实验5

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 }

Q1:获取数组x[ ]中的最大值和最小值

Q2:都指向数组x[ ]中的第一个的地址

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 }

Q1:返回数组x[ ]中最大值的地址

Q2:可以

task2.1

 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 }

Q1:80；sizeof计算s1所占字节的大小；strlen统计s1内字符串的长度

Q2:不能；s1作为数组，赋值时应带有[ ]

Q3:是

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 }

Q1:存放的是字符串的地址；sizeof存放该地址的字节长度；strlen统计该字符串的长度

Q2:可以；2.1中存放的是字符串的本体；2.2中只是存放了字符串的地址

Q3:交换的是s1和s2中存放的地址，内存中并没有交换

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; 
 7     int(*ptr2)[4];
 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 }

Q1:指针

Q2:数组名

task4

 1 #include <stdio.h>
 2 #define N 80
 3 
 4 void replace(char *str, char old_char, char new_char);
 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', '*');
13 
14     printf("处理后文本: \n");
15     printf("%s\n", text);
16 
17     return 0;
18 }
19 
20 void replace(char *str, char old_char, char new_char) {
21     int i;
22 
23     while(*str) {
24         if(*str == old_char)
25             *str = new_char;
26         str++;
27     }
28 }

Q1:将数组中的特定元素用另一元素替代

Q2:不行；'\0'没有独立的地址，不能用指针

task5

 1 #include<stdio.h>
 2 #define N 80
 3 
 4 char *str_trunc(char *str,char x);
 5 
 6 int main()
 7 {
 8     char str[N];
 9     char ch;
10     
11     while(printf("输入字符串："),gets(str)!=NULL)
12     {
13         printf("输入一个字符：");
14         ch = getchar();
15         
16         printf("截断处理...\n");
17         str_trunc(str,ch);
18         
19         printf("截断处理后的字符串：%s\n\n",str);
20         getchar();
21     }
22     
23     return 0;
24 }
25 
26 char *str_trunc(char *str,char x)
27 {
28     int i,j;
29     char m[N];
30     
31     while(str[i]!=x)
32     {
33         i++;
34     }
35     while(str[j]!='\0')
36     {
37         j++;
38     }
39     str[j]=0;
40     for(;j>i;j--)
41     {
42         str[j-1]=str[j];
43     }
44     
45     return str;
46 }

task6

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

task7

 1 #include <stdio.h>
 2 #define N 80
 3 void encoder(char *str, int n);
 4 void decoder(char *str, int n);
 5 
 6 int main() 
 7 {
 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 {
30     int i = 0;
31     while (str[i]!= '\0') 
32     {
33         if ((str[i] >= 'a' && str[i] <= 'z') || (str[i] >= 'A' && str[i] <= 'Z')) 
34         {
35             char base = (str[i] >= 'a' && str[i] <= 'z')? 'a' : 'A';
36             str[i] = ((str[i] - base + n) % 26) + base;
37         }
38         i++;
39     }
40 }
41 
42 void decoder(char *str, int n) 
43 {
44     int i = 0;
45     while (str[i]!= '\0') 
46     {
47         if ((str[i] >= 'a' && str[i] <= 'z') || (str[i] >= 'A' && str[i] <= 'Z')) 
48         {
49             char base = (str[i] >= 'a' && str[i] <= 'z')? 'a' : 'A';
50             int code = str[i] - base;
51             if (code - n < 0) 
52             {
53                 code += 26;
54             }
55             str[i] = (code - n) % 26 + base;
56         }
57         i++;
58     }
59 }

task8

 

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

 

posted @ 2024-12-05 21:57  yueTO  阅读(2)  评论(0)  编辑  收藏  举报