题目

        请使用互斥锁 和 信号量分别实现5个线程之间的同步

代码：信号量实现

void* task1(void* arg);
void* task2(void* arg);
void* task3(void* arg);
void* task4(void* arg);
void* task5(void* arg);sem_t sem[5];					//信号量变量int main(int argc, const char *argv[])
{//信号量初始化for(int i=0;i<5;i++){sem_init(&sem[i],0,0);//初始化信号量数量}sem_post(&sem[0]);//线程初始化pthread_t id[4];pthread_create(&id[0],0,task1,NULL);pthread_create(&id[1],0,task2,NULL);pthread_create(&id[2],0,task3,NULL);pthread_create(&id[3],0,task4,NULL);pthread_detach(id[0]);pthread_detach(id[1]);pthread_detach(id[2]);pthread_detach(id[3]);while(1){task5(NULL);}return 0;
}void* task1(void* arg)
{while(1){sem_wait(&sem[0]);		//获取信号量0printf("报数:1\n");sleep(1);sem_post(&sem[1]);		//释放信号量1}
}void* task2(void* arg)
{while(1){sem_wait(&sem[1]);	//获取信号量0printf("报数:2\n");sleep(1);sem_post(&sem[2]);		//释放信号量1}
}void* task3(void* arg)
{while(1){sem_wait(&sem[2]);	//获取信号量0printf("报数:3\n");sleep(1);sem_post(&sem[3]);		//释放信号量1}
}void* task4(void* arg)
{while(1){sem_wait(&sem[3]);	//获取信号量0printf("报数:4\n");sleep(1);sem_post(&sem[4]);		//释放信号量1}
}void* task5(void* arg)
{while(1){sem_wait(&sem[4]);	//获取信号量0printf("报数:5\n\n");sleep(1);sem_post(&sem[0]);		//释放信号量1}
}

效果

代码：互斥锁实现

void* task1(void* arg);
void* task2(void* arg);
void* task3(void* arg);
void* task4(void* arg);
void* task5(void* arg);pthread_mutex_t mutex[5];		//互斥锁变量int main(int argc, const char *argv[])
{//互斥锁初始化pthread_mutexattr_t mutex_attr; //锁属性pthread_mutexattr_init(&mutex_attr); //锁属性初始化for(int i=0;i<5;i++){pthread_mutex_init(&mutex[i],&mutex_attr);		//默认锁pthread_mutex_lock(&mutex[i]);                  //上锁}pthread_mutex_unlock(&mutex[0]);                    //解锁0//线程初始化pthread_t id[4];pthread_create(&id[0],0,task1,NULL);pthread_create(&id[1],0,task2,NULL);pthread_create(&id[2],0,task3,NULL);pthread_create(&id[3],0,task4,NULL);pthread_detach(id[0]);pthread_detach(id[1]);pthread_detach(id[2]);pthread_detach(id[3]);while(1){task5(NULL);}return 0;
}void* task1(void* arg)
{while(1){pthread_mutex_lock(&mutex[0]);      //上锁0printf("报数:1\n");sleep(1);pthread_mutex_unlock(&mutex[1]);    //开锁1}
}void* task2(void* arg)
{while(1){pthread_mutex_lock(&mutex[1]);      //上锁0printf("报数:2\n");sleep(1);pthread_mutex_unlock(&mutex[2]);    //开锁1}
}void* task3(void* arg)
{while(1){pthread_mutex_lock(&mutex[2]);      //上锁2printf("报数:3\n");sleep(1);pthread_mutex_unlock(&mutex[3]);    //开锁3}
}void* task4(void* arg)
{while(1){pthread_mutex_lock(&mutex[3]);      //上锁3printf("报数:4\n");sleep(1);pthread_mutex_unlock(&mutex[4]);    //开锁4}
}void* task5(void* arg)
{while(1){pthread_mutex_lock(&mutex[4]);      //上锁4printf("报数:5\n\n");sleep(1);pthread_mutex_unlock(&mutex[0]);    //开锁0}
}

效果