问题 单项选择题

Speaker A: The native American Craft Exhibit closed this afternoon.Speaker B: ______

A.Oh, no! I've wanted to see that for months.

B.Well, why didn't you take me to see the exhibit before it closed

C.Thank you. When will there be such an exhibit again

D.If I were you, I would have gone to see it.

答案

参考答案:A

解析:A方向B方传递了展览结束的信息,A项表达了B方未看到展览的遗憾之情,是正确反应。B项带有责备之意,不得体。C项后半部分不符合这一情景下的交际惯例,D项与A方的话不一致。

解答题
问答题

针对以下C语言程序,请按要求回答问题。

已知link. c源程序如下:

/*link. c程序对单向链表进行操作,首先建立一个单向链表,然后根据用户的选择可以对其进行插入结点、删除结点和链表反转操作*/

#include<stdio. h>

#include<stdlib. h>

typedef struct list_node * list_pointer; //定义链表指针

typedef struct list_node{ //定义链表结构

int data;

list_pointer link;

}list_node;

//用到的操作函数

list_pointer create(); //建立一个单向链表

void insert(list_pointer * p_ptr,list_pointer node); //在node后加入一个新的结点

void delete_node(list_pointer * p_ptr,list_pointer trail,list_pointer node);

//删除前一个结点是trail的当前结点node

void print(list_pointer * p_ptr); //打印链表结点中的值

list_pointer invert(list_pointer lead); //反转链表

int main()

{

list_pointer ptr=NULL;

list_pointer node,trail;

list_pointer * P=&ptr;

int choose,location,i;

printf("you should create a link first:\n");

//建立一个单向链表

prt=create(); //ptr指向链表的第一个结点

print(ptr);

//根据用户的不同选择进行相应的操作:

printf("input number 0,you can quit the program\n");

printf("input number 1,you can insert a new node to link\n"):

printf("input number 2,you can delete a node from the link\n");

printf("input number 3,you can invert the link\n"):

printf("please input you choice\n");

scanf("%d",&choose);

while(choose!=0){

switch(choose){

case 1:

i=1:

while(i<location){

node=node->link;

i++:

}

insert(p,node); //p为指向ptr的指针

print(ptr);

break;

case 2:

printf("you will delete a node from the link\n");

printf("please input the location of the node:\n");

scanf("%d",&location):

node=ptr;

if(location==1)

trail=NULL;

trail=ptr;

i=1:

while(i<location){

trail=trail->link:

i++:

}

node=trail->link;

delete_node(p,trail,node);

print(ptr);

break;

case 3:

printf("you will invert the link\n");

ptr=invert(ptr);

print(ptr);

break;

default;

break;

return -1;

}

printf("please input you choice\n");

scanf("%d". &choose):

}

return 0;

//根据用户的输入值建立一个新的单向链表:

list_pointer create()

{

int i,current,length;

list_pointer p1,p2,head;

printf("please input the node number of the link:\n");

scanf("%d". &length):

printf("the number of the link is:%d",length);

printf("please input the data for the link node:\n");

i=0;

p1=p2=(list_pointer)malloc(sizeof(list_node));

head=p1;

for(i=1;i<length;i++){

scanf("%d",&current);

p1->data=current;

p2->link=p1;

p2=p1;

p1=(list_pointer)malloc(sizeof(list_node));

}

p2->link=NULL;

return head;

}

设计一组测试用例,尽量使main函数的语句覆盖率能达到100%。如果认为该函数的语句覆盖率无法达到100%,需要说明原因。