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发布时间:2024-04-02 19:11:48

[简答题]

已知C源程序如下:
/*Input today’s date,output tomorrow’s date * /
/* version 2 * /
#include<stdio. h>
struct ydate
int day; int month; int year;;
int leap(struct ydate d)
if((d. year%4==0&&d. year%100 ! =0)||(d. year%400==0))
return 1;
else
return 0;

int numdays(struct ydate d)
int day;
static int daytab[]=
31,28,31,30,31,30,3l,31,30,31,30,31;
if(1eap(d)&&d. month==2)
day=29;
else
day=daytabEd. month-1];
return day;

int main(void)
struct ydate today,tomorrow;printf("format of date is:year,month,day输入的年、月、日之间应用逗号隔开/n);
printf(" today is:");
scanf(“%d,%d.%",&today.year,&today.month,&today.day);
while(0>=today. year
|| today. year>65535||0>=today. month||today. month>12)||
0>=today. day||today. day>numdays(today))
printf("input date error!reenter the day!/n");<

更多"已知C源程序如下: /*Input today&rsquo;s dat"的相关试题:

[简答题]已知C源程序如下:
/*Input today’s date,output tomorrow’s date*/
/*version 2*/
#include<stdio.h>
struct ydate
int day;int month;int year;;
int leap(struct ydate d)
if((d.year%4==0&&d.year%100!=0)||(d.year%400==0))
    return 1;
   else
    return 0;

int numdays(struct ydate d)
int day;
   static int daytab[]=
    31,28,31,30,31,30,31,31,30,31,30,31);
   if(leap(d)&&d.month==2)
    day=29;
   else
    day=daytab[d.month-1];
   return day;

int main(void)
struct ydate today,tomorrow;
   printf("format of date is:year,month,day 输入的年、月、日之间应用逗号隔开/n");
   printf("today is:");
   scanf("%d,%d,%d",&today.year,&today.month,&today.day);
   while(0>=today.year
|| today.year>65535 || 0>=today.month || today.month>12) ||
0>=today.day || today.day>numdays(today))
     printf("input date error!reenter the day!/n");
printf("today is:");
scanf("
[简答题]

已知C源程序如下:
/*Input today’s date,output tomorrow’s date*/
/*version 2*/
#include<stdio.h>
struct ydate
int day;int month;int year;;
int leap(struct ydate d)
if((d.year%4==0&&d.year%100!=0)||(d.year%400==0))
    return 1;
   else
    return 0;

int numdays(struct ydate d)
int day;
   static int daytab[]=
    31,28,31,30,31,30,31,31,30,31,30,31);
   if(leap(d)&&d.month==2)
    day=29;
   else
    day=daytab[d.month-1];
   return day;

int main(void)
struct ydate today,tomorrow;
   printf("format of date is:year,month,day 输入的年、月、日之间应用逗号隔开/n");
   printf("today is:");
   scanf("%d,%d,%d",&today.year,&today.month,&today.day);
   while(0>=today.year
|| today.year>65535 || 0>=today.month || today.mon

[简答题]已知C源程序如下:
/*A simple mailing list example using an array of structures. */
# include<stdion.h>
# include<stdlib.h>
define MAX 4
struct addr
char name[30];
char street[40];
char city[20];
unsigned long int zip;
addr_list[MAX];
void init_list(void),enter(void);
void deleteAddr(void),list(void);
int menu_select(void),find_free(void);
int main(void)

char choice;
init_list( );/*initialize the structure array*/
for(;;)
choice=menu_select( );
switch(choice)
case 1:enter( );
break;
case 2:deleteAddr( );
break;
case 3:list( );
break;
case 4:exit(0);


return 0;

/*Initialize the list.*/
void initlist(void)

register int t;
for(t=0;t<MAX; ++t)addr_list[t], name[0]=/0;

/*Get a menu selection. */
int menu_select(void)
char s[80. ;
int c;
prin
[简答题]
已知C源程序如下:
/*A simple mailing list example using an array of structures. */
# include<stdion.h>
# include<stdlib.h>
define MAX 4
struct addr{
char name[30];
char street[40];
char city[20];
unsigned long int zip;
}addr_list[MAX];
void init_list(void),enter(void);
void deleteAddr(void),list(void);
int menu_select(void),find_free(void);
int main(void)
{
char choice;
init_list( );/*initialize the structure array*/
for(;;){
choice=menu_select( );
switch(choice){
case 1:enter( );
break;
case 2:deleteAddr( );
break;
case 3:list( );
break;
case 4:exit(0);
}
}
return 0;
}
/*Initialize the list.*/
void initlist(void)
{
register int t;
for(t=0;t<MAX; ++t)addr_list[t], name[0]=/0;
}
/*Get a menu selection. */
int menu_select(void)
char s[80. ;
[简答题]
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");
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 c
[简答题]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节点、删除节点和链表反转操作*/
#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后加入一个新的节点
void delete_node (list_pointer *p_ptr, list_pointer trail, list_pointer node);
//删除前一个节点是trail的当前节点node
void print (list_pointer *p ptr) ; //打印链表节点中的值
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指向链表的第一个节点* /
print (ptr);
//根据用户的不同选择进行相应的操作
printf("inpu
[多项选择]针对以下C语言程序,请按要求回答问题。
已知weekday.c源程序如下:
#inelude<sgidio.h>
#include<conio.h>
/*主函数*/
Int main ( )

Char letter;
Printf("please input the first letter,"Y"to exit!/n");
While ((letter=getch ( )) !="Y") //当输入字母为Y是结束

Switch (letter)

Case ’S’:
Printf ("%c/n", letter);
Printf("please input second letter/n"); //输入第二个字母
If((letter=getch( ))==’a’)
Printf ("Saturday/n");
Else if (letter==’u’)
Printf ("Sunday/n");
Else printf("data error/n");
Break;
Case ’F’:
Printf ("fridaykn");
Break;
Case’M’:
Printf ("mondayha");
Break;
Case’T’:
Printf("%c/n",letter);
Printf("please input second letter/a"); //输入第二个字母
If ((letter=getch ( )) ==’u’)
Printf("Tuesday/n");
Else if(letter==’h’)
Printf ("Thursday/n");
Break;
Case’W’:
Printf ("Wednesday/n");


Return 0;

设计一组测试用例,使main函数的语句覆盖率尽量达到100%。

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