consumer_threads[i], NULL);
    }

    sem_destroy(&empty);   // 销毁信号量
    sem_destroy(&full);
    pthread_mutex_destroy(&mutex);   // 销毁互斥锁

    return 0;
}