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--->s3 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

#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)
        return 0;
        
    while(*str != 0)
    {
        if(*str>='0'&& *str<='9' || *str=='X')
        
           i=1;
        else
           i=0;
           
    str++;
    }
    
    return i;
      
}

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,k;
    j=strlen(s);
    for(i=0;i<j/2;i++)
    {
        if(s[i]==s[j-i-1])
          k=1;
        else
          return 0;
    }
    return k;
    
}

task6

#include <stdio.h>
#include <string.h>
#define N 80
void encoder(char *s);  
void decoder(char *s); 

int main()
{
    char words[N];
    
    printf("Enter English text: ");
    gets(words);
    
    printf("Encoded English text: ");
    encoder(words);  
    printf("%s\n", words);
    
    printf("Decode the encoded English text: ");
    decoder(words);  
    printf("%s\n", words);
    
    return 0;
}

void encoder(char *s)
{
    int i,j;
    j==strlen(s);
    for(i=0;i<=j;i++)
    {
        if((s[i]>='a'&&s[i]<'z')||(s[i]>='A'&&s[i]<'Z'))
        {
            s[i]=s[i]+1;
        }
        else if(s[i]=='z'||s[i]=='Z')
        {
            s[i]=s[i]-25;
        }
    }
}

void decoder(char *s)
{
    int i,j;
    j==strlen(s);
    for(i=0;i<=j;i++)
    {
        if((s[i]>'a'&&s[i]<='z')||(s[i]>'A'&&s[i]<='Z'))
        {
            s[i]=s[i]-1;
        }
        else if(s[i]=='a'||s[i]=='A')
        {
            s[i]=s[i]+25;
        }
    }
}

 

posted @ 2022-06-14 10:33  尚黑  阅读(4)  评论(1编辑  收藏  举报