实验5
任务1
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 }
问题1:函数的功能是找出最大值和最小值
问题2:指向数组中的第一个元素
task1_ 2.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 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 }
问题1:返回最大值的地址
问题2:可以
任务2
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
1 #include <stdio.h> 2 #define N 80 3 4 char *str_trunc(char *str, char ch) 5 { 6 char *ptr = str; 7 while (*ptr) 8 { 9 if (*ptr == ch) 10 { 11 *ptr = '\0'; 12 break; 13 } 14 ptr++; 15 } 16 return str; 17 } 18 19 int main() { 20 char str[N]; 21 char ch; 22 23 while(printf("输入字符串: "), gets(str) != NULL) { 24 printf("输入一个字符: "); 25 ch = getchar(); 26 27 printf("截断处理...\n"); 28 str_trunc(str, ch); 29 30 printf("截断处理后的字符串: %s\n\n", str); 31 getchar(); 32 } 33 34 return 0; 35 }
27 return 0; 28 }
问题1:数组s1的大小是80个字节,sizeof(s1)计算的是数组s1的长度,strlen(s1)统计的是s1字符长度
问题2:不可以,数组名本质上是地址,不能将字符串赋值给数组名
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 }
问题1:指针变量s1存放的是字符串起始地址,sizeof(s1)计算的是指针占用的内存,strlen(s1)统计的是字符串长度
问题2:可以一个是地址常量,一个是指针变量
问题3:交换的是变量的值,没有交换字符串常量
任务3
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 }
问题1:ptr表示的是指针
问题2:ptr表示的是数组
任务4
1 #define N 80 2 void replace(char* str,char old_char,char new_char) 3 { 4 int i; 5 while(*str) 6 { 7 if(*str==old_char) 8 *str=new_char; 9 str++; 10 } 11 } 12 int main() 13 { 14 char text[N]="Programming is difficult or not ,it is a queston."; 15 printf("原始文本:\n"); 16 printf("%s\n",text); 17 18 replace(text,'i','*'); 19 printf("处理后文本:\n"); 20 printf("%s\n",text); 21 return 0; 22 }
问题1：函数功能是将i全部转换为*
问题2：可以
任务5
1 #include <stdio.h> 2 #define N 80 3 4 char *str_trunc(char *str, char ch) 5 { 6 char *ptr = str; 7 while (*ptr) 8 { 9 if (*ptr == ch) 10 { 11 *ptr = '\0'; 12 break; 13 } 14 ptr++; 15 } 16 return str; 17 } 18 19 int main() { 20 char str[N]; 21 char ch; 22 23 while(printf("输入字符串: "), gets(str) != NULL) { 24 printf("输入一个字符: "); 25 ch = getchar(); 26 27 printf("截断处理...\n"); 28 str_trunc(str, ch); 29 30 printf("截断处理后的字符串: %s\n\n", str); 31 getchar(); 32 } 33 34 return 0; 35 }
任务6
1 #include <stdio.h> 2 #include <string.h> 3 #define N 5 4 5 int check_id(char *str) 6 { 7 if(strlen(str)!=18) 8 return 0; 9 int i; 10 for(i=0;i<17;i++) 11 if (str[i]>'9') 12 return 0; 13 if((str[17]<='9'&&str[17]>='0')||str[17]=='X') 14 return 1; 15 else 16 return 0; 17 } 18 19 int main() 20 { 21 char *pid[N] = {"31010120000721656X", 22 "3301061996X0203301", 23 "53010220051126571", 24 "510104199211197977", 25 "53010220051126133Y"}; 26 int i; 27 28 for (i = 0; i < N; ++i) 29 if (check_id(pid[i])) // 函数调用 30 printf("%s\tTrue\n", pid[i]); 31 else 32 printf("%s\tFalse\n", pid[i]); 33 34 return 0; 35 }
任务7
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 { 28 while (*str) 29 { 30 if (*str >= 'a' && *str <= 'z') 31 { 32 *str = (*str - 'a' + n + 26) % 26 + 'a'; 33 } 34 if (*str >= 'A' && *str <= 'Z') 35 { 36 *str = (*str - 'A' + n + 26) % 26 + 'A'; 37 } 38 str++; 39 } 40 } 41 42 void decoder(char *str, int n) 43 { 44 while (*str) 45 { 46 if (*str >= 'a' && *str <= 'z') 47 { 48 *str = (*str - 'a' - n + 26) % 26 + 'a'; 49 } 50 if (*str >= 'A' && *str <= 'Z') 51 { 52 *str = (*str - 'A' - n + 26) % 26 + 'A'; 53 } 54 str++; 55 } 56 }
任务8
1 #include <stdio.h> 2 #include <stdlib.h> 3 #include <string.h> 4 void bubbleSort(char *names[], int n) { 5 int i, j; 6 for (i = 0; i < n - 1; i++) { 7 for (j = 0; j < n - i - 1; j++) { 8 if (strcmp(names[j], names[j + 1]) > 0) { 9 char *temp = names[j]; 10 names[j] = names[j + 1]; 11 names[j + 1] = temp; 12 } 13 } 14 } 15 } 16 int main(int argc, char *argv[]) { 17 18 if (argc < 2) { 19 printf("请在命令行输入至少一个姓名作为参数\n"); 20 return 1; 21 } 22 bubbleSort(argv + 1, argc - 1); 23 int i; 24 for (i=1; i < argc; ++i) { 25 printf("hello, %s\n", argv[i]); 26 } 27 return 0; 28 }
