实验五

task1_1

#include<stdio.h> 
#define N 4

int main()
{
    int x[N]={1,9,8,4};
    int i;
    int *p;
    
    for(i=0;i<N;++i)
    printf("%d",x[i]);
    printf("\n");
    
    for(p=x;p<x+N;++p)
    printf("%d",*p);
    printf("\n");
    
    p=x;
    for(i=0;i<N;++i)
    printf("%d",*(p+i));
    printf("\n");
    
    p=x;
    for(i=0;i<N;++i)
    printf("%d",p[i]);
    printf("\n");
    
    return 0;
}

 

 task1_2

#include<stdio.h>

int main()
{
    int x[2][4]={{1,9,8,4},{2,0,4,9}};
    int i,j;
    int *p;
    int(*q)[4];
    
    for(i=0;i<2;++i)
    {
        for(j=0;j<4;++j)
        printf("%d",x[i][j]);
        printf("\n");    
    }
    
    for(p=&x[0][0],i=0;p<&x[0][0]+8;++p,++i)
    {
        printf("%d",*p);
        if((i+1)%4==0)
        printf("\n");
    }
    
    for(q=x;q<x+2;++q)
    {
        for(j=0;j<4;++j)
        printf("%d",*(*q+j));
        printf("\n");
    }
    
    return 0;
}

 

 task2_1

#include<stdio.h>
#include<string.h>
#define N 80

int main() 
{
    char s1[]="Learning makes me happy";
    char s2[]="Learning makes me sleepy";
    char tmp[N];
    
    printf("sizeof(s1) vs. strlen(s1)；\n");
    printf("sizeof(s1)=%d\n",sizeof(s1));
    printf("strlen(s1)=%d\n",strlen(s1));
    
    printf("\nbefore swap:\n");
    printf("s1:%s\n",s1);
    printf("s2:%s\n",s2);
    
    printf("\nswapping...\n");
    strcpy(tmp,s1);
    strcpy(s1,s2) ;
    strcpy(s2,tmp);
    
    printf("\nafter swap:\n");
    printf("s1:%s\n",s1);
    printf("s2:%s\n",s2);
    
    return 0;
}

1.  23

    sizeof(s1)计算数组s1被分配的空间的大小包括‘\0’

    strlen(s1)统计数组s1的长度且返回的长度大小中不包括‘\0’

2.  不能，s1是一个地址，其并不能当作数组来储存字符串

3.  完成了交换

 

 task2_2

#include<stdio.h>
#include<string.h>
#define N 80

int main() 
{
    char *s1="Learning makes me happy";
    char *s2="Learning makes me sleepy";
    char *tmp;
    
    printf("sizeof(s1) vs. strlen(s1)；\n");
    printf("sizeof(s1)=%d\n",sizeof(s1));
    printf("strlen(s1)=%d\n",strlen(s1));
    
    printf("\nbefore swap:\n");
    printf("s1:%s\n",s1);
    printf("s2:%s\n",s2);
    
    printf("\nswapping...\n");
    tmp=s1;
    s1=s2;
    s2=tmp;
    
    printf("\nafter swap:\n");
    printf("s1:%s\n",s1);
    printf("s2:%s\n",s2);
    
    return 0;
}

1.  s1存放字符串所对应的地址

    计算s1作为地址变量存放一个地址所需要的字节数

    计算在s1地址下字符串的长度

2.  可以，该处意为s1指向此字符串，上一次的s1是地址常量不能存放字符串

3.  交换的是两个字符串所对应的地址，内存空间没有改变

 

 task3

#include<stdio.h>

void str_cpy(char *target, const char *source) ;
void str_cat(char *str1,char *str2);

int main()
{
    char s1[80], s2[20]="1984";
    
    str_cpy(s1,s2);
    puts(s1);
    
    str_cat(s1,"Animal Farm");
    puts(s1);
    
    return 0;    
}

void str_cpy(char *target, const char *source)
{
    while(*target++=*source++)
    ;
}

void str_cat(char *str1,char *str2)
{
    while(*str1)
    str1++;
    
    while(*str1++=*str2++)
    ;
}

 

 

task4

#include<stdio.h>
#define N 80
int func(char *);

int main()
{
    char str[80];
    
    while(gets(str)!=NULL)
    {
        if(func(str))
        printf("yes\n");
        else
        printf("no\n");
    }
    
    return 0;
}

int func(char *str)
{
    char *begin,*end;
    
    begin=end=str;
    
    while(*end)
    end++;
    
    end--;
    
    while(begin<end)
    {
        if(*begin!=*end)
        return 0;
        else
        {
            begin++;
            end--;
        }
    }
    
    return 1;
}

 

 

task5

#include<stdio.h>
#define N 80

void func(char*);

int main()
{
    char s[N];
    
    while(scanf("%s",s)!=EOF)
    {
        func(s);
        puts(s);
    }
        return 0;
}

void func(char *str)
{
    int i;
    char*p1,*p2,*p;
    
    p1=str;
     while(*p1=='*')
      p1++;
    p2=str;
     while(*p2)
      p2++;
     p2--;
    while(*p2=='*')
     p2--;
     
     p=str;
     i=0;
     while(p<p1)
     {
         str[i]=*p;
         p++;
         i++;    
     }
     
     while(p<=p2)
     {
         if(*p!='*')
         {
             str[i]=*p;
             i++;
         }
         p++;
     }
     
     while(*p!='\0')
     {
         str[i]=*p;
         p++;
         i++;
     }
     
    str[i]='\0';
}

 

 

 

task6_1

#include<stdio.h>
#include<string.h>
void sort(char *name[],int n);

int main()
{
    char *course[4]={"C Program",
                     "C++ Object Oriented Program",
                     "Operating System",
                     "Date structure and Algorithms"};
    int i;
    
    sort(course,4);
    
    for(i=0;i<4;i++)
    printf("%s\n",course[i]);
    
    return 0;
}

void sort(char *name[],int n)
{ 
    int i,j;
    char *tmp;
    
    for(i=0;i<n-1;++i)
      for(j=0;j<n-1-i;++j)
        if(strcmp(name[j],name[j+1])>0)
        {
            tmp=name[j];
            name[j]=name[j+1];
            name[j+1]=tmp;
        }

}

 

 

task6_2

#include<stdio.h>
#include<string.h>
void sort(char *name[],int n);

int main()
{
    char *course[4]={"C Program",
                     "C++ Object Oriented Program",
                     "Operating System",
                     "Date structure and Algorithms"};
    int i;
    
    sort(course,4);
    
    for(i=0;i<4;i++)
    printf("%s\n",course[i]);
    
    return 0;
}

void sort(char *name[],int n)
{ 
    int i,j,k;
    char *tmp;
    
    for(i=0;i<n-1;i++)
    {
      k=i;
      for(j=i+1;j<n;j++)
    
        if(strcmp(name[j],name[k])<0)
        k=j;
        if(k!=i)
        {
            tmp=name[i];
            name[i]=name[k];
            name[k]=tmp;
        }
    }
}

交换的是指针变量的值，而不是内存中字符串储存的位置。

 

task7

 

#include<stdio.h>
#include<string.h>
#define N 5

int check_id(char *str);

int main()
{
    char*pid[N]={"31010120000721656X",
                 "330106199609203301",
                 "53010220051126571",
                 "510104199211197977",
                 "53010220051126133Y"};
    int i;
    
    for(i=0;i<N;++i)
        if(check_id(pid[i]))
           printf("%s\tTrue\n",pid[i]);
        else
           printf("%s\tFalse\n",pid[i]);
           
    return 0;
}

int check_id(char *str)
{
    int i;
    for(i=0;i<17;++i)
    {
    if(!(47<str[i]&&str[i]<58))
        return 0;
    }
    if((47<str[17]&&str[17]<58)||str[17]=='X')
        return 1;
    else
        return 0;
}

 

 

task8

 

#define N 80
void encoder(char *s);
void decoder(char *s);

int main()
{
    char words[N];
    
    printf("输入英文文本：");
    gets(words) ;
    
    printf("编码后的英文文本：");
    encoder(words) ;
    printf("%s\n",words);
    
    printf("对编码后的英文文本解码：");
    decoder(words) ;
    printf("%s\n",words);
    
    return 0;
}
void encoder (char *s)
{
    int i=0;
    for(;s[i]!='\0';i++)
    {
    if(s[i]==90)
    s[i]=65;
    if(s[i]==122)
    s[i]=97;    
    if((65<=s[i]&&s[i]<90)||(97<=s[i]&&s[i]<122))
    s[i]=s[i]+1;
    }
}

void decoder (char *s)
{
    int i=0;
    for(;s[i]!='\0';i++)
    {
    if(s[i]==65)
    s[i]=90;
    if(s[i]==97)
    s[i]=122;    
    if((65<s[i]&&s[i]<=90)||(97<s[i]&&s[i]<=122))
    s[i]=s[i]-1;
    }
}