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0
		printf("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0
		printf("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0
		printf("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0
		printf("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0
		printf("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0
		printf("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0
		printf("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2
		printf("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3
		printf("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4
		printf("报数:5\n\n");
        sleep(1);
		pthread_mutex_unlock(&mutex[0]);    //开锁0
	}
}

效果