实验任务1（1）
1 #include <stdio.h> 2 #define N 5 3 void input(int x[], int n); 4 void output(int x[], int n); 5 void find_min_max(int x[], int n, int *pmin, int *pmax); 6 int main() { 7 int a[N]; 8 int min, max; 9 printf("录入%d个数据:\n", N); 10 input(a, N); 11 printf("数据是: \n"); 12 output(a, N); 13 printf("数据处理...\n"); 14 find_min_max(a, N, &min, &max); 15 printf("输出结果:\n"); 16 printf("min = %d, max = %d\n", min, max); 17 return 0; 18 } 19 void input(int x[], int n) { 20 int i; 21 for(i = 0; i < n; ++i) 22 scanf("%d", &x[i]); 23 } 24 void output(int x[], int n) { 25 int i; 26 for(i = 0; i < n; ++i) 27 printf("%d ", x[i]); 28 printf("\n"); 29 } 30 void find_min_max(int x[], int n, int *pmin, int *pmax) { 31 int i; 32 *pmin = *pmax = x[0]; 33 for(i = 1; i < n; ++i) 34 if(x[i] < *pmin) 35 *pmin = x[i]; 36 else if(x[i] > *pmax) 37 *pmax = x[i]; 38 }
结果演示
答：（1）找出5个数据中的最大数和最小数
（2）指向max和min
实验任务1（2）
1 #include <stdio.h> 2 #define N 5 3 void input(int x[], int n); 4 void output(int x[], int n); 5 int *find_max(int x[], int n); 6 int main() { 7 int a[N]; 8 int *pmax; 9 printf("录入%d个数据:\n", N); 10 input(a, N); 11 printf("数据是: \n"); 12 output(a, N); 13 printf("数据处理...\n"); 14 pmax = find_max(a, N); 15 printf("输出结果:\n"); 16 printf("max = %d\n", *pmax); 17 return 0; 18 } 19 void input(int x[], int n) { 20 int i; 21 for(i = 0; i < n; ++i) 22 scanf("%d", &x[i]); 23 } 24 void output(int x[], int n) { 25 int i; 26 for(i = 0; i < n; ++i) 27 printf("%d ", x[i]); 28 printf("\n"); 29 } 30 31 int *find_max(int x[], int n) { 32 int max_index = 0; 33 int i; 34 for(i = 1; i < n; ++i) 35 if(x[i] > x[max_index]) 36 max_index = i; 37 return &x[max_index]; 38 }
结果演示
答：（1）返回的是最大数的地址
（2）可以
实验任务2（1）
1 #include <stdio.h> 2 #include <string.h> 3 #define N 80 4 int main() { 5 char s1[] = "Learning makes me happy"; 6 char s2[] = "Learning makes me sleepy"; 7 char tmp[N]; 8 printf("sizeof(s1) vs. strlen(s1): \n"); 9 printf("sizeof(s1) = %d\n", sizeof(s1)); 10 printf("strlen(s1) = %d\n", strlen(s1)); 11 12 printf("\nbefore swap: \n"); 13 printf("s1: %s\n", s1); 14 printf("s2: %s\n", s2); 15 16 printf("\nswapping...\n"); 17 strcpy(tmp, s1); 18 strcpy(s1, s2); 19 strcpy(s2, tmp); 20 21 printf("\nafter swap: \n"); 22 printf("s1: %s\n", s1); 23 printf("s2: %s\n", s2); 24 return 0; 25 }
结果演示
答：（1）存放的是“Learning.......happy\0”，sizeof计算的是s1占的总空间大小，strlen统计的是字符串的长度
（2）不能，没有定义s1大小，会报错
（3）会交换
实验任务2(2)
1 #include <stdio.h> 2 #include <string.h> 3 #define N 80 4 int main() { 5 char *s1 = "Learning makes me happy"; 6 char *s2 = "Learning makes me sleepy"; 7 char *tmp; 8 9 printf("sizeof(s1) vs. strlen(s1): \n"); 10 printf("sizeof(s1) = %d\n", sizeof(s1)); 11 printf("strlen(s1) = %d\n", strlen(s1)); 12 13 printf("\nbefore swap: \n"); 14 printf("s1: %s\n", s1); 15 printf("s2: %s\n", s2); 16 17 printf("\nswapping...\n"); 18 tmp = s1; 19 s1 = s2; 20 s2 = tmp; 21 22 printf("\nafter swap: \n"); 23 printf("s1: %s\n", s1); 24 printf("s2: %s\n", s2); 25 return 0; 26 }
结果演示
答：（1）存放的是s1的地址，计算的是s1地址的内存大小，统计的是s1字符串的长度
（2）可以
（3）交换的是地址，没有交换
实验任务3
1 #include <stdio.h> 2 int main() { 3 int x[2][4] = {{1, 9, 8, 4}, {2, 0, 4, 9}}; 4 int i, j; 5 int *ptr1; 6 int(*ptr2)[4]; 7 printf("输出1: 使用数组名、下标直接访问二维数组元素\n"); 8 for (i = 0; i < 2; ++i) { 9 for (j = 0; j < 4; ++j) 10 printf("%d ", x[i][j]); 11 printf("\n"); 12 } 13 14 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"); 15 for (ptr1 = &x[0][0], i = 0; ptr1 < &x[0][0] + 8; ++ptr1, ++i) { 16 printf("%d ", *ptr1); 17 if ((i + 1) % 4 == 0) 18 printf("\n"); 19 } 20 21 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"); 22 for (ptr2 = x; ptr2 < x + 2; ++ptr2) { 23 for (j = 0; j < 4; ++j) 24 printf("%d ", *(*ptr2 + j)); 25 printf("\n"); 26 } 27 28 return 0; 29 }
结果演示
实验任务4
1 #include <stdio.h> 2 #define N 80 3 void replace(char *str, char old_char, char new_char); 4 int main() { 5 char text[N] = "c programming is difficult or not, it is a question."; 6 printf("原始文本: \n"); 7 printf("%s\n", text); 8 9 replace(text, 'i', '*'); 10 11 printf("处理后文本: \n"); 12 printf("%s\n", text); 13 return 0; 14 } 15 16 void replace(char *str, char old_char, char new_char) { 17 int i; 18 while(*str) { 19 if(*str == old_char) 20 *str = new_char; 21 str++; 22 } 23 }
结果演示
答：（1）将语句中的某个字母换成另外一个字母
（2）可以
实验任务4（2）
1 #include <stdio.h> 2 #define N 80 3 void str_trunc(char *str, char x); 4 int main() { 5 char str[N]; 6 char ch; 7 printf("输入字符串: "); 8 gets(str); 9 printf("输入一个字符: "); 10 ch = getchar(); 11 12 printf("截断处理...\n"); 13 str_trunc(str, ch); 14 15 printf("截断处理后的字符串: %s\n", str); 16 17 } 18 void str_trunc(char *str, char x) { 19 while(*str) { 20 if(*str == x) 21 break; 22 23 else str++; 24 } 25 26 *str='\0'; 27 }
结果演示
实验任务5
1 #include <stdio.h> 2 #include <string.h> 3 void sort(char *name[], int n); 4 int main() { 5 char *course[4] = {"C Program", 6 "C++ Object Oriented Program", 7 "Operating System", 8 "Data Structure and Algorithms"}; 9 int i; 10 11 sort(course, 4); 12 13 for (i = 0; i < 4; i++) 14 printf("%s\n", course[i]); 15 return 0; 16 } 17 18 void sort(char *name[], int n) { 19 int i, j; 20 char *tmp; 21 22 for (i = 0; i < n - 1; ++i) 23 for (j = 0; j < n - 1 - i; ++j) 24 if (strcmp(name[j], name[j + 1]) > 0) { 25 tmp = name[j]; 26 name[j] = name[j + 1]; 27 name[j + 1] = tmp; 28 } 29 }
结果演示
1 #include <stdio.h> 2 #include <string.h> 3 void sort(char *name[], int n); 4 int main() { 5 char *course[4] = {"C Program", 6 "C++ Object Oriented Program", 7 "Operating System", 8 "Data Structure and Algorithms"}; 9 int i; 10 11 sort(course, 4); 12 for (i = 0; i < 4; i++) 13 printf("%s\n", course[i]); 14 15 return 0; 16 } 17 void sort(char *name[], int n) { 18 int i, j, k; 19 char *tmp; 20 21 for (i = 0; i < n - 1; i++) { 22 k = i; 23 for (j = i + 1; j < n; j++) 24 if (strcmp(name[j], name[k]) < 0) 25 k = j; 26 27 if (k != i) { 28 tmp = name[i]; 29 name[i] = name[k]; 30 name[k] = tmp; 31 } 32 } 33 }
结果演示
答：交换了指针变量的值
实验任务6
1 #include <stdio.h> 2 #include <string.h> 3 #define N 5 4 int check_id(char *str); 5 int main() { 6 char *pid[N] = {"31010120000721656X", 7 "330106199609203301", 8 "53010220051126571", 9 "510104199211197977", 10 "53010220051126133Y"}; 11 int i; 12 for (i = 0; i < N; ++i) 13 if (check_id(pid[i])) 14 printf("%s\tTrue\n", pid[i]); 15 else 16 printf("%s\tFalse\n", pid[i]); 17 18 return 0; 19 } 20 int check_id(char *str) { 21 int i=0; 22 while(*str){ 23 if(*str>'9'&&*str!='X'||*str<'0'&&*str!='X') return 0; 24 str++; 25 i++; 26 } 27 if(i!=18) return 0; 28 return 1; 29 }
结果演示
实验任务7
1 #include <stdio.h> 2 #define N 80 3 void encoder(char *str); 4 void decoder(char *str); 5 int main() { 6 char words[N]; 7 8 printf("请输入英文文本："); 9 gets(words); 10 printf("编码后的英文文本："); 11 encoder(words); 12 printf("%s\n",words); 13 14 printf("对编码后的英文文本解码："); 15 decoder(words); 16 printf("%s\n",words); 17 18 return 0; 19 } 20 void encoder(char *str){ 21 while(*str){ 22 if(*str=='z'){ 23 *str='a'; 24 }else if(*str=='Z'){ 25 *str='A'; 26 27 }else{ 28 (*str)++; 29 } 30 str++; 31 } 32 } 33 34 void decoder(char *str){ 35 while(*str){ 36 if(*str=='a'){ 37 *str='z'; 38 }else if(*str=='A'){ 39 *str='Z'; 40 41 }else{ 42 (*str)++; 43 } 44 str++; 45 } 46 }
结果演示
实验任务8