实验五

实验任务1

程序代码:

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;
}

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 (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;
}

 

运行结果:

 

实验任务2

程序代码:

 

#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;
}
#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;
}

 

运行结果:

 

讨论:

  1、s1的大小是24,sizeof (s1) 计算的是数组中存放的字节数,strlen (s1) 计算的是字符串的有效字符数(除空字符以外的字符个数)。

  2、不能,s1是数组名,存放的是地址名。

  3、s1与s2中的内容进行了交换。

  

  1、s1中存放的是地址,sizeof (s1) 计算的是指针变量占用的字节大小为8个,strlen (s1) 统计的是有效字符数。

  2、不能换,task2_1中是将字符串赋值给数组名或者说是起始地址,在2_1中这是一个常量 因此赋值是非法的;而2_2中的s1是指针变量,存放的应是地址。

  3、交换的是指针变量的名称,二者在内存存储单元中没有交换。

实验任务3

程序代码:

 

#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++)
        ;
}

 

运行结果:

 

实验任务4

程序代码:

 

#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;
}

 

运行结果:

 

实验任务5

程序代码:

 

#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';
    
}

 

运行结果:

 

实验任务6

程序代码:

 

#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;
            }
}
#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;
        }
    }
}

 

运行结果:

 

实验任务7

程序代码:

 

#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)
{
    char *p;
    int i;
 
    p = str;
 
    while((*p >= '0' && *p <= '9')||*p=='X')
            p++;
 
    if(*p == '\0' && strlen(str) == 18)
        return 1;
    else
        return 0;
    // 补足函数实现
    // ...
}

 

运行结果:

 

 

实验任务8

程序代码:

 

#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;
}

/*函数定义
功能:对s指向的字符串进行编码处理
编码规则:
对于a~z或A~Z之间的字母字符,用其后的字符替换; 其中,z用a替换,Z用A替换
其它非字母字符,保持不变
*/
void encoder(char *s)
{
    
    char *p;
    int i=0;
    
    p=s;
    
    while(*p)
    {
        if (*p=='z') 
            s[i++]='a',p++;
        else if 
            (*p=='Z') s[i++]='A',p++;
        else
        {
            s[i++]=*p+1;
            p++;
        }
    }
    // 补足函数实现
}

/*函数定义
功能:对s指向的字符串进行解码处理
解码规则:
对于a~z或A~Z之间的字母字符,用其前面的字符替换; 其中,a用z替换,A用Z替换
其它非字母字符,保持不变
*/
void decoder(char *s)
{
    
    char *p;
    int i=0;
    
    p=s;
    
    while(*p)
    {
        if (*p=='a') 
            s[i++]='z',p++;
        else if 
            (*p=='A') s[i++]='Z',p++;
        else
        {
            s[i++]=*p-1;
                p++;
        }
    }
    // 补足函数实现
    // ×××
}

 

运行结果:

 

 

posted @ 2023-05-04 18:12  lzque  阅读(25)  评论(2编辑  收藏  举报