实验五

TASK 1.1

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#include <stdio.h>
#define N 5

void input(int x[], int n);
void output(int x[], int n);
void find_min_max(int x[], int n, int *pmin, int *pmax);

int main() {
    int a[N];
    int min, max;

    printf("录入%d个数据:\n", N);
    input(a, N);

    printf("数据是: \n");
    output(a, N);

    printf("数据处理...\n");
    find_min_max(a, N, &min, &max);

    printf("输出结果:\n");
    printf("min = %d, max = %d\n", min, max);

    return 0;
}

void input(int x[], int n) {
    int i;

    for(i = 0; i < n; ++i)
        scanf("%d", &x[i]);
}

void output(int x[], int n) {
    int i;

    for(i = 0; i < n; ++i)
        printf("%d ", x[i]);
    printf("\n");
}

void find_min_max(int x[], int n, int *pmin, int *pmax) {
    int i;

    *pmin = *pmax = x[0];

    for(i = 1; i < n; ++i)
        if(x[i] < *pmin)
            *pmin = x[i];
        else if(x[i] > *pmax)
            *pmax = x[i];
}
![](https://img2024.cnblogs.com/blog/3405472/202405/3405472-20240523100754684-1112008966.png) 1.找到最大值和最小值 2.指向x【0】的地址

TASK 1.2

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#include <stdio.h>
#define N 5

void input(int x[], int n);
void output(int x[], int n);
int *find_max(int x[], int n);

int main() {
    int a[N];
    int *pmax;

    printf("录入%d个数据:\n", N);
    input(a, N);

    printf("数据是: \n");
    output(a, N);

    printf("数据处理...\n");
    pmax = find_max(a, N);

    printf("输出结果:\n");
    printf("max = %d\n", *pmax);

    return 0;
}

void input(int x[], int n) {
    int i;

    for(i = 0; i < n; ++i)
        scanf("%d", &x[i]);
}

void output(int x[], int n) {
    int i;
    
    for(i = 0; i < n; ++i)
        printf("%d ", x[i]);
    printf("\n");
}

int *find_max(int x[], int n) {
    int max_index = 0;
    int i;

    for(i = 1; i < n; ++i)
        if(x[i] > x[max_index])
            max_index = i;
    
    return &x[max_index];
}
![](https://img2024.cnblogs.com/blog/3405472/202405/3405472-20240523101840484-1669488123.png) 1.返回数据中最大值的地址 2.可以

TASK 2.1

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#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;
}
![](https://img2024.cnblogs.com/blog/3405472/202405/3405472-20240523101944042-724595083.png) 1.大小为24 sizeof(s1)计算的是s1占用内存字节数 strlen(s1)计算的是字符串长度 2.不能,定义会出错 3.发生了交换

TASK 2.2

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#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;
}
![](https://img2024.cnblogs.com/blog/3405472/202405/3405472-20240523102837060-342765621.png)

1.指针变量s1中存放的是字符串地址
sizeof(s1) 计算的是单词数
strlen(s1) 统计的是字符串长度
2.不能
3.没有交换,交换的是地址

TASK 3

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#include <stdio.h>

#include <stdio.h>

int main() {
    int x[2][4] = {{1, 9, 8, 4}, {2, 0, 4, 9}};
    int i, j;
    int *ptr1;     // 指针变量,存放int类型数据的地址
    int(*ptr2)[4]; // 指针变量,指向包含4个int元素的一维数组

    printf("输出1: 使用数组名、下标直接访问二维数组元素\n");
    for (i = 0; i < 2; ++i) {
        for (j = 0; j < 4; ++j)
            printf("%d ", x[i][j]);
        printf("\n");
    }

    printf("\n输出2: 使用指向元素的指针变量ptr1间接访问二维数组元素\n");
    for (ptr1 = &x[0][0], i = 0; ptr1 < &x[0][0] + 8; ++ptr1, ++i) {
        printf("%d ", *ptr1);

        if ((i + 1) % 4 == 0)
            printf("\n");
    }

    printf("\n输出3: 使用指向一维数组的指针变量ptr2间接访问二维数组元素\n");
    for (ptr2 = x; ptr2 < x + 2; ++ptr2) {
        for (j = 0; j < 4; ++j)
            printf("%d ", *(*ptr2 + j));
        printf("\n");
    }

    return 0;
}
![](https://img2024.cnblogs.com/blog/3405472/202405/3405472-20240523103320121-1071903040.png) int (*ptr)[4];中,标识符ptr表示的语义是指针 int *ptr[4];中,标识符ptr表示的语义是指针数组名

TASK4.1

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#include <stdio.h>
#include<stdlib.h>
#define N 80

void replace(char *str, char old_char, char new_char); // 函数声明

int main() {
    char text[N] = "c programming is difficult or not, it is a question.";

    printf("原始文本: \n");
    printf("%s\n", text);

    replace(text, 'i', '*'); // 函数调用 注意字符形参写法,单引号不能少

    printf("处理后文本: \n");
    printf("%s\n", text);

    system("pause");
    return 0;
}

// 函数定义
void replace(char *str, char old_char, char new_char) {
    int i;

    while(*str) {
        if(*str == old_char)
            *str = new_char;
        str++;
    }
}
![](https://img2024.cnblogs.com/blog/3405472/202405/3405472-20240523103505011-885779381.png) 1.replace的功能是将'i'改成'*' 2.line24,圆括号里循环条件可以改写成 *str != '\0' TASK4.2
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#include <stdio.h>
#include<stdlib.h>
#define N 80

void str_trunc(char *str, char x);

int main() {
    char str[N];
    char ch;

    printf("输入字符串: ");
    gets(str);

    printf("输入一个字符: ");
    ch = getchar();

    printf("截断处理...\n");
    str_trunc(str, ch);

    printf("截断处理后的字符串: %s\n", str);

    system("pause");
    return 0;
}

void str_trunc(char *str, char x) {
    while(*str) {
        if(*str == x)
            *str='\0';     // blank1

             str++;   // blank2
    }

    str--;// blank3
}
![](https://img2024.cnblogs.com/blog/3405472/202405/3405472-20240523104027561-1773301825.png) TASK 5.1
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#include <stdio.h>
#include<stdlib.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]);

    system("pause");
    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;
            }
}
![](https://img2024.cnblogs.com/blog/3405472/202405/3405472-20240523104456356-1350198246.png)

TASK5.2

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#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;
        }
    }
}

![](https://img2024.cnblogs.com/blog/3405472/202405/3405472-20240523104738323-784278912.png) 交换两个指针变量的值

TASK 6

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#include <stdio.h>
#include<stdlib.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]);

    system("pause");
    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) {
    int len=strlen(str);
    int i = 0;
    if (len!=18)
        return 0;
    while (*str) {
        for (; i < len; i++) {
            if (*(str + i) < '0' || (*(str + i) > '9' && *(str + i) != 'X'))

                return 0;
        }
        return 1;
    }
}
![](https://img2024.cnblogs.com/blog/3405472/202405/3405472-20240523105045529-241484426.png)

TASK 7

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#include <stdio.h>
#define N 80
void encoder(char* str); // 函数声明
void decoder(char* str); // 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* str) {
    while (*str) {
        if (*str == 'z')
            *str = 'a';
        else if (*str == 'Z')
            *str == 'A';
        else if ((*str >= 'a' && *str <= 'y') || (*str >= 'A' && *str <= 'Y'))
            (*str)++;
        str++;
    }

}

/*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* str) {
    while (*str) {
        if (*str == 'a')
            *str = 'z';
        else if (*str == 'A')
            *str == 'z';
        else if ((*str >= 'b' && *str <= 'z') || (*str >= 'B' && *str <= 'Z'))
            (*str)--;
        str++;
    }

}
![](https://img2024.cnblogs.com/blog/3405472/202405/3405472-20240523105304764-2089306149.png) ![](https://img2024.cnblogs.com/blog/3405472/202405/3405472-20240523105312377-764755310.png) ![](https://img2024.cnblogs.com/blog/3405472/202405/3405472-20240523105318387-475135236.png)
posted @ 2024-05-23 10:53  MINXZBC  阅读(7)  评论(0编辑  收藏  举报