实验五

实验1

1
#include <stdio.h> 2 #define N 5 3 4 void input(int x[], int n); 5 void output(int x[], int n); 6 void find_min_max(int x[], int n, int *pmin, int *pmax); 7 8 int main() { 9 int a[N]; 10 int min, max; 11 12 printf("录入%d个数据:\n", N); 13 input(a, N); 14 15 printf("数据是: \n"); 16 output(a, N); 17 18 printf("数据处理...\n"); 19 find_min_max(a, N, &min, &max); 20 21 printf("输出结果:\n"); 22 printf("min = %d, max = %d\n", min, max); 23 24 return 0; 25 } 26 27 void input(int x[], int n) { 28 int i; 29 30 for(i = 0; i < n; ++i) 31 scanf("%d", &x[i]); 32 } 33 34 void output(int x[], int n) { 35 int i; 36 37 for(i = 0; i < n; ++i) 38 printf("%d ", x[i]); 39 printf("\n"); 40 } 41 42 void find_min_max(int x[], int n, int *pmin, int *pmax) { 43 int i; 44 45 *pmin = *pmax = x[0]; 46 47 for(i = 0; i < n; ++i) 48 if(x[i] < *pmin) 49 *pmin = x[i]; 50 else if(x[i] > *pmax) 51 *pmax = x[i]; 52 }

捕获

 

作用:历遍数组 找到最大和最小值

pmin指向主函数中min的指针

pmax指向主函数中max的指针

#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 = 0; i < n; ++i)
        if(x[i] > x[max_index])
            max_index = i;
    
    return &x[max_index];
}

1

 功能:历遍数组中所有元素 找到最大值并返回其地址

可以 历遍数组后ptr更新为最大元素的地址

实验2

 1 #include <stdio.h>
 2 #include <string.h>
 3 #define N 80
 4 
 5 int main() {
 6     char s1[N] = "Learning makes me happy";
 7     char s2[N] = "Learning makes me sleepy";
 8     char tmp[N];
 9 
10     printf("sizeof(s1) vs. strlen(s1): \n");
11     printf("sizeof(s1) = %d\n", sizeof(s1));
12     printf("strlen(s1) = %d\n", strlen(s1));
13 
14     printf("\nbefore swap: \n");
15     printf("s1: %s\n", s1);
16     printf("s2: %s\n", s2);
17 
18     printf("\nswapping...\n");
19     strcpy(tmp, s1);
20     strcpy(s1, s2);
21     strcpy(s2, tmp);
22 
23     printf("\nafter swap: \n");
24     printf("s1: %s\n", s1);
25     printf("s2: %s\n", s2);
26 
27     return 0;
28 }

2

 

s1大小:80

sizeof(s1):计算字符组s1所占的总字节数

strlen(s1):字符串中实际有效的字节数

不能  s1是地址常量,不能被赋值

是  

 1 #include <stdio.h>
 2 #include <string.h>
 3 #define N 80
 4 
 5 int main() {
 6     char *s1 = "Learning makes me happy";
 7     char *s2 = "Learning makes me sleepy";
 8     char *tmp;
 9 
10     printf("sizeof(s1) vs. strlen(s1): \n");
11     printf("sizeof(s1) = %d\n", sizeof(s1));
12     printf("strlen(s1) = %d\n", strlen(s1));
13 
14     printf("\nbefore swap: \n");
15     printf("s1: %s\n", s1);
16     printf("s2: %s\n", s2);
17 
18     printf("\nswapping...\n");
19     tmp = s1;
20     s1 = s2;
21     s2 = tmp;
22 
23     printf("\nafter swap: \n");
24     printf("s1: %s\n", s1);
25     printf("s2: %s\n", s2);
26 
27     return 0;
28 }

3

 

s1:字符串在内存中的首地址

sizeof(s1):字符指针变量本身的大小

strlen(s1):到‘\0’结束的字符个数

可以 原来是定义指针变量的同时初始化 现在是先定义指针变量 再单独赋值

s1和s2存储的地址值

不会

 

实验3

 1 #include <stdio.h>
 2 
 3 int main() {
 4     int x[2][4] = {{1, 9, 8, 4}, {2, 0, 4, 9}};
 5     int i, j;
 6     int *ptr1;     
 7     int(*ptr2)[4];
 8     printf("输出1: 使用数组名、下标直接访问二维数组元素\n");
 9     for (i = 0; i < 2; ++i) {
10         for (j = 0; j < 4; ++j)
11             printf("%d ", x[i][j]);
12         printf("\n");
13     }
14 
15     printf("\n输出2: 使用指针变量ptr1(指向元素)间接访问\n");
16     for (ptr1 = &x[0][0], i = 0; ptr1 < &x[0][0] + 8; ++ptr1, ++i) {
17         printf("%d ", *ptr1);
18 
19         if ((i + 1) % 4 == 0)
20             printf("\n");
21     }
22                          
23     printf("\n输出3: 使用指针变量ptr2(指向一维数组)间接访问\n");
24     for (ptr2 = x; ptr2 < x + 2; ++ptr2) {
25         for (j = 0; j < 4; ++j)
26             printf("%d ", *(*ptr2 + j));
27         printf("\n");
28     }
29 
30     return 0;
31 }

4

 

包含四个int数据类型元素的一维数组的指针

包含四个int数据类型元素的一维数组

 

实验4

 1 #include <stdio.h>
 2 #define N 80
 3 
 4 void replace(char *str, char old_char, char new_char); 
 5 int main() {
 6     char text[N] = "Programming is difficult or not, it is a question.";
 7 
 8     printf("原始文本: \n");
 9     printf("%s\n", text);
10 
11     replace(text, 'i', '*');
12 
13     printf("处理后文本: \n");
14     printf("%s\n", text);
15 
16     return 0;
17 }
18 
19 void replace(char *str, char old_char, char new_char) {
20     int i;
21 
22     while(*str) {
23         if(*str == old_char)
24             *str = new_char;
25         str++;
26     }
27 }

5

 replace:将所有的old_char 换成 new_char

可以

 

实验5

 

#include <stdio.h>
#define N 80


char *str_trunc(char *str, char x) {
    char *p = str;  
    
    while (*p != '\0') {
        
        if (*p == x) {
            *p = '\0';  
            break;      
        }
        p++;  
    }
    return str;  句)
}

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

   
    while (1) {
        printf("输入字符串:");
        scanf("%s", str);  
        printf("输入一个字符:");
        getchar(); 
        ch = getchar();

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

        printf("截断处理后的字符串:%s\n\n", result);

        
        printf("是否继续?(y/n):");
        ch = getchar();
        getchar();
        if (ch != 'y' && ch != 'Y') {
            break;
        }
    }

    return 0;
}

屏幕截图 2025-12-11 230057

跳过输入字符的步骤  存放\n

 

实验六

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

int check_id(char *str);

int main() {
    char *pid[N] = {"31010120000721656X",
                    "3301061996X0203301",
                    "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) {
    int len = strlen(str);
    char *p = str;
    if (len != 18) return 0;
    while (p < str + 17) {
        if (*p < '0' || *p > '9') return 0;
        p++;
    }
    if ((*p >= '0' && *p <= '9') || *p == 'X' || *p == 'x' || *p == 'Y' || *p == 'y')
        return 1;
    else
        return 0;
}

屏幕截图 2025-12-11 230741

 

实验七

 1 #include <stdio.h>
 2 #define N 80
 3 
 4 void encoder(char *str, int n);
 5 void decoder(char *str, int n);
 6 
 7 int main() {
 8     char words[N];
 9     int n;
10 
11     printf("输入英文文本: ");
12     gets(words);
13 
14     printf("输入n: ");
15     scanf("%d", &n);
16 
17     printf("编码后的英文文本: ");
18     encoder(words, n);
19     printf("%s\n", words);
20 
21     printf("对编码后的英文文本解码: ");
22     decoder(words, n);
23     printf("%s\n", words);
24 
25     return 0;
26 }
27 
28 void encoder(char *str, int n) {
29     char *p = str;
30     n = n % 26;
31     while (*p != '\0') {
32         if (*p >= 'a' && *p <= 'z') {
33             *p = (*p - 'a' + n) % 26 + 'a';
34         } else if (*p >= 'A' && *p <= 'Z') {
35             *p = (*p - 'A' + n) % 26 + 'A';
36         }
37         p++;
38     }
39 }
40 
41 void decoder(char *str, int n) {
42     char *p = str;
43     n = n % 26;
44     while (*p != '\0') {
45         if (*p >= 'a' && *p <= 'z') {
46             *p = (*p - 'a' - n + 26) % 26 + 'a';
47         } else if (*p >= 'A' && *p <= 'Z') {
48             *p = (*p - 'A' - n + 26) % 26 + 'A';
49         }
50         p++;
51     }
52 }

屏幕截图 2025-12-11 231448

屏幕截图 2025-12-11 231639

屏幕截图 2025-12-11 231754

 

实验八

#include <stdio.h>
#include <string.h>

int main(int argc, char *argv[]) {
    int i, j;
    char *tmp;
    if (argc < 2) {
        printf("请输入参数!\n");
        return 1;
    }
    for (i = 1; i < argc - 1; ++i) {
        for (j = i + 1; j < argc; ++j) {
            if (strcmp(argv[i], argv[j]) > 0) {
                tmp = argv[i];
                argv[i] = argv[j];
                argv[j] = tmp;
            }
        }
    }
    for (i = 1; i < argc; ++i) {
        printf("hello, %s\n", argv[i]);
    }
    return 0;
}

 

posted @ 2025-12-11 23:22  cchheenn  阅读(6)  评论(0)    收藏  举报