实验五

1.c

1-1.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("¼Èë%d¸öÊý¾Ý:\n", N);
13     input(a, N);
14 
15     printf("Êý¾ÝÊÇ: \n");
16     output(a, N);
17 
18     printf("Êý¾Ý´¦Àí...\n");
19     find_min_max(a, N, &min, &max);
20 
21     printf("Êä³ö½á¹û:\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 = 1; i < n; ++i)
48         if(x[i] < *pmin)
49             *pmin = x[i];
50         else if(x[i] > *pmax)
51             *pmax = x[i];
52 }

回答：1：找出最大值和最小值。2：都指向数组的第一个元素的地址

1-2.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 = 1; i < n; ++i)
47         if(x[i] > x[max_index])
48             max_index = i;
49     
50     return &x[max_index];
51 }

回答：1：返回数组中最大值的地址。2：可以

2.c

2-1.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[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 }

回答：1：s1的大小是24字节。sizeof(s1)计算的是数据在内存中所占用的存储空间。strlen(s1)统计的是字符串的长度。

           2：不能，字符数组整体赋值只能在初始化时使用，字符数组的赋值只能对元素一一赋值

           3：交换了。

2-2.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 }

回答：1：s1存放的是数组中每个元素的的地址。sizeof(s1)计算的是指针数组的大小。strlen(s1)统计的是字符串长度。

           2：可以。区别：前者不能动态分配内存空间，后者可以。

           3：交换的是地址，两个字符串常量没有交换。

3.c

 1 #include <stdio.h>
 2 
 3 #include <stdio.h>
 4 
 5 int main() {
 6     int x[2][4] = {{1, 9, 8, 4}, {2, 0, 4, 9}};
 7     int i, j;
 8     int *ptr1;     
 9     int(*ptr2)[4]; 
10 
11     printf("输出1: 使用数组名、下标直接访问二维数组元素\n");
12     for (i = 0; i < 2; ++i) {
13         for (j = 0; j < 4; ++j)
14             printf("%d ", x[i][j]);
15         printf("\n");
16     }
17 
18     printf("\n输出2: 使用指向元素的指针变量p间接访问二维数组元素\n");
19     for (ptr1 = &x[0][0], i = 0; ptr1 < &x[0][0] + 8; ++ptr1, ++i) {
20         printf("%d ", *ptr1);
21 
22         if ((i + 1) % 4 == 0)
23             printf("\n");
24     }
25                          
26     printf("\n输出3: 使用指向一维数组的指针变量q间接访问二维数组元素\n");
27     for (ptr2 = x; ptr2 < x + 2; ++ptr2) {
28         for (j = 0; j < 4; ++j)
29             printf("%d ", *(*ptr2 + j));
30         printf("\n");
31     }
32 
33     return 0;
34 }

回答：1：指针变量指向一维数组。2：指针变量指向数据地址。

4.c

4-1. c

 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] = "c 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 
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 }

回答：1：用新字符替代指定的旧字符。2：可以。

4-2.c

 1 #include <stdio.h>
 2 #define N 80
 3 
 4 void str_trunc(char *str, char x);
 5 
 6 int main() {
 7     char str[N];
 8     char ch;
 9 
10     printf("输入字符串: ");
11     gets(str);
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", str);
20 
21 }
22 
23 void str_trunc(char *str, char x) {
24     while(*str) {
25         if(*str == x)
26             break;     // blank1
27 
28         str++;   // blank2
29     }
30 
31     *str='\0';    // blank3
32 }

5.c

5-1.c

 1 #include <stdio.h>
 2 #include <string.h>
 3 void sort(char *name[], int n);
 4 
 5 int main() {
 6     char *course[4] = {"C Program",
 7                        "C++ Object Oriented Program",
 8                        "Operating System",
 9                        "Data Structure and Algorithms"};
10     int i;
11 
12     sort(course, 4);
13 
14     for (i = 0; i < 4; i++)
15         printf("%s\n", course[i]);
16 
17     return 0;
18 }
19 
20 void sort(char *name[], int n) {
21     int i, j;
22     char *tmp;
23 
24     for (i = 0; i < n - 1; ++i)
25         for (j = 0; j < n - 1 - i; ++j)
26             if (strcmp(name[j], name[j + 1]) > 0) {
27                 tmp = name[j];
28                 name[j] = name[j + 1];
29                 name[j + 1] = tmp;
30             }
31 }

5-2.c

 1 #include <stdio.h>
 2 #include <string.h>
 3 void sort(char *name[], int n);
 4 
 5 int main() {
 6     char *course[4] = {"C Program",
 7                        "C++ Object Oriented Program",
 8                        "Operating System",
 9                        "Data Structure and Algorithms"};
10     int i;
11 
12     sort(course, 4);
13     for (i = 0; i < 4; i++)
14         printf("%s\n", course[i]);
15 
16     return 0;
17 }
18 
19 void sort(char *name[], int n) {
20     int i, j, k;
21     char *tmp;
22 
23     for (i = 0; i < n - 1; i++) {
24         k = i;
25         for (j = i + 1; j < n; j++)
26             if (strcmp(name[j], name[k]) < 0)
27                 k = j;
28 
29         if (k != i) {
30             tmp = name[i];
31             name[i] = name[k];
32             name[k] = tmp;
33         }
34     }
35 }

回答：交换的是指针变量的值

6.c

 1 #include <stdio.h>
 2 #include <string.h>
 3 #define N 5
 4 
 5 int check_id(char *str); 
 6 
 7 int main() {
 8     char *pid[N] = {"31010120000721656X",
 9                     "330106199609203301",
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     int i,s;
26     char list[]={"0123456789X"};
27     if(strlen(str)!=18)
28         return 0;
29     for(;*str!='\0';str++){
30         s=0;
31     for(i=0;list[i]!='\0';++i){
32             if(*str==list[i]){
33                s=1;
34                break;}
35             }}
36     if(s==0)
37         return 0;
38     else 
39         return 1;        
40     }

 

7.c

 

 1 #include <stdio.h>
 2 #define N 80
 3 void encoder(char *str);
 4 void decoder(char *str);
 5 
 6 int main() {
 7     char words[N];
 8 
 9     printf("输入英文文本: ");
10     gets(words);
11 
12     printf("编码后的英文文本: ");
13     encoder(words);
14     printf("%s\n", words);
15 
16     printf("对编码后的英文文本解码: ");
17     decoder(words);
18     printf("%s\n", words);
19 
20     return 0;
21 }
22 
23 void encoder(char *str) {
24     int i;
25     for(i=0;i<N;++i){
26         if(97<=*str&&*str<=121)
27            *str=*str+1;
28         else if(65<=*str&&*str<=89)
29            *str=*str+1;
30         else if(*str==122||*str==90)
31            *str=*str-25;
32         str++;
33     } 
34 }
35 
36 void decoder(char *str) {
37     int i;
38     for(i=0;i<N;++i){
39         if(98<=*str&&*str<=122)
40            *str=*str-1;
41         else if(66<=*str&&*str<=90)
42            *str=*str-1;
43         else if(*str==65||*str==97)
44            *str=*str+25;
45         str++;
46     }     
47 }

 

posted @ 2023-11-27 19:19  If-You  阅读(16)  评论(0)    收藏  举报