实验4 C语言数组应用编程

实验任务1

代码task1_1.c

 1 #include<stdio.h>
 2 #define N 4
 3 
 4 void test1(){
 5     int a[N] = {1, 9, 8, 4};
 6     int i;
 7     
 8     printf("sizeof(a) = %d\n", sizeof(a));
 9     
10     for(i = 0; i < N; ++i)
11         printf("%p: %d\n",&a[i], a[i]);
12         
13     printf("a = %p\n", a);
14 } 
15 
16 void test2(){
17     char b[N] = {'1', '9', '8', '4'};
18     int i;
19     
20     printf("sizeof(b) = %d\n", sizeof(b));
21     
22     for(i = 0; i < N; ++i)
23         printf("%p: %c\n",&b[i], b[i]);
24         
25     printf("b = %p\n", b);
26 } 
27 
28 int main(){
29     printf("测试1: int类型一维数据\n");
30     test1();
31     
32     printf("\n测试2: char类型一维数据\n");
33     test2();
34     
35     return 0;
36 }

运行结果

 问题1

int型数组a，在内存中是连续存放的，每个元素占用4个内存字节单元，数组名a对应的值，和&a[0]是一样的

问题2

char型数组b，在内存中是连续存放的，每个元素占用1个内存字节单元，数组名b对应的值，和&b[0]是一样的

代码task1_2.c

 1 #include<stdio.h>
 2 #define N 2
 3 #define M 4
 4 
 5 void test1(){
 6     int a[N][M] = {{1, 9, 8, 4},{2, 0, 4, 9}};
 7     int i, j;
 8     
 9     printf("sizeof(a) = %d\n", sizeof(a));
10     
11     for(i = 0; i < N; ++i)
12         for(j = 0; j < M; ++i)
13             printf("%p: %d\n",&a[i][j], a[i][j]);
14     printf("\n");
15     
16     printf("a = %p\n", a);
17     printf("a[0] = %p\n", a[0]);
18     printf("a[1] = %p\n", a[1]);
19     printf("\n");
20 } 
21 
22 void test2(){
23     char b[N][M] = {{1, 9, 8, 4},{2, 0, 4, 9}};
24     int i, j;
25     
26     printf("sizeof(b) = %d\n", sizeof(b));
27     
28     for(i = 0; i < N; ++i)
29         for(j = 0; j < M; ++i)
30             printf("%p: %c\n",&b[i][j], b[i][j]);
31     printf("\n");
32     
33     printf("b = %p\n", b);
34     printf("b[0] = %p\n", b[0]);
35     printf("b[1] = %p\n", b[1]);
36 } 
37 
38 int main(){
39     printf("测试1: int类型两维数据\n");
40     test1();
41     
42     printf("\n测试2: char类型两维数据\n");
43     test2();
44     
45     return 0;
46 }

运行结果

 问题1

int型二维数组a，在内存中是"按行连续存放"的,每个元素占用4个内存字节单元，int型二维数组a, 数组名a的值和&a[0][0]的值，在数字字面值上，是一样的

问题2

char型二维数组b，在内存中是"按行连续存放"的，每个元素占用1个内存字节单元,char型二维数组b, 数组名b的值&b[0][0]的值，在数字字面值上，是一样的

问题3

a[0], a[1]的值相差12，也就是三个int的内存大小，a数组的二维恰好有3个

b[0], b[1]的值相差3， 是三个char的内存大小 ，b数组的二维恰好3个

 

实验任务2

代码task2.c

 1 #include <stdio.h>
 2 #include <string.h>
 3 
 4 #define N 80
 5 
 6 void swap_str(char s1[N], char s2[N]);
 7 void test1();
 8 void test2();
 9 
10 int main() {
11     printf("测试1: 用两个一维char数组，实现两个字符串交换\n");
12     test1();
13 
14     printf("\n测试2: 用二维char数组，实现两个字符串交换\n");
15     test2();
16 
17     return 0;
18 }
19 
20 void test1() {
21     char views1[N] = "hey, C, I hate u.";
22     char views2[N] = "hey, C, I love u.";
23 
24     printf("交换前: \n");
25     puts(views1);
26     puts(views2);
27 
28     swap_str(views1, views2);
29 
30     printf("交换后: \n");
31     puts(views1);
32     puts(views2);
33 }
34 
35 void test2() {
36     char views[2][N] = {"hey, C, I hate u.", 
37                         "hey, C, I love u."};
38 
39     printf("交换前: \n");
40     puts(views[0]);
41     puts(views[1]);
42 
43     swap_str(views[0], views[1]);
44 
45     printf("交换后: \n");
46     puts(views[0]);
47     puts(views[1]);
48 }
49 
50 void swap_str(char s1[N], char s2[N]) {
51     char tmp[N];
52 
53     strcpy(tmp, s1);
54     strcpy(s1, s2);
55     strcpy(s2, tmp);
56 }

运行结果

 总结：test1中，views1和views2是两个一维数组；test2中，views[0], views[1]是二维数组第一维的两个数组,所以后者加中括号，传入swap_str的实际上都是一维数组

 

实验任务3

代码task3_1.c

 1 #include <stdio.h>
 2 
 3 #define N 80
 4 
 5 int count(char x[]);
 6 
 7 int main() {
 8     char words[N+1];
 9     int n;
10 
11     while(gets(words) != NULL) {
12         n = count(words);
13         printf("单词数: %d\n\n", n);
14     }
15 
16     return 0;
17 }
18 
19 int count(char x[]) {
20     int i;
21     int word_flag = 0;  
22     int number = 0;  
23 
24     for(i = 0; x[i] != '\0'; i++) {
25         if(x[i] == ' ')
26             word_flag = 0;
27         else if(word_flag == 0) {
28             word_flag = 1;
29             number++;
30         }
31     }
32 
33     return number;
34 }

运行结果

 代码task3_2.c

 1 #include <stdio.h>
 2 #define N 1000
 3 
 4 int main() {
 5     char line[N];
 6     int word_len;   
 7     int max_len;    
 8     int end;        
 9     int i;
10 
11     while(gets(line) != NULL) {
12         word_len = 0;
13         max_len = 0;
14         end = 0;
15 
16         i = 0;
17         while(1) {
18             while(line[i] == ' ') {
19                 word_len = 0; 
20                 i++;
21             }
22 
23             while(line[i] != '\0' && line[i] != ' ') {
24                 word_len++;
25                 i++;
26             }
27         
28             if(max_len < word_len) {
29                 max_len = word_len;
30                 end = i;   
31             }
32 
33             if(line[i] == '\0')
34                 break;
35         }
36 
37         printf("最长单词: ");
38         for(i = end - max_len; i < end; ++i)
39             printf("%c", line[i]);
40         printf("\n\n");
41     }
42 
43     return 0;
44 }

运行结果

 

实验任务4

代码task4.c

 1 #include <stdio.h>
 2 #define N 100
 3 void dec_to_n(int x, int n); 
 4 
 5 int main() {
 6     int x;
 7 
 8     printf("输入一个十进制整数: ");
 9     while(scanf("%d", &x) != EOF) {
10         dec_to_n(x, 2);  
11         dec_to_n(x, 8);  
12         dec_to_n(x, 16); 
13 
14         printf("\n输入一个十进制整数: ");
15     }
16 
17     return 0;
18 }
19 
20 void dec_to_n(int x, int n){
21     char map[] = {"0123456789ABCDEF"};
22     char ans[N];
23     int r;
24     int cnt = 0, i;
25     
26     do{
27         r = x %n;
28         ans[cnt++] = map[r];
29         x = x / n;
30     }while(x != 0);
31     
32     for(i = cnt - 1;i >= 0; --i)
33         printf("%c", ans[i]);
34         
35         printf("\n");
36 }

 

运行结果

 

实验任务5

代码task5.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 double average(int x[], int n);
 7 void bubble_sort(int x[], int n);
 8 
 9 int main() {
10     int scores[N];
11     double ave;
12     
13     printf("录入%d个分数:\n", N);
14     input(scores, N);
15     
16     printf("\n输出课程分数: \n");
17     output(scores, N);
18     
19     printf("\n课程分数处理: 计算均分、排序...\n");
20     ave = average(scores, N);
21     bubble_sort(scores, N);
22     
23     printf("\n输出课程均分: %.2f\n", ave);
24     printf("\n输出课程分数(高->低):\n");
25     output(scores, N);
26     
27     return 0;
28 }
29  
30 void input(int x[], int n) {
31     int i;
32     
33     for(i = 0; i < n; ++i)
34         scanf("%d", &x[i]); 
35 }
36 
37 void output(int x[], int n) {
38     int i;
39     
40     for(i = 0; i < n; ++i)
41         printf("%d ", x[i]);
42     printf("\n");
43 }
44 
45 double average(int x[], int n){
46     int i;
47     double t, s = 0;
48     
49     for(i = 0; i < n; ++i)
50         s = s + x[i];        
51     t = s / n;
52     return t;
53 }  
54 
55 void bubble_sort(int x[], int n){
56     int i, a, b, c;
57     
58     for(i = 1; i <= n - 1; ++i){
59         for(a = 0; a <= n - 1 - i; ++a){
60             if(x[a] < x[a + 1]){
61                 b = x[a];
62                 x[a] = x[a + 1];
63                 x[a + 1] = b;
64             }
65         }
66     }
67     for(c = 0; c < n; ++c)
68         printf("&d", x[c]);
69 }

运行结果

 

实验任务6

代码task6.c

 1 #include <stdio.h>
 2 #include <string.h>
 3 
 4 #define N 5
 5 #define M 20
 6 
 7 void output(char str[][M], int n);
 8 void bubble_sort(char str[][M], int n);
 9 
10 int main() {
11     char name[][M] = {"Bob", "Bill", "Joseph", "Taylor", "George"};
12     int i;
13 
14     printf("输出初始名单:\n");
15     output(name, N);
16 
17     printf("\n排序中...\n");
18     bubble_sort(name, N);  // 函数调用 
19 
20     printf("\n按字典序输出名单:\n");
21     output(name, N);
22 
23     return 0;
24 }
25 
26 void output(char str[][M], int n) {
27     int i;
28 
29     for(i = 0; i < n; ++i)
30         printf("%s\n", str[i]);
31 }
32 
33 void bubble_sort(char str[][M], int n) {
34     char temp[M];
35     int i, j;
36     
37     for(i = 0; i < n - 1; i++){
38         for(j = 0; j < n - i - 1; j++) {
39             if (str[j][0] > str[j + 1][0]){
40                 strcpy(temp, str[j]);
41                 strcpy(str[j], str[j + 1]);
42                 strcpy(str[j + 1], temp);
43             }
44         }
45     }
46 }

运行结果

 

实验任务7

代码task7.c

 1 #include <stdio.h>
 2 #include <string.h>
 3 int main(){
 4     char a[10000];
 5     int i, s;
 6     
 7     while (scanf("%s", a) != EOF) {
 8         int x = 0;
 9         for(i = 0; i < strlen(a); i++) {
10             for(s = i + 1; s < strlen(a); s++){
11                 if(a[i] == a[s]){
12                     x = 1;
13                     break;
14                 }
15             }
16             if (x == 1){
17                 break;
18             }
19         }
20         if(x == 1)
21             printf("YES\n\n");        
22         else if(x == 0)
23             printf("NO\n\n");       
24     }
25     return 0;
26 }

运行结果

 

实验任务8

代码task8.c

 1 #include <stdio.h>
 2 #define N 100
 3 #define M 4
 4 
 5 void output(int x[][N], int n);          
 6 void rotate_to_right(int x[][N], int n); 
 7 
 8 
 9 int main() {
10     int t[][N] = {{21, 12, 13, 24},
11                   {25, 16, 47, 38},
12                   {29, 11, 32, 54},
13                   {42, 21, 33, 10}};
14 
15     printf("原始矩阵:\n");
16     output(t, M);
17 
18     rotate_to_right(t, M);
19 
20     printf("变换后矩阵\n");
21     output(t, M);
22 
23     return 0;
24 }
25 
26 void output(int x[][N], int n) {
27     int i, j;
28 
29     for (i = 0; i < n; ++i) {
30         for (j = 0; j < n; ++j)
31             printf("%4d", x[i][j]);
32 
33         printf("\n");
34     }
35 }
36 
37 void rotate_to_right(int x[][N], int n){        
38     int i,t;
39     
40     for(i=0;i<4;++i){
41         t=x[i][0];
42         x[i][0]=x[i][3];
43         x[i][3]=t;
44         t=x[i][1];
45         x[i][1]=x[i][3];
46         x[i][3]=t;
47         t=x[i][3];
48         x[i][3]=x[i][2];
49         x[i][2]=t;
50     }
51 }

运行结果

 

posted @ 2023-11-19 15:58  臣子民  阅读(28)  评论(0)  编辑  收藏  举报