实验六


#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;
}

2004

2001

 

 

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;
}

2004 2016

2001 2004

 

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;
}

问题1:s1大小为13;

sizeof(s1)计算的是字符数组的大小,strlen(s1)计算的是字符串的长度。

问题2:不能

问题3:交换了

 

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;
}

问题1:字符串的首地址;计算字符串首地址的存储大小;统计字符串长度。

问题2:能。

问题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)
{
    char *p,*temp;
    for(p=str;p<str+5;p++) 
    {
        if(strlen(p) != 18)
            {
               return 0;
               break;
            }           
        else
            {
                temp = p;
                for(;temp<p+18;temp++)
                {
                    if(*temp>='0'&&*temp<='9'||*temp == 'X');
                    else
                        {
                            return 0;
                            break;
                        }
                }
                if(temp = p+18)
                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,n;
    n=strlen(s);
        for(i=0;i<=(n/2);++i)
        {
            if(s[i]==s[n-i-1])
            {
            return 1;
        }
            if(s[i]!=s[n-1-i])
            return 0;
        }
}

 

task6.

#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;
    for(i=0;i<N;++i)
    {
    if(s[i]!='z'&&s[i]!='Z'&&s[i]<='z'&&s[i]>='A')
    {
        s[i]=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;
    for(i=0;i<N;++i)
    {
    if(s[i]!='a'&&s[i]!='A'&&s[i]<='z'&&s[i]>='A')
    {
        s[i]=s[i]-1;
    }
    else if(s[i]=='a')
    s[i]='z';
    else if(s[i]=='A')
    s[i]='Z';
}     
}

 

posted @ 2022-06-14 11:53  Ares-Snape  阅读(12)  评论(1编辑  收藏  举报