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 }
task1_1

task1

find_min_max的作用是找到最大值与最小值

pmin和pmax都指向：x[0],即第一个元素

tak1_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 }
task1_2

task1_2

find_max的作用是找到最大值，返回的是最大元素的地址；

改变后代码可以实现；

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 }
task2_1

task2_1

数组s1大小为80，sizeof(s1)计算的是数组s1所占的字节数，strlen(s1)统计的是数组s1的字符数，其中‘\0'不计入；

line7不能更换，s1表示的是数组s1的第一个元素地址，是不可改变的值；

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 }
task2_2

屏幕截图 2025-12-06 073857

s1存放第一个字符串首元素地址，sizeof(s1)计算首元素地址所占用字节数，strlen(s1)计算字符串长度，不包含‘\0’;

可以替换；task2_1中s1是数组，存放的是字符，task2_2中s1是指针变量，存放第一个元素的地址；

交换的是指针的指向，元素的地址不变；

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 }
task3

屏幕截图 2025-12-06 075157

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 }
task4

屏幕截图 2025-12-06 075601

replace:将所有i替换成*

不可以改写，'\0'是字符常量，表示空字符，0表示逻辑假；

task5

 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     return 0;
21 }
22 
23 char *str_trunc(char *str, char x){
24     int i=0;
25     while(str[i]!='\0'){
26         if(str[i]==x){
27             str[i]='\0';
28             break;
29         }
30         ++i;
31     }
32     return str;
33 }
task5

屏幕截图 2025-12-07 081633

line18去掉后无法输入新的字符串，getchar作用是制造缓冲区，防止换行被读取为字符串；

task6

 1 #include <stdio.h>
 2 #include <string.h>
 3 #define N 5
 4 
 5 int check_id(char *str);
 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])) 
18             printf("%s\tTrue\n", pid[i]);
19         else
20             printf("%s\tFalse\n", pid[i]);
21             
22     return 0;
23 }
24 
25 int check_id(char *str) {
26     int len=strlen(str),i;
27     
28     if(len!=18){
29         return 0;
30     } 
31     for(i=0;i<17;++i){
32         if(!('0'<=str[i]&&str[i]<='9')){
33             return 0;
34         } 
35     }
36     if(!((str[17]>='0'&&str[17]<='9')||str[17]=='x'||str[17]=='X')){
37             return 0;
38     }
39     return 1; 
40 }
task6

屏幕截图 2025-12-07 134224

 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     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("%d", &n);
15 
16     printf("编码后的英文文本: ");
17     encoder(words, n);      
18     printf("%s\n", words);
19 
20     printf("对编码后的英文文本解码: ");
21     decoder(words, n); 
22     printf("%s\n", words);
23 
24     return 0;
25 }
26 
27 void encoder(char *str, int n) {
28     int i=0;
29     while(str[i]!='\0'){
30         if(str[i]>='a'&&str[i]<='z'){
31             str[i]=(str[i]-'a'+n)%26+'a';
32         }
33         else if(str[i]>='A'&&str[i]<='Z'){
34             str[i]=(str[i]-'A'+n)%26+'A';
35         }
36         ++i;
37     }
38 
39 }
40 
41 void decoder(char *str, int n) {
42     int i=0;
43     while(str[i]!='\0'){
44         if(str[i]>='a'&&str[i]<='z'){
45             str[i]=(str[i]-'a'-n+26)%26+'a';
46         }
47         else if(str[i]>='A'&&str[i]<='Z'){
48             str[i]=(str[i]-'A'-n+26)%26+'A';
49         }
50         ++i;
51     }
52     
53 }
task7

屏幕截图 2025-12-07 205832

屏幕截图 2025-12-07 205934

屏幕截图 2025-12-07 210006

task8

 1 #include<stdio.h>
 2 
 3 int main(int argc,char *argv[]){
 4     int i;
 5     
 6     for(i=1;i<argc;++i){
 7         printf("hello,%s\n",argv[i]);
 8     }
 9     return 0;
10 }
task8.1

屏幕截图 2025-12-08 193820

 1 #include <stdio.h>
 2 #include<string.h>
 3 
 4 void sort(int n,char *s[]);
 5   
 6 int main(int argc, char *argv[]) {
 7     int i;
 8     sort(argc-1,argv+1);
 9     for(i = 1; i < argc; ++i)
10         printf("hello, %s\n", argv[i]);
11     return 0;
12 }
13 void sort(int n,char *s[]){
14     int i,j;
15     char *tmp;
16     for(i=0;i<n-1;++i){
17         for(j=0;j<n-i-1;++j){
18             if(strcmp(s[j],s[j+1])>0){
19                 tmp=s[j];
20                 s[j]=s[j+1];
21                 s[j+1]=tmp;
22             }
23         }
24     }     
25 }
task8.2

屏幕截图 2025-12-08 193338

 

posted @ 2025-12-08 19:40  ZL425  阅读(0)  评论(0)    收藏  举报