信號量對比互斥鎖實現生產者消費者模型Linux C語言

/*
*mutex=1;
*lock mutex=0;
*unlock mutex=1;
*車位：共享資源 [管理員：互斥鎖：串行---信號量：並行 ]車：線程
*互斥鎖：每次只能夠放一輛車
*信號量：可以運行多個線程同時訪問共享資源
*sem_init();
*sem_wait();===
*sem_post(sem,**)//解鎖
*/

#include <stdio.h>
#include <pthread.h>
#include <string.h>
#include <stdlib.h>
#include <unistd.h>
#include <signal.h>
#include <errno.h>
#include <limits.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>
#include <semaphore.h>

sem_t produce_sem;
sem_t consume_sem;


//線程同步

pthread_mutex_t mylock=PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t cond=PTHREAD_COND_INITIALIZER;


typedef struct node
{
  int data;
  struct node *next;

}Node;
//指針
Node *head=NULL;
/*
producer:
consumer:
*/
void *consumer(void *arc){

  while(1){

  

    sem_wait(&consume_sem);//初始化爲0則阻塞
    //阻塞到post時候進行喚醒
    //吃東西訪問共享數據刪除節點

    Node *pdel= head;
    head = head->next;
    printf("-------consumer：%#X spend %d\n",pthread_self(),pdel->data);
    free(pdel);
    //將佔有的資源還給生產者？？？
    sem_post(&produce_sem);//++
    sleep(rand()% 3);
  }
  pthread_exit((void *)0);
}
void *producer(void *arc){


  while(1){

    sem_wait(&produce_sem);//相當於資源--
     //創建節點
     Node *pnew =(Node *)malloc(sizeof(Node));

     //插入節點初始化
     pnew->data = rand() % 1000; //0-999

     //頭插法
     pnew->next = head; 
     head = pnew;
     printf("-------producer:%#X produce %d\n",pthread_self(),pnew->data);

     pthread_mutex_unlock(&mylock);
     //通知消費者喚醒一個線程
     sem_post(&consume_sem);//++
     //pthread_cond_broadcast(&cond);//喚醒所有線程

    // pthread_cond_signal(&cond);
    //printf("lmj\n");
     sleep(rand()%3);
     
  }

  //通知消費者
  pthread_exit((void *)1);

}


 int main(int argc, char const *argv[])
{
  
    pthread_t tid1,tid2;
  
    //消費者資源爲0
    sem_init(&consume_sem,0,0);
    sem_init(&produce_sem,0,4);//最多5個

    pthread_create(&tid1,NULL,producer,NULL);
    pthread_create(&tid2,NULL,consumer,NULL);


    pthread_join(tid1,NULL);
    pthread_join(tid2,NULL);

    sem_destory(consume_sem);
    sem_destory(&produce_sem)
  

  //sleep(10);
  return 0;
}