生产者消费者

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

#define N 30   // 消费者或者生产者的数目
#define M 30 // 缓冲数目

int in = 0;   // 生产者放置产品的位置,下一个空位
int out = 0; // 消费者取产品的位置，第一个非空位
int k = 0, m = 0, count = 0, a[30], products_sum = 1, count_rnd = 0, j = 0;

char buffer[M];// 缓冲初始化为0， 开始时没有产品
char consumer_store[1000];

sem_t empty_sem; // 同步信号量
sem_t full_sem;   // 同步信号量
pthread_mutex_t mutex; // 互斥信号量

int producer_num = 1;   //生产者id
int consumer_num = 1; //消费者id

/* 打印缓冲情况 */
void print(){
    int i;
    printf("\033[33m缓冲区中剩余产品: \033[0m\n");
    for(i = 0; i < M; i++)
        printf("\033[33m%c \033[0m", buffer[i]);
    printf("\n");
}

void print_consumer(){
    int i;
    for(i = 0; i < M; i++)
        printf("%c", consumer_store[i]);
    printf("\n");
}

/* 生产者方法 */
void *product(){  
     sem_wait(&empty_sem);
     pthread_mutex_lock(&mutex);
     buffer[in] = 'A' + count;
     count++;
     printf("生产者 %d 生产产品 %c 放入缓冲区 %d \n",producer_num, buffer[in], in); 
     in++;
     in=in%M;
     producer_num++;
     pthread_mutex_unlock(&mutex);
     sem_post(&full_sem);
}
/* 消费者方法 */
void *prochase(){
    sleep(1);
     sem_wait(&full_sem);           //获取信号量
     pthread_mutex_lock(&mutex);    //加锁
     int n = 0, l = 0, p = 0;
     j=(rand() % 30) + 1;
     if(count_rnd == 0){
         a[m] = j;
         m++;
     }
     else{
         while(n < m){
             if(a[n] != j)
                 n++;
             else{
                 j=(rand() % 30) + 1;
                 n = 0;
             }
         }
     }
     a[m] = j;
     m++;
     if(consumer_num % 2 != 0) {         //奇数号消费者
         if(consumer_num == 29) {  //所有奇数号消费者都消费之后，1，3，5，7，9号产品从缓冲区撤销
             buffer[0] = ' ';
             buffer[2] = ' ';
             buffer[4] = ' ';
             buffer[6] = ' ';
             buffer[8] = ' ';
         }
         if(j == 1 || j == 3 || j == 5 || j == 7 || j == 9)  { //1，3，5，7，9号生产者供奇数号消费者消费
             printf("\033[35;4m消费者 %d 消费由生产者  %d 生产的 %c\033[0m\n", consumer_num, j, buffer[j-1]);
         }
         else if((j>1&&j<11&&j%2 == 0) || (j>10&&j<21&&consumer_num != j)) { 
             printf("消费者 %d 不能消费由生产者 %d 生产的产品!\n", consumer_num, j);
             j=(rand() % 30) + 1;
             while((j>1&&j<11&&j%2 == 0) || (j>10&&j<21&&consumer_num != j)){
                 j=(rand() % 30) + 1;
             }
             while(buffer[j-1] == ' '){
                 j=(rand() % 30) + 1;
             }
             a[m] = j;
             printf("\033[35;4m消费者 %d 消费由生产者  %d 生产的 %c\033[0m\n", consumer_num, j, buffer[j-1]);
             buffer[j-1] = ' ';
         }
         else if(j>10&&j<21&&consumer_num == j){//11-20号生产者只供对应编号相同的消费者消费
         
             printf("\033[35;4m消费者 %d 消费由生产者  %d 生产的 %c\033[0m\n", consumer_num, j, buffer[j-1]);
             buffer[j-1] = ' ';
         }
         else{   //其他任意编号的生产者被随意消费
            printf("\033[35;4m消费者 %d 消费由生产者  %d 生产的 %c\033[0m\n", consumer_num, j, buffer[j-1]);
            buffer[j-1] = ' ';
         }
     }
     else{   //偶数号消费者
        if(j == 2 || j == 4 || j == 6 || j == 8 || j == 10){ //2，4，6，8，10号生产者被所有偶数号消费者消费
            printf("\033[35;4m消费者 %d 消费由生产者  %d 生产的 %c\033[0m\n", consumer_num, j, buffer[j-1]);
            buffer[j-1] = ' ';
         }
         else if((j > 0 && j < 10 && j % 2 != 0) || (j > 10 && j < 21 && consumer_num != j)){
             printf("消费者 %d 不能消费由生产者 %d 生产的产品!\n",consumer_num,j);
             j=(rand() % 30) + 1;
             while((j > 0 && j < 10 && j % 2 != 0) || (j > 10 && j < 21 && consumer_num != j)) {
                j=(rand() % 30) + 1;
             }
             while(buffer[j-1] == ' '){
                 j=(rand() % 30) + 1;
             }
             a[m] = j;
             printf("\033[35;4m消费者 %d 消费由生产者  %d 生产的 %c\033[0m\n", consumer_num, j, buffer[j-1]);
             buffer[j-1] = ' ';
         }
         else if(j>10&&j<21&&consumer_num == j) {    //11-20编号的生产者供对应编号相同的消费者消费
            printf("\033[35;4m消费者 %d 消费由生产者  %d 生产的 %c\033[0m\n", consumer_num, j, buffer[j-1]);
             buffer[j-1] = ' ';
         }
         else{
             printf("\033[35;4m消费者 %d 消费由生产者  %d 生产的 %c\033[0m\n", consumer_num, j, buffer[j-1]);
             buffer[j-1] = ' ';
         }
     }
     buffer[j-1] = ' ';
     print();
     out++;
     out=out % M;
     consumer_num++;
     count_rnd++;
     pthread_mutex_unlock(&mutex);
     sem_post(&empty_sem);
}

int main(){
    pthread_t id1[N];
    pthread_t id2[N];
    int i;
    int ret[N];
    srand((unsigned)time(0));
    // 初始化同步信号量
     int ini1 = sem_init(&empty_sem, 0, M);
     int ini2 = sem_init(&full_sem, 0, 0);  
     if(ini1 && ini2 != 0){
        printf("sem init failed \n");
        exit(1);
     }
     //初始化互斥信号量
     int ini3 = pthread_mutex_init(&mutex, NULL);
     if(ini3 != 0){
        printf("mutex init failed \n");
        exit(1);
     }
     // 每个生产者生产30个消息后结束运行
     while(products_sum <= 30){
         //创建N个生产者线程
         for(i = 0; i < N; i++){
            ret[i] = pthread_create(&id1[i], NULL, product, (void *)(&i));
            if(ret[i] != 0){
                printf("product %d creation failed \n", i);
                exit(1);
            }
        }
     //创建N个消费者线程
        for(i = 0; i < N; i++){
             ret[i] = pthread_create(&id2[i], NULL, prochase, NULL);
             if(ret[i] != 0){
                 printf("prochase %d creation failed \n", i);
                 exit(1);
            }
        }
         for(i = 0; i < N; i++){
             pthread_join(id1[i], NULL);
             pthread_join(id2[i], NULL);
        }
         
         printf("++++++++++++++++++++++++++++++++++++++++++++++++++++++\n");
         printf("完成第 %d 次生产\n",products_sum);
         printf("++++++++++++++++++++++++++++++++++++++++++++++++++++++\n\n");
         products_sum++;
         in = 0;
         out = 0;
         count = 0;
         consumer_num = 1;
         producer_num = 1;
         count_rnd = 0;
         m = 0;
         a[30] = ' ';
     }
     sem_destroy(&empty_sem);
     sem_destroy(&full_sem);
     exit(0);    
}