2012校园招聘笔试题-基础部分.docx

C语言试题1：请给出以下程序的运行结果int main(int argc, char *argv) int a22 = 1,2,2;int b10 = 1; printf(a01=%d n, a01);printf(b1=%d n, b1);return 0;2：请给出以下程序的运行结果 (运行于32 位 Linux系统)int main(int argc, char *argv) int a22 = 1,2,2; int b10 = 1; char buf=sizeof(a); char *p = buf; printf(sizeof(a)=%d n, sizeof a); printf(sizeof(b)=%d n, sizeof b); printf(sizeof(buf)=%d n, sizeof buf); printf(sizeof(buf)=%d n, sizeof p); return 0;3：请给出以下程序的运行结果char *alloc_buf(char *ptr) ptr = hello; return ptr;int main(int argc, char *argv) char *p = (char*)malloc (32); snprintf(p, 32, %s, end of world, 2012); char *q = alloc_buf(p); printf(q = %s p=%sn, q, p); return 0;4：请给出以下程序的运行结果void free_buf(char *ptr) free(ptr); ptr = NULL; return ;int main(int argc, char *argv) char *p = (char*)malloc (32); char *q = p; snprintf(p, 32, %s, end of world, 2012); free_buf(p); q = NULL; printf(%sn, p=q?p=q:p!=q); return 0;5：请给出以下程序的运行结果int multi(int y) static int x = 3; x *=y; return x;int main(int argc, char *argv) int i = 1; int sum = 0; for (; i member ) *_mptr = (ptr); (type *)( (char *)_mptr - offsetof(type,member) );)#define LIST_HEAD_INIT(name) &(name), &(name) #define LIST_HEAD(name) struct list_head name = LIST_HEAD_INIT(name)#define list_entry(ptr, type, member) container_of(ptr, type, member)#define list_for_each_entry_safe(pos, n, head, member) for (pos = list_entry(head)-next, typeof(*pos), member), n = list_entry(pos-member.next, typeof(*pos), member); &pos-member != (head); pos = n, n = list_entry(n-member.next, typeof(*n), member)static inline void _list_add(struct list_head *new, struct list_head *prev, struct list_head *next) next-prev = new; new-next = next; new-prev = prev; prev-next = new;static inline void list_add_tail(struct list_head *new, struct list_head *head) _list_add(new, head-prev, head);LIST_HEAD(radix_head);int main(int argc, char *argv) int i = 1; struct radix *p = NULL; struct radix *q = NULL; for (i = 5; i; i-) p = (struct radix *)malloc(sizeof (struct radix); p-key = i; list_add_tail(&p-list, &radix_head); list_for_each_entry_safe(p, q, &radix_head, list) printf(key = %dn, p-key); list_for_each_entry_safe(p, q, &radix_head, list) free(p); return 0;9：填空题 程序功能：在不移动数组元素的情况下实现数组排序（从小到大），程序原理1：在数组元素中增加一个成员，index，用于指示下一个更大的元素的索引；2：增加一个全局变量head，指示最小的元素的索引；3：最大的元素的索引置为-1，表示序列中最后一个元素；4：如下图所示，将每个数组元素的index成员指向下一个更大的成员，全局变量head指向最小的元素索引，这样通过遍历head和各index成员，即从小到大的遍历数组，右则箭头为从小到大的示意图。struct array_list int key; int index;int head = 0;struct array_list array8 = 25,-1, 4,-1, 13,-1, 19, -1, 7, -1, 4, -1, 1, -1, 23, -1;int array_sort(struct array_list arr, int len) int i = 0; int j = i; int min = 0; int tmp = INT_MAX; int last_min = 0; for (i = 0; i len; i+) tmp = INT_MAX; for(j = 0; j len; j+) if (arrj.index != -1) _( 1 )_; if (arrj.key = tmp) min = j; _( 2 )_; if (i = 0) _( 3 )_; else arrlast_min.index = min; arrmin.index = 0; _( 4 )_; _( 5 )_; return 0;int main(int argc, char *argv) int index = head; array_sort(array, 8);index = head;/遍历数组，将数组从小到大打印出来 do printf(array%d: %d , index, arrayindex.key); _( 6 )_; while (arrayindex.index != -1); printf(n); return 0;C语言试题答案1：a01=0 b1=02：sizeof(a)=16 sizeof(b)=40 sizeof(buf)=10 sizeof(buf)=43：q = hello p=end of world, 20124：p！=q5：sum=996：xyzy=x7：offset i = 0offset c = 4offset s = 6offs