实验六

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>
#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;
}

1.2004

2.2001

3.字符与整型变量存储空间大小不同

task2-1：

#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：

#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;

 } 

1.2004   2017

2.2001 2005

task3-1：

#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("s2:%s\n",s2);
    
    return 0;
    
}

 

1.s1-->13 sizeof(s1)计算的是字符数组的大小，strlen()统计的是字符串的长度
2.no
3.yes

task3-2：

#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("s2:%s\n",s2);
    
    return 0;
}

1.字符串的首地址.   字符串首地址的存储大小     字符串的长度

2. 可以
3.s1和s2的地址，没有

task4-1：

#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 k,i,c=0,flag=1;
    for(k=0;str[k]!='\0';k++)
    {
        c+=1;
        if(str[k]<'0'||str[k]>'9')
            if(str[k]!='X')
                flag=0;
    }    
    if(c!=18||flag==0)
    {
        return 0;
    }
    else 
    return 1;
}

task5：

#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,j,flag=1,n;
    n=strlen(s);
    if(n%2==0)
    {
        for(i=0;i<n/2;i++)
        {
            if(*(s+i)!=*(s+n-1-i))
                {
                flag=0;
                return 0;
                }
        }
    }
    else
    {
        for(i=0;i<(n-1)/2;i++)
        {
            if(*(s+i)!=*(s+n-i-1))
            {
                flag=0;
                return 0;
            }
        }
    }
    if(flag==1)
    {
        return 1;
    }
}

#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);  // 函数调用
    printf("%s\n", words);
    
    printf("对编码后的英文文本解码: ");
    decoder(words);  // 函数调用
    printf("%s\n", words);
    
    return 0;
}



void encoder(char *s)
{
    int i;
    for(i=0;s[i]!='\0';i++)
    {
        if(s[i]>='a'&&s[i]<='z')
        {
            if(s[i]!='z')
            {
                s[i]=s[i]+1;
            }
            else
            {
                s[i]='a';
            }
        }
        if(s[i]>='A'&&s[i]<='Z')
        {
            if(s[i]!='Z')
            {
                s[i]=s[i]+1;
            }
            else
            {
                s[i]='A';
            }
        }
    }
}



void decoder(char *s)
{
    int i;
    for(i=0;s[i]!='\0';i++)
    {
        if(s[i]>='a'&&s[i]<='z')
        {
            if(s[i]!='a')
            {
                s[i]=s[i]-1;
            }
            else
            {
                s[i]='z';
            }
        }
        if(s[i]>='A'&&s[i]<='Z')
        {
            if(s[i]!='A')
            {
                s[i]=s[i]-1;
            }
            else
            {
                s[i]='Z';
            }
        }
    } 
}