答复: 迅雷亲历面经：JAVA 笔试+上机+面试（完整面试题大讨论）

引用

有三个线程ID分别是A、B、C,请有多线编程实现，在屏幕上循环打印10次ABCABC…

引申了一下：

有n个线程，ID为0...n-1，在屏幕上循环打印m次012..n-1

用 c/pthread 实现

//@Author      : [email protected]
#include <stdio.h>
#include <stdlib.h>
#include <errno.h>
#include <pthread.h>

#define GROUP_COUNT 100
#define GROUP_SIZE 4 

typedef struct {
	pthread_mutex_t mutex;
	pthread_cond_t cond;
	int index;
} syn_obj_t;

syn_obj_t syn_obj = {PTHREAD_MUTEX_INITIALIZER, 
	PTHREAD_COND_INITIALIZER, 0};

typedef struct {
	int flag;
} elem_t; 

void *
thread_routine(void *arg);

int
main(int argc, char** argv)
{
	elem_t elems[GROUP_SIZE];
	pthread_t pds[GROUP_SIZE];
	int i;

	printf("syn_obj.index = %d\n", syn_obj.index);

	for (i = 0; i < GROUP_SIZE; i++) {
		elems[i].flag = i;
		if ( (pthread_create(&pds[i], NULL, thread_routine, &elems[i])) != 0 ) {
			perror("pthread create");
			exit(-1);
		}
	}

	for (i = 0; i < GROUP_SIZE; i++) {
		pthread_join(pds[i], NULL);
	}

	pthread_mutex_destroy(&syn_obj.mutex);
	pthread_cond_destroy(&syn_obj.cond);

	printf("\nsyn_obj.index = %d\n", syn_obj.index);

	return 0;
}

void *
thread_routine(void *arg) {
	elem_t *elem = (elem_t *)arg;
	int i;
	for (i = 0; i < GROUP_COUNT; i++) {
		pthread_mutex_lock(&syn_obj.mutex);
		while ( (syn_obj.index % GROUP_SIZE) != elem->flag ) {
			pthread_cond_wait(&syn_obj.cond, &syn_obj.mutex);
	 	}
		printf("%d", elem->flag);
		if ( 0 == (syn_obj.index+1) % GROUP_SIZE ) {
			printf("\t");
		}
		syn_obj.index++;
		pthread_cond_broadcast(&syn_obj.cond);
		// may be cause deadlock 
		// pthread_cond_signal(&syn_obj.cond);
		pthread_mutex_unlock(&syn_obj.mutex);
		// sleep(1);
	}
	return NULL;
}