实验五
task1_1
1 #include <stdio.h> 2 #define N 5 3 #include<stdlib.h> 4 5 void input(int x[], int n); 6 void output(int x[], int n); 7 void find_min_max(int x[], int n, int *pmin, int *pmax); 8 9 int main() { 10 int a[N]; 11 int min, max; 12 13 printf("录入%d个数据:\n", N); 14 input(a, N); 15 16 printf("数据是: \n"); 17 output(a, N); 18 19 printf("数据处理...\n"); 20 find_min_max(a, N, &min, &max); 21 22 printf("输出结果:\n"); 23 printf("min = %d, max = %d\n", min, max); 24 25 system("pause"); 26 return 0; 27 } 28 29 void input(int x[], int n) { 30 int i; 31 32 for(i = 0; i < n; ++i) 33 scanf("%d", &x[i]); 34 } 35 36 void output(int x[], int n) { 37 int i; 38 39 for(i = 0; i < n; ++i) 40 printf("%d ", x[i]); 41 printf("\n"); 42 } 43 44 void find_min_max(int x[], int n, int *pmin, int *pmax) { 45 int i; 46 47 *pmin = *pmax = x[0]; 48 49 for(i = 0; i < n; ++i) 50 if(x[i] < *pmin) 51 *pmin = x[i]; 52 else if(x[i] > *pmax) 53 *pmax = x[i]; 54 }
问题1：在整数数组中找出最大值和最小值
问题2：pmin指向函数中min的地址；pmax指向函数中max的地址
task1_2
1 #include <stdio.h> 2 #define N 5 3 #include<stdlib.h> 4 5 void input(int x[], int n); 6 void output(int x[], int n); 7 int *find_max(int x[], int n); 8 9 int main() { 10 int a[N]; 11 int *pmax; 12 13 printf("录入%d个数据:\n", N); 14 input(a, N); 15 16 printf("数据是: \n"); 17 output(a, N); 18 19 printf("数据处理...\n"); 20 pmax = find_max(a, N); 21 22 printf("输出结果:\n"); 23 printf("max = %d\n", *pmax); 24 25 system("pause"); 26 return 0; 27 } 28 29 void input(int x[], int n) { 30 int i; 31 32 for(i = 0; i < n; ++i) 33 scanf("%d", &x[i]); 34 } 35 36 void output(int x[], int n) { 37 int i; 38 39 for(i = 0; i < n; ++i) 40 printf("%d ", x[i]); 41 printf("\n"); 42 } 43 44 int *find_max(int x[], int n) { 45 int max_index = 0; 46 int i; 47 48 for(i = 0; i < n; ++i) 49 if(x[i] > x[max_index]) 50 max_index = i; 51 52 return &x[max_index]; 53 }
问题1：找到最大值元素的地址
问题2：可以
```
task2
1 #include <stdio.h> 2 #include <string.h> 3 #include<stdlib.h> 4 #define N 80 5 6 int main() { 7 char s1[N] = "Learning makes me happy"; 8 char s2[N] = "Learning makes me sleepy"; 9 char tmp[N]; 10 11 printf("sizeof(s1) vs. strlen(s1): \n"); 12 printf("sizeof(s1) = %d\n", sizeof(s1)); 13 printf("strlen(s1) = %d\n", strlen(s1)); 14 15 printf("\nbefore swap: \n"); 16 printf("s1: %s\n", s1); 17 printf("s2: %s\n", s2); 18 19 printf("\nswapping...\n"); 20 strcpy(tmp, s1); 21 strcpy(s1, s2); 22 strcpy(s2, tmp); 23 24 printf("\nafter swap: \n"); 25 printf("s1: %s\n", s1); 26 printf("s2: %s\n", s2); 27 system("pause"); 28 return 0; 29 }
问题1：s1数组大小是23；sizeof计算数组在内存中占用的总字节数；strlen统计字符串中实际字节数
问题2：S1是一个地址不能被赋值，并且如果是字符串赋值需要用到函数strcpy
问题3：交换了
task2_2
1 #include <stdio.h> 2 #include <string.h> 3 #include <stdlib.h> 4 #define N 80 5 6 int main() { 7 char *s1 = "Learning makes me happy"; 8 char *s2 = "Learning makes me sleepy"; 9 char *tmp; 10 11 printf("sizeof(s1) vs. strlen(s1): \n"); 12 printf("sizeof(s1) = %d\n", sizeof(s1)); 13 printf("strlen(s1) = %d\n", strlen(s1)); 14 15 printf("\nbefore swap: \n"); 16 printf("s1: %s\n", s1); 17 printf("s2: %s\n", s2); 18 19 printf("\nswapping...\n"); 20 tmp = s1; 21 s1 = s2; 22 s2 = tmp; 23 24 printf("\nafter swap: \n"); 25 printf("s1: %s\n", s1); 26 printf("s2: %s\n", s2); 27 28 system("pause"); 29 return 0; 3
问题1：存放字符串在内存中的首地址；sizeof计算s1占用的内存字节数；strlen统计\0之前的字节数
问题2：不能。指针储存的是字符串的首地址,这是一个字符指针指向字符串；数组是储存字符串，给数组初始化。
问题3：交换的是数组内容，在内存中没有交换
```
task3
1 #include <stdio.h> 2 #include <stdlib.h> 3 4 int main() { 5 int x[2][4] = {{1, 9, 8, 4}, {2, 0, 4, 9}}; 6 int i, j; 7 int *ptr1; 8 int(*ptr2)[4]; 9 10 printf("输出1: 使用数组名、下标直接访问二维数组元素\n"); 11 for (i = 0; i < 2; ++i) { 12 for (j = 0; j < 4; ++j) 13 printf("%d ", x[i][j]); 14 printf("\n"); 15 } 16 17 printf("\n输出2: 使用指针变量ptr1(指向元素)间接访问\n"); 18 for (ptr1 = &x[0][0], i = 0; ptr1 < &x[0][0] + 8; ++ptr1, ++i) { 19 printf("%d ", *ptr1); 20 21 if ((i + 1) % 4 == 0) 22 printf("\n"); 23 } 24 25 printf("\n输出3: 使用指针变量ptr2(指向一维数组)间接访问\n"); 26 for (ptr2 = x; ptr2 < x + 2; ++ptr2) { 27 for (j = 0; j < 4; ++j) 28 printf("%d ", *(*ptr2 + j)); 29 printf("\n"); 30 } 31 32 system("pause"); 33 return 0; 34 }
```
task4
1 #include <stdio.h> 2 #define N 80 3 #include<stdlib.h> 4 5 void replace(char *str, char old_char, char new_char); 6 7 int main() { 8 char text[N] = "Programming is difficult or not, it is a question."; 9 10 printf("原始文本: \n"); 11 printf("%s\n", text); 12 13 replace(text, 'i', '*'); 14 15 printf("处理后文本: \n"); 16 printf("%s\n", text); 17 18 system("pause"); 19 return 0; 20 } 21 23 void replace(char *str, char old_char, char new_char) { 24 int i; 25 26 while(*str) { 27 if(*str == old_char) 28 *str = new_char; 29 str++; 30 } 31 }
问题1：replace的作用是将字符串中的old_char替换成new_charbu
问题2：可以
```
task5:
#include <stdio.h> #define N 80 #include<stdlib.h>. char *str_trunc(char *str, char x); int main() { char str[N]; char ch; while(printf("输入字符串: "), gets(str) != NULL) { printf("输入一个字符: "); ch = getchar(); printf("截断处理...\n"); str_trunc(str, ch); printf("截断处理后的字符串: %s\n\n", str); getchar(); } system("pause"); return 0; } char *str_trunc(char *str, char x){ char *b=str; while(*b!='\0'&&*b!=x){ b++; } if(*b ==x){ *b='\0'; }; return str; }
```
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 26 int check_id(char *str) { 27 int i; 28 if(strlen(str)!=18) 29 return 0; 30 31 for(i=0;i<17;i++){ 32 if(str[i]<'0'||str[i]>'9') 33 return 0; 34 } 35 if (str[17]>='0'&&str[17]<='9'||str[17]=='X') 36 return 1; 37 else 38 return 0; 39 }
```
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; 29 for(i=0;str[i]!='\0';i++){ 30 if(str[i]>='a'&&str[i]<='z') 31 str[i]='a'+(str[i]+n-'a')%26; 32 if(str[i]>='A'&&str[i]<='Z') 33 str[i]='A'+(str[i]+n-'A')%26; 34 } 35 } 36 37 void decoder(char *str, int n) { 38 int i; 39 for(i=0;str[i]!='\0';i++){ 40 if(str[i]>='a'&&str[i]<='z') 41 str[i]='a'+(str[i]+(26-n)-'a')%26; 42 if(str[i]>='A'&&str[i]<='Z') 43 str[i]='A'+(str[i]+(26-n)-'A')%26; 44 } 45 }
```
task8
1 #include <stdio.h> 2 #include <string.h> 3 int main(int argc, char *argv[]) { 4 int i,j; 5 for (i=1;i<argc;i++){ 6 for(j=1;j<argc-i;j++){ 7 8 if(strcmp(argv[j],argv[j+1])>0){ 9 char *temp=argv[j]; 10 argv[j]=argv[j+1]; 11 argv[j+1]=temp; 12 } 13 } 14 } 15 for(i = 1; i < argc; ++i) 16 printf("hello, %s\n", argv[i]); 17 18 return 0; 19 }
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