实验五

task1.1

 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 }

View Code

函数 find_min_max 功能是计算数组a[N]中的最大数和最小数

指针变量pmin、pmax分别指向&min, &max

task1.2

 1 #include <stdio.h>
 2 #define N 5
 3 
 4 void input(int x[], int n);
 5 void output(int x[], int n);
 6 int *find_max(int x[], int n);
 7 
 8 int main() {
 9     int a[N];
10     int *pmax;
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     pmax = find_max(a, N);
20 
21     printf("输出结果:\n");
22     printf("max = %d\n", *pmax);
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 int *find_max(int x[], int n) {
43     int max_index = 0;
44     int i;
45 
46     for(i = 0; i < n; ++i)
47         if(x[i] > x[max_index])
48             max_index = i;
49     
50     return &x[max_index];
51 }

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函数 find_max 功能是计算数组a[N]中的最大数；返回的是最大数的地址

函数 find_max 实现写成以下代码，仍然可以

task2.1

 1 #define  _CRT_SECURE_NO_WARNINGS
 2 #include <stdio.h>
 3 #include <string.h>
 4 #define N 80
 5 
 6 int main() {
 7     char s1[N] = "Learning makes me happy";
 8     char s2[N] = "Learning makes me sleepy";
 9     char tmp[N];
10 
11     printf("sizeof(s1) vs. strlen(s1): \n");
12     printf("sizeof(s1) = %d\n", sizeof(s1));
13     printf("strlen(s1) = %d\n", strlen(s1));
14 
15     printf("\nbefore swap: \n");
16     printf("s1: %s\n", s1);
17     printf("s2: %s\n", s2);
18 
19     printf("\nswapping...\n");
20     strcpy(tmp, s1);
21     strcpy(s1, s2);
22     strcpy(s2, tmp);
23 
24     printf("\nafter swap: \n");
25     printf("s1: %s\n", s1);
26     printf("s2: %s\n", s2);
27 
28     return 0;
29 }

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数组s1的大小是80； sizeof(s1) 计算的是数组s1所占的内存； strlen(s1) 统计的是数组s1中的有效字符数

line7代码，不能替换成以下写法，因为数组名是指针常量，不能对其进行赋值

line19-21执行后，字符数组s1和s2中的内容交换

task2.2

 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 }

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指针变量s1中存放的是字符串的首字符的地址； sizeof(s1) 计算的是指针变量s1占的内存； strlen(s1) 统计的是指针变量s1指向的字符串的有效字符数

line6代码能替换成下面的写法；数组名是不可修改的地址常量，不可对其赋值；指针是变量，可以对其赋值

line20-line21，交换的是两个指针指向的地址；字符串常量"Learning makes me happy"和字符串常量"Learning makes me sleepy"在内存中没有交换

task3

 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;     // 指针变量，存放int类型数据的地址
 7     int(*ptr2)[4]; // 指针变量，指向包含4个int元素的一维数组
 8 
 9     printf("输出1: 使用数组名、下标直接访问二维数组元素\n");
10     for (i = 0; i < 2; ++i) {
11         for (j = 0; j < 4; ++j)
12             printf("%d ", x[i][j]);
13         printf("\n");
14     }
15 
16     printf("\n输出2: 使用指针变量ptr1(指向元素)间接访问\n");
17     for (ptr1 = &x[0][0], i = 0; ptr1 < &x[0][0] + 8; ++ptr1, ++i) {
18         printf("%d ", *ptr1);
19 
20         if ((i + 1) % 4 == 0)
21             printf("\n");
22     }
23                          
24     printf("\n输出3: 使用指针变量ptr2(指向一维数组)间接访问\n");
25     for (ptr2 = x; ptr2 < x + 2; ++ptr2) {
26         for (j = 0; j < 4; ++j)
27             printf("%d ", *(*ptr2 + j));
28         printf("\n");
29     }
30 
31     return 0;
32 }

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int (*ptr)[4]; 中，标识符ptr表示的语义是“指向含 4 个 int 的数组” 的指针

int *ptr[4]; 中，标识符ptr表示的语义是含 4 个元素的数组，本质来说ptr是数组名

task4

 1 #include <stdio.h>
 2 #define N 80
 3 
 4 void replace(char *str, char old_char, char new_char); // 函数声明
 5 
 6 int main() {
 7     char text[N] = "Programming is difficult or not, it is a question.";
 8 
 9     printf("原始文本: \n");
10     printf("%s\n", text);
11 
12     replace(text, 'i', '*'); // 函数调用 注意字符形参写法，单引号不能少
13 
14     printf("处理后文本: \n");
15     printf("%s\n", text);
16 
17     return 0;
18 }
19 
20 // 函数定义
21 void replace(char *str, char old_char, char new_char) {
22     int i;
23 
24     while(*str) {
25         if(*str == old_char)
26             *str = new_char;
27         str++;
28     }
29 }

View Code

函数 replace 的功能是把字符数组text[N]中的'i'换成'*'

line24, 圆括号里循环条件可以改写成 *str != '\0'

task5

 1 #define  _CRT_SECURE_NO_WARNINGS
 2 #include <stdio.h>
 3 #define N 80
 4 
 5 char* str_trunc(char* str, char x);
 6 
 7 int main() {
 8     char str[N];
 9     char ch;
10 
11     while (printf("输入字符串: "), gets(str) != NULL) {
12         printf("输入一个字符: ");
13         ch = getchar();
14 
15         printf("截断处理...\n");
16         str_trunc(str, ch);         // 函数调用
17 
18         printf("截断处理后的字符串: %s\n\n", str);
19         getchar();
20     }
21 
22     return 0;
23 }
24 
25 char* str_trunc(char* str, char x) {
26     int i;
27     for (i = 0; i < strlen(str); i++) {
28         if(str[i]==x)
29             {
30             str[i]='\0';
31             break;
32         }
33         
34     }
35     return str;
36 }

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main函数line18 getchar()的作用是读取前一个字符输入时的'\n'

task6

 1 #define  _CRT_SECURE_NO_WARNINGS
 2 #include <stdio.h>
 3 #include <string.h>
 4 #define N 5
 5 
 6 int check_id(char *str); // 函数声明
 7 
 8 int main()
 9 {
10     char *pid[N] = {"31010120000721656X",
11                     "3301061996X0203301",
12                     "53010220051126571",
13                     "510104199211197977",
14                     "53010220051126133Y"};
15     int i;
16 
17     for (i = 0; i < N; ++i)
18         if (check_id(pid[i])) // 函数调用
19             printf("%s\tTrue\n", pid[i]);
20         else
21             printf("%s\tFalse\n", pid[i]);
22 
23     return 0;
24 }
25 
26 
27 int check_id(char *str) {
28     int len = strlen(str);
29     if ( len != 18) {
30         return 0; 
31     }
32     for (int i = 0; i < 18; ++i) {
33         if (i < 17) {
34             if (str[i] < '0' || str[i] > '9') {
35                 return 0;
36             }
37         } 
38         else {
39             if ((str[i] <'0' || str[i] > '9') && str[i] != 'X') {
40                 return 0; 
41             }
42         }
43     }
44     return 1; 
45 }

View Code

task7

 1 #define  _CRT_SECURE_NO_WARNINGS
 2 #include <stdio.h>
 3 #define N 80
 4 void encoder(char* str, int n); // 函数声明
 5 void decoder(char* str, int n); // 函数声明
 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);      // 函数调用
19     printf("%s\n", words);
20 
21     printf("对编码后的英文文本解码: ");
22     decoder(words, n); // 函数调用
23     printf("%s\n", words);
24 
25     return 0;
26 }
27 
28 /*函数定义
29 功能：对str指向的字符串进行编码处理
30 编码规则：
31 对于a~z或A~Z之间的字母字符，用其后第n个字符替换; 其它非字母字符，保持不变
32 */
33 void encoder(char* str, int n) {
34     int i = 0;
35     char ch;
36     while (str[i] != '\0') {
37         ch = str[i];
38         if (ch >= 'a' && ch <= 'z') {
39             str[i] = (ch - 'a' + n) % 26 + 'a';
40         } else if (ch >= 'A' && ch <= 'Z') {
41             str[i] = (ch - 'A' + n) % 26 + 'A';
42         }
43         i++;
44     }
45     
46 }
47 
48 void decoder(char* str, int n) {
49     int i = 0;
50     char ch;
51     while (str[i] != '\0') {
52         ch = str[i];
53         if (ch >= 'a' && ch <= 'z') {
54             str[i] = (ch - 'a' - n + 26) % 26 + 'a';
55         } else if (ch >= 'A' && ch <= 'Z') {
56             str[i] = (ch - 'A' - n + 26) % 26 + 'A';
57         }
58         i++;
59     }
60     
61 }

View Code

task8

 1 #define  _CRT_SECURE_NO_WARNINGS
 2 #include <stdio.h>
 3 #include <string.h>
 4 
 5 int main(int argc, char* argv[]) {
 6     
 7 
 8     int i, j;
 9     char* temp;
10 
11     
12     for (i = 1; i < argc - 1; i++) {
13         for (j = 1; j < argc - i; j++) {
14             if (strcmp(argv[j], argv[j + 1]) > 0) {
15                 
16                 temp = argv[j];
17                 argv[j] = argv[j + 1];
18                 argv[j + 1] = temp;
19             }
20         }
21     }
22 
23 
24     for (i = 1; i < argc; i++) {
25         printf("hello, %s\n", argv[i]);
26     }
27 
28     return 0;
29 }

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posted @ 2025-12-06 16:43 张天宁阅读(0) 评论(0) 收藏举报

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