发一下牢骚和主题无关：

考参他人的方法，自己做了简略修改，现实一个通用的线程队列。

#include <stdio.h>
#include <pthread.h>
#include <unistd.h>#define POOL_SIZE 10
#define QUEUE_LEN 20
pthread_mutex_t queue_lock; //队列锁
pthread_cond_t  queue_cond; //条件量变int head = 0, tail = 0;
int queue[QUEUE_LEN];void delay(int x)
{int i,j;for(i = 0; i < x; i++)for(j = 0; j <x; j++);
}void fun1(){printf("fun1 start\n");delay(1000);printf("fun1 end\n");
}void fun2(){printf("fun2 start\n");delay(2000);printf("fun2 end\n");
}void fun3(){printf("fun3 start\n");delay(3000);printf("fun3 end\n");
}void fun4(){printf("fun4 start\n");delay(4000);printf("fun4 end\n");
}int dequeue()//出对列函数
{int y;pthread_mutex_lock(&queue_lock);while(head == tail)//队列中 无情求时，待等{pthread_cond_wait(&queue_cond, &queue_lock);//动自释放锁        }y = queue[++head];//需做循环处置  queue[head] is NULLif(head >= QUEUE_LEN)head = 0;pthread_mutex_unlock(&queue_lock);return y;
}void * process_queue(void *arg)
{int x = 0;for(;;){x = dequeue();printf("%d dequeue \n",x);switch(x){case 1:fun1(); break;case 2:fun2(); break;case 3:fun3(); break;default:fun4(); break;}     }
}int enqueue(int x)//插入求请
{int next;pthread_mutex_lock(&queue_lock);next = tail + 1;if(next >= QUEUE_LEN)next = 0;if(next == head){pthread_cond_signal(&queue_cond);//pthread_mutex_unlock(&queue_lock);   return 0;// 示表队列已满}queue[next] = x;tail = next; pthread_cond_signal(&queue_cond);//激活pthread_mutex_unlock(&queue_lock);return 1;//入队胜利
}int main()
{int i;int num = 0;pthread_t tid[POOL_SIZE];pthread_mutex_init(&queue_lock, NULL);pthread_cond_init(&queue_cond, NULL);for(i = 0; i < POOL_SIZE; i++)pthread_create(&tid[i], NULL, process_queue, NULL);while(1){scanf("%d",&num);if(0 == num)// break;return 0;printf("%d ",num);while(!enqueue(num))//队列满时 待等{printf("queue is full\n");sleep(1);}}//  for(i = 0; i < POOL_SIZE; i++)
//      pthread_join(tid[i], NULL);//   pthread_mutex_destory(&queue_lock);
//  pthread_cond_destroy(&queue_cond);return 0;
}

还有一些不足之处，随后补上。

测试方法：

键入 ：

4 3 2 1

结果：
4 3 2 1 4 dequeue
fun4 start
3 dequeue
fun3 start
2 dequeue
fun2 start
1 dequeue
fun1 start
fun1 end
fun2 end
fun3 end
fun4 end

键入：

0

退出函数

文章结束给大家分享下程序员的一些笑话语录： 古鸽是一种搜索隐禽，在中国快绝迹了…初步的研究表明，古鸽的离去，很可能导致另一种长着熊爪，酷似古鸽，却又习性不同的猛禽类——犤毒鸟