实验五

任务1

task1_1.c

 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 }

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问题1：找到一维数组的最大最小值；

问题2：&x[0]；

task1_2.c

 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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问题1：一维数组的最大值的地址；

问题2：可以；

任务2

task2_1.c

 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 }

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问题1：80；sizeof(s1)计算一维数组s1所占的字节数；strlen(s1)统计s1字符串的长度；

问题2：不能；char s1[ ]没有指定数组大小；数组名是一个指针常量，指向数组的第一个元素，不能被重新赋值；

问题3：s1与s2交换；

task2_2.c

 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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问题1：字符串"Learning makes me happy"首字符的地址；指针变量 s1在内存中所占的字节数；统计s1字符串的长度；

问题2：可以；task2_1.c中定义一个数组s1,将字符串存储在数组中;char *s1;声明指针变量s1(野指针)，s1="Learning makes me happy";将指针s1指向字符串常量"Learning makes me happy";

问题3：交换的是指针变量s1和s2所指向的地址；没有交换;

任务3

task3.c

 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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问题：int (*ptr)[4]表示一个指向数组的指针，ptr是一个指针；int *ptr[4]表示一个指针数组，ptr是一个数组；

任务4

task4.c

 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 }

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问题1：将文本中的i替换为*；

问题2：可以；

任务5

task5.c

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

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问题：getchar()读取回车；

任务6

task6.c

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

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任务7

task7.c

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

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任务8

task8.c

 1 #include <stdio.h>
 2 #include <string.h>
 3 
 4 void compare(int n, char* temp[]);
 5 
 6 int main(int argc, char* argv[]) {
 7 
 8     int i;
 9     compare(argc, argv);
10 
11     for (i = 1; i < argc ; ++i)
12         printf("hello,%s\n", argv[i]);
13 
14     return 0;
15 }
16 
17 void compare(int n, char* names[]) {
18     int i, j;
19     char* temp;
20 
21     for (i = 1; i < n; ++i){
22         for (j = 1; j < n - i; ++j) {
23             if (strcmp(names[j], names[j + 1]) > 0){
24             temp = names[j];
25             names[j] = names[j + 1];
26             names[j + 1] = temp;
27             }
28         }
29     }
30 }

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posted @ 2025-05-18 22:20 tingying 阅读(29) 评论(0) 收藏举报

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