创建多线程做减法

师父给出了另外一道题：

给一个数，然后开5个线程对它进行相减，直到这个数为0或小于0为止；

我用多线程实现如下：

// methods.c
#include <stdio.h>
#include <pthread.h>
#include <stdlib.h>

int sum;

static pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER;
static pthread_cond_t cond = PTHREAD_COND_INITIALIZER;

void * thread1(void *);
void * thread2(void *);
void * thread3(void *);
void * thread4(void *);
void * thread5(void *);

int main(int argc, char * argv[])
{
    pthread_t tid1, tid2, tid3, tid4, tid5;
    int rc1 = 0, rc2 = 0, rc3 = 0, rc4 = 0, rc5 = 0;
    printf("enter main\n");
    printf("Please input a number : \n");
    scanf("%d", &sum);
    while (sum >= 0)
    {
	rc1 = pthread_create(&tid1, NULL, thread1, NULL);
    	if (rc1 != 0)
		printf("The thread1-create is failed!\n");
    
    	rc2 = pthread_create(&tid2, NULL, thread2, NULL);
    	if (rc2 != 0)
		printf("The thread2-create is failed!\n");
    
    	rc3 = pthread_create(&tid3, NULL, thread3, NULL);
    	if (rc3 != 0)
		printf("The thread3-create is failed!\n");
    
    	rc4 = pthread_create(&tid4, NULL, thread4, NULL);
    	if (rc4 != 0)
		printf("The thread4-create is failed!\n");
    
    	rc5 = pthread_create(&tid5, NULL, thread5, NULL);
    	if (rc5 != 0)
		printf("The thread5-create is failed!\n");
    

    	pthread_cond_wait(&cond, &mutex);
    }
    printf("leave main\n");
    exit(0);
}

void * thread1(void * arg)
{
    printf("enter thread1\n");
    
    pthread_mutex_lock(&mutex);
    if (sum <= 0)
	exit(0);
    else
    	printf("this is thread1, sum : %d, thread id is %u\n", sum,
	    (unsigned int)pthread_self());
    	pthread_cond_signal(&cond);
	sum -= 1;
    printf("this is thread1, sum : %d, thread id is %u\n", sum,
	    (unsigned int)pthread_self());
    pthread_mutex_unlock(&mutex);
    printf("leave thread1\n");
    pthread_exit(0);
}

void * thread2(void * arg)
{
    printf("enter thread2\n");
    
    pthread_mutex_lock(&mutex);
    if (sum <= 0)
	exit(0);
    else
    	printf("this is thread2, sum : %d, thread id is %u\n", sum,
	    (unsigned int)pthread_self());
	pthread_cond_signal(&cond);
	sum -= 2;
    printf("this is thread2, sum : %d, thread id is %u\n", sum,
	    (unsigned int)pthread_self());
    pthread_mutex_unlock(&mutex);
    printf("leave thread2\n");
    pthread_exit(0);
}

void * thread3(void * arg)
{
    printf("enter thread3\n");
    
    pthread_mutex_lock(&mutex);
    if (sum <= 0)
	exit(0);
    else
	printf("this is thread3, sum : %d, thread id is %u\n", sum,
	    (unsigned int)pthread_self());
	pthread_cond_signal(&cond);
	sum -= 3;
    printf("this is thread3, sum : %d, thread id is %u\n", sum,
	    (unsigned int)pthread_self());
    pthread_mutex_unlock(&mutex);
    printf("leave thread3\n");
    pthread_exit(0);
}

void * thread4(void * arg)
{
    printf("enter thread4\n");
    
    pthread_mutex_lock(&mutex);
    if (sum <= 0)
	exit(0);
    else
	printf("this is thread4, sum : %d, thread id is %u\n", sum,
	    (unsigned int)pthread_self());
	pthread_cond_signal(&cond);
	sum -= 4;
    printf("this is thread4, sum : %d, thread id is %u\n", sum,
	    (unsigned int)pthread_self());
    pthread_mutex_unlock(&mutex);
    printf("leave thread4\n");
    pthread_exit(0);
}

void * thread5(void * arg)
{
    printf("enter thread5\n");
    
    pthread_mutex_lock(&mutex);
    if (sum <= 0)
	exit(0);
    else
	printf("this is thread5, sum : %d, thread id is %u\n", sum,
	    (unsigned int)pthread_self());
	pthread_cond_signal(&cond);
	sum -= 5;
    printf("this is thread5, sum : %d, thread id is %u\n", sum,
	    (unsigned int)pthread_self());
    pthread_mutex_unlock(&mutex);
    printf("leave thread5\n");
    pthread_exit(0);
}

 以下是运行结果，只截取了部分，因为结果过长：