tgwz

实验5

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 }

image

 A1：功能是找出数组中的最小值和最大值

A2：pmin指向主函数中min的内存地址、pmax指向主函数中max的内存地址

 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 }

image

 A1：功能是找出数组中的最大值，返回的是该最大元素所对应的指针

A2：可以

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 }

image

 A1：数组s1的大小是80，sizeof(s1)计算的是数组s1所占用的内存空间大小，strlen(s1)统计的是数组s1的长度

A2：不能，因为Learning makes me happy是一个字符串常量，在内存中有自己的地址，数组名不能重新指向另一个地址

A3：是

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 }

image

 A1：指针变量s1中存放的是字符串“Learning makes me happy"的内存地址；sizeof(s1)计算的是数组s1所占用的内存大小；strlen(s1)统计的是数组s1的长度

A2：能；task2_1.c中s1[N]表示数组，将字符串直接存入数组中；task2_2.c中用指针s1指向字符串的内存地址

A3：交换的是s1和s2的值；没有

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;     
 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 }

image

 task4.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] = "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 }

image

A1：功能是将原字符串中所有指定的字符都替换成新字符

A2：可以

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);         
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 char *str_trunc(char *str,char x){
24     char *p=str;
25     while(*p!='\0'&&*p!=x){
26         p++;
27     }
28     if (*p==x){
29         *p='\0';
30     }
31     return str;
32 }

image

 A：会直接读取上一次输入字符后的换行符；拿走上一次输入字符后的换行符

task6.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 {
 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 }

image

 task7.c

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

image

 task8.c

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

image

 

posted on 2025-12-10 23:34  添罐望仔  阅读(4)  评论(0)    收藏  举报

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