实验5

实验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 = 1; i < n; ++i)
48         if(x[i] < *pmin)
49             *pmin = x[i];
50         else if(x[i] > *pmax)
51             *pmax = x[i];
52 }

函数find_max的功能是找出几个数中的最大值和最小值。

*pmin,pmax都指向x[0]

 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 }

函数fid_max的功能是找到x[0]中最大值的地址,返回的是地址。

可以

实验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[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 }

大小是23字节,sizeof(s1)计算s1占用了多少字节,strlen则是统计s1字符串的长度。

不可以。没有给s1赋初始长度

发生了交换。

 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 }

字符串第一个字符的地址,计算的是地址所占字节数,统计字符串长度。

能,一中的是给数组赋初始值,二中是将字符串地址赋给s1。

交换的是地址,储存单元交换了。

实验3

 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 }

 

实验4

#include <stdio.h>
#define N 80

void replace(char *str, char old_char, char new_char); 

int main() {
    char text[N] = "c programming is difficult or not, it is a question.";

    printf("原始文本: \n");
    printf("%s\n", text);

    replace(text, 'i', '*'); 

    printf("处理后文本: \n");
    printf("%s\n", text);

    return 0;
}


void replace(char *str, char old_char, char new_char) {
    int i;

    while(*str) {
        if(*str == old_char)
            *str = new_char;
        str++;
    }
}

将文本中的'i'替换成'*'。

可以。

 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             *str='\0';     // blank1
27 
28         str++;   // blank2
29     }
30 
31     free(*str);    // blank3
32 }

 

验5

 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 }

 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 }

储存位置发生了变化。

实验6

 

 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     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])) // 函数调用
17             printf("%s\tTrue\n", pid[i]);
18         else
19             printf("%s\tFalse\n", pid[i]);
20 
21     return 0;
22 }
23 
24 // 函数定义
25 // 功能: 检查指针str指向的身份证号码串形式上是否合法。
26 // 形式合法,返回1,否则,返回0
27 int check_id(char *str) {
28     int i, j;
29     char a;
30     i = strlen( str );
31     if( i != 18){
32         return 0;
33     }
34     for(j = 0; j < i; j++){
35         a = str[j];
36         if( a < 48){
37             return 0;
38         }
39         if( a > 57 && a != 88){
40             return 0;
41         }
42     } 
43     return 1;  
44 }

 

实验7

 

 1 #include <stdio.h>
 2 #include <string.h>
 3 #define N 80
 4 void encoder(char *str); 
 5 void decoder(char *str); 
 6 
 7 int main() {
 8     char words[N];
 9 
10     printf("输入英文文本: ");
11     gets(words);
12 
13     printf("编码后的英文文本: ");
14     encoder(words); 
15     printf("%s\n", words);
16 
17     printf("对编码后的英文文本解码: ");
18     decoder(words); 
19     printf("%s\n", words);
20 
21     return 0;
22 }
23 
24 
25 void encoder(char *str) {
26     int i, j;
27     i = strlen( str);
28     for(j = 0; j < i; j++){
29         if( str[j] >= 65 && str[i] <= 122){
30             if( str[j] == 122){
31                 str[j] = 'a';
32             }
33             else if( str[j] == 90){
34                 str [j] = 'A';
35             }
36             else{
37                 str[j] = str[j] + 1;
38             }
39         }
40     }
41 }
42 
43 
44 void decoder(char *str) {
45     int i, j;
46     i = strlen( str);
47     for(j = 0; j < i; j++){
48         if( str[j] >= 65 && str[j] <= 144){
49         if( str[j] == 'a'){
50             str[j] = 'z';
51         }
52         else if( str[j] == 'A'){
53             str [j] = 'Z';
54         }
55         else{
56             str[j] = str[j] - 1;
57         }
58         }
59     }
60 }

 

posted @ 2023-12-04 10:41  夏夜#  阅读(10)  评论(0)    收藏  举报