实验六

task1_1.c

#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.c

#include<stdio.h>
#define N 4

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

 

 1)p地址是2000，++p后p的地址是2004

2)p地址是2000，++p后p的地址是2001

不同的原因是一个是指向整形数组的地址变量，一个是指向字符数组的指针变量。

task2_1.c

#include<stdio.h>

int main()
{
    int x[2][4] = {{1,9,8,4},{2,0,2,2}};
    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_2.c

#include<stdio.h>

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

 

 1)++p存放的地址2004，++q存放的地址2004

2)++p存放的地址2001，++q存放的地址2001

task3_1.c

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

int main()
{
    char s1[]="C, I love u.";
    char s2[]="C, I hate u.";
    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("s1:%s\n",s2);
    
    return 0;
}

 

 1）数组s1的大小是12，sizeof计算的是数组所占字节数，strlen统计的是数组长度

2）不能替换

3）内容交换了

task3_2.c

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

int main()
{
    char *s1="C, I love u.";
    char *s2="C, I hate u.";
    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("s1:%s\n",s2);
    
    return 0;
}

 

 1)指针变量s1中存放的是C，I love u. sizeof(s1)计算的是所占字节 strlen统计的是数组的长度

2）可以替换

3）交换的是地址，二者在内存存储单元中没有交换

task4_1.c

#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;
    if(strlen(str) == 18)
    {
        for(i=0;i<18;++i)
        {
            if((str[i]>='0'&&str[i]<='9')||str[i]=='X')
            return 1;
        }
    }
    else
    return 0;
}

 

 task5.c

#include <stdio.h>
#include <string.h>
#define N 80
int is_palindrome(char *s); 
int main()
{
    char str[N];
    int flag;
    printf("Enter a string:\n");
    gets(str);
    flag = is_palindrome(str); 
    if (flag)
        printf("YES\n");
    else
        printf("NO\n");
    return 0;
}
int is_palindrome(char *s)
{
    int i;
       int n=strlen(s);
       for(i=0;i<n/2;i++)
       {
       if(s[i]!=s[n-i-1])
       {
          return 0;
        break;
       
       }
   }
   return 1;
}

 

 task6.c

#include <stdio.h>
#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,j;
    while(s[i])
    i++;
    
    j=i;
    for(i=0;i<j;i++)
    {
        if((s[i]>='a'&&s[i]<='y')||(s[i]>='A'&&s[i]<='Y'))
        s[i]+=1;
        
        else if(s[i]=='z')
         s[i]='a';
        else if(s[i]=='Z')
         s[i]=='A';
    }
}
void decoder(char *s)
{
    int i=0,j;
    while(s[i])
    i++;
    
    j=i;
    for(i=0;i<j;i++)
    {
        if((s[i]>='a'&&s[i]<='y')||(s[i]>='A'&&s[i]<='Y'))
        s[i]-=1;
        
        else if(s[i]=='a')
         s[i]='z';
        else if(s[i]=='A')
         s[i]=='Z';
    }
}