实验五

实验任务1

源代码1-1

#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;
    }

源代码1-2

#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 (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;
    }

实验任务2

源代码2-2

#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[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;
   }

回答问题1

①数组s1的大小是24

②sizeof（s1）计算的是s1所有常量（数字常亮，表达式，基本类型，数组，结构体）等的字节数

③strlen(s1)统计的是数组s1的长度

回答问题2

不能，原因是此时数组名表示的是一个指针，指向数组首元素地址，这样的赋值就等于尝试修改地址

回答问题3

是交换了

源代码2-2

#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;
}

回答问题1

①存放s1对应的地址型变量

②izeof(s1)计算的是s1中存放数据所占的字节数

③strlen(s1)统计的是s1所指向的字符串的长度

回答问题2

可以

回答问题3

交换的是s1和s2的指向，两个字符串常量在内存存储单元中没有交换

实验任务3

源代码

#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++);
    }

实验任务4

源代码

#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;
   }

试验任务5

源代码

#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';
    }

实验任务6

源代码6-1

#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",
                       "Data 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;
    }
}

源代码6-2

#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",
                       "Data 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;
       }
}
}

回答问题

交换的是指针变量的值，即交换了指针的指向，内存中字符串的存储位置没有发生交换

实验任务7

源代码

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

int check_id(char *str); // 函数声明

int main()
{
    char *pid[N] = {"31010120000721656X",
                    "330106199609203301",
                    "53010220051126571",
                    "510104199211197977",
                    "53010220051126133Y"};
    int i;

    for (i = 0; i < N; ++i)
        if (check_id(pid[i])) // 函数调用
            printf("%s\tTrue\n", pid[i]);
        else
            printf("%s\tFalse\n", pid[i]);

    return 0;
}

// 函数定义
// 功能: 检查指针str指向的身份证号码串形式上是否合法。
// 形式合法，返回1，否则，返回0
int check_id(char *str)
{

    if(strlen(str) != 18)
        return 0;
        
    if((str[17] >= 48 && str[17] <= 57) || str[17] == 'X')
        return 1;
        
    return 0;

}

实验任务8

源代码

#include <stdio.h>
#define N 80

void encoder(char *s); // 函数声明
void decoder(char *s); // 函数声明

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)
{
    
    char *ln;
    int i=0;
    
    ln=s;
    
    while(*ln)
    {
        if (*ln=='z') 
            s[i++]='a',ln++;
        else if 
            (*ln=='Z') s[i++]='A',ln++;
        else
        {
            s[i++]=*ln+1;
            ln++;
        }
    }
 
}


void decoder(char *s)
{
    
    char *ln;
    int i=0;
    
    ln=s;
    
    while(*ln)
    {
        if (*ln=='a') 
            s[i++]='z',ln++;
        else if 
            (*ln=='A') s[i++]='Z',ln++;
        else
        {
            s[i++]=*ln-1;
                ln++;
        }
    }

}

posted @ 2023-05-07 21:52 郑文仪阅读(16) 评论(0) 收藏举报

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