实验5

task1

task1-1

程序源码

#include <stdio.h>
#define N 4

int main()
{
    int x[N] = {1, 9, 8, 4};
    int i;
    int *p;
    
    //方式1:通过数组名和下标遍历输出数组元素
    for (i = 0; i < N; ++i)
        printf("%d", x[i]);
    printf("\n");
    
    //方式2:通过指针变量遍历输出数组元素(写法1)
    for (p = x; p < x + N; ++p)
        printf("%d", *p);
    printf("\n");
    
    //方式2:通过指针变量遍历输出数组元素(写法2)
    p = x;
    for (i = 0; i < N; ++i)
        printf("%d", *(p + 1));
    printf("\n");
    
    //方式2:通过指针变量遍历输出数组元素(写法3)
    p = x;
    for (i = 0; i < N; ++i)
        printf("%d", p[i]);
    printf("\n");
    
    return 0; 
}

程序运行截图

task1-2

程序源码

#include <stdio.h>

int main()
{
    int x[2][4] = {{1, 9, 8, 4}, {2, 0, 4, 9}};
    int i, j;
    int *p;      //指针变量,存放int类型数据的地址
    int(*q)[4];  //指针变量,指向包含4个int型元素的一维数组
    
    //使用数组名、下标访问二维数组元素
    for (i = 0; i < 2; ++i)
    {
        for (j = 0; j < 4; ++j)
            printf("%d", x[i][j]);
        printf("\n");
     } 
     
    //使用指针变量p间接访问二维数组元素
    for (p = &x[0][0], i = 0; p < &x[0][0] + 8; ++p, ++i)
    {
        printf("%d", *p);
        if ((i + 1) % 4 == 0)
            printf("\n");
     }
     
    //使用指针变量q间接访问二维数组元素
    for (q = x; q < x + 2; ++q)
    {
        for (j = 0; j < 4; ++j)
            printf("%d", *(*q + j));
        printf("\n");
     }
     
    return 0; 
}

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task2

task2-1

程序源码

#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;
}

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问题回答

1.数组s1的大小是24;sizeof(s1)计算的是数组s1的大小;strlen(s1)统计的是字符串的长度。

2.line7代码,不能替换成以下写法;因为char s1[]没有定义长度。

3.line20-22执行后,字符数组s1和s2中的内容交换了。

task2-2

程序源码

#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;
    
    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;
}

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问题回答

1.指针变量s1中存放的是地址;sizeof(s1)计算的是地址所占大小;strlen(s1)统计的是地址内容的长度。

2.line7代码能替换成下面的写法;char s[] 是定义数组,char *s 是定义指针。

3.line20-line22,交换的是内容;字符串常量"Learning makes me happy"和字符串常 量"Learning makes me sleepy"在内存存储单元中没有交换。

 

task3

程序源码

#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++)
        ;
}

程序运行截图

 

task4

程序源码

#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;
}

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task5

程序源码

#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';
}

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task6

task6-1

程序源码

#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;
            }
}

程序运行截图

task6-2

程序源码

#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, 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;
            }
    }
}

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问题回答

这两种算法实现中,交换的是指针变量的值。

 

task7

程序源码

#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; 
  } 

程序运行截图

 

task8

程序源码

#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)
{
   while(*s)
   {
        if(*s >= 65 && *s <= 90 || *s >= 97 && *s <= 122)
        {
            *s += 1;
            s++;
            continue;
        }
        
        else if(*s == 90 || *s == 122)
        {
            *s -= 25;
            s++;
            continue;
        }
        
    s++;
    
   }
}

 /*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)
{
    while(*s)
    {
        if(*s >= 66 && *s <= 90 || *s >= 98 && *s <= 122)
        {
            *s -= 1;
            s++;
            continue;
        }
        
        else if(*s == 65 || *s == 97)
        {
            *s += 25;
            s++;
            continue;
        }
        
    s++;
    
    }
}

程序运行截图

 

posted @ 2023-05-05 22:28  邱奕萱  阅读(8)  评论(0编辑  收藏  举报