实验5
实验任务1
源代码
1 #include <stdio.h> 2 #include <stdlib.h> 3 #define N 5 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 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 void find_min_max(int x[], int n, int *pmin, int *pmax) { 46 int i; 47 48 *pmin = *pmax = x[0]; 49 50 for(i = 0; i < n; ++i) 51 if(x[i] < *pmin) 52 *pmin = x[i]; 53 else if(x[i] > *pmax) 54 *pmax = x[i]; 55 }
运行截图
问题
1）找输入数据的最小值和最大值
2）指向数组x的第一个元素
实验任务2
源代码
1 #define _CRT_SECURE_NO_WARNINGS 2 #include <stdio.h> 3 #include <string.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("sizeof(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("\nswappping...\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 28 return 0; 29 }
运行结果
问题
1）s1的大小为80个字节；sizeof(s1)计算的是s1所占空间的内存大小；strlen(s1)统计的是s1中含有的字符数量
2）不能，不能将字符串作为整体输入给数组
3）交换
源代码
1 #define _CRT_SECURE_NO_WARNINGS 2 #include <stdio.h> 3 #include <string.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("sizeof(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("\nswappping...\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 return 0; 29 }
运行结果
问题
1）指针变量s1存放的是s1[0]的地址；sizeof(s1)计算的是指针变量s1所占空间的内存大小；strlen(s1)统计的是s1内字符数量
2）能；原代码是定义一个指向数组的指针变量并将字符串赋值给此变量；给出代码是先定义一个指针变量后进行赋值
3)交换的是地址；字符串常量未进行交换
实验任务3
源代码
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）直接访问\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）直接访问\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 }
运行结果
实验任务4
源代码
1 #include <stdio.h> 2 #define N 80 3 4 void replace(char* str, char old_char, char new_char); 5 6 int main() { 7 char text[N] = "Programming is difficult or not,it is a question"; 8 9 printf("原始文本:\n"); 10 printf("%s\n", text); 11 12 replace(text, 'i', '*'); 13 14 printf("处理后文本：\n"); 15 printf("%s\n", text); 16 17 return 0; 18 } 19 20 21 void replace(char* str, char old_char, char new_char) { 22 int i; 23 24 while (*str) { 25 if (*str == old_char) 26 *str = new_char; 27 str++; 28 } 29 30 }
运行结果
问题
1）replace的作用是将字符串中的i全部替换为*
2）可以
实验任务5
源代码
1 #include <stdio.h> 2 #define N 80 3 4 char *str_trunc(char* str, char x); 5 6 int main() { 7 char str[N]; 8 char ch; 9 10 while (printf("输入字符串："), gets(str) != NULL) { 11 printf("输入一个字符："); 12 ch = getchar(); 13 14 printf("截断处理...\n"); 15 str_trunc(str, ch); 16 17 printf("截断处理后的字符串：%s\n\n", str); 18 getchar(); 19 } 20 21 return 0; 22 23 } 24 25 char *str_trunc(char* str, char x) { 26 char* p = str; 27 28 while(*p != x) { 29 p++; 30 } 31 32 *p = '\0'; 33 34 return str; 35 36 }
运行结果
问题
去掉后，接下来不能重新输入一个字符，line18的作用为重置ch，以便接下来输入字符
实验任务6
源代码
1 #include <stdio.h> 2 #include <string.h> 3 #define N 5 4 5 int check_id(char* str); 6 7 int main() { 8 char* pid[N] = { 9 "31010120000721656x", 10 "3301061996x0203301", 11 "53010220051126571 ", 12 "510104199211197977", 13 "53010220051126133y" 14 }; 15 16 int i; 17 18 for (i = 0; i < N; ++i) 19 if (check_id(pid[i])) 20 printf("%s\tTrue\n", pid[i]); 21 else 22 printf("%s\tFalse\n", pid[i]); 23 24 return 0; 25 } 26 27 int check_id(char* str) { 28 if (strlen(str) != 18) 29 return 0; 30 31 int i; 32 33 for (i = 0; i < 17; i++) { 34 if (str[i] < '0' || str[i]>'9') 35 return 0; 36 } 37 38 if (!(str[17]>='0' && str[17]<='9' || str[17] == 'x')) 39 return 0; 40 41 return 1; 42 }
运行结果
实验任务7
源代码
1 #include <stdio.h> 2 #define N 80 3 4 void encoder(char* str, int n); 5 void decoder(char* str, int n); 6 7 int main() { 8 char words[N]; 9 int n; 10 11 printf("输入英文文本:"); 12 gets(words); 13 14 printf("输入n:"); 15 scanf_s("%d", &n); 16 17 printf("编码后的英文文本:"); 18 encoder(words, n); 19 printf("%s\n", words); 20 21 printf("对编码后的英文文本解码:"); 22 decoder(words, n); 23 printf("%s\n", words); 24 25 return 0; 26 27 } 28 29 void encoder(char* str, int n) { 30 int i; 31 32 for (i = 0; str[i] != '\0'; ++i) { 33 if (str[i] >= 'a' && str[i] <= 'z') { 34 str[i] = 'a' + (str[i] - 'a' + n) % 26; 35 } 36 37 else if (str[i] >= 'A' && str[i] <= 'Z') { 38 str[i] = 'A' + (str[i] - 'A' + n) % 26; 39 } 40 } 41 42 } 43 44 void decoder(char* str, int n) { 45 int i; 46 47 for (i = 0; str[i] != '\0'; ++i) { 48 if (str[i] >= 'a' && str[i] <= 'z') { 49 str[i] = 'a' + (str[i] - 'a' - n) % 26; 50 } 51 52 else if (str[i] >= 'A' && str[i] <= 'Z') { 53 str[i] = 'A' + (str[i] - 'A' - n) % 26; 54 } 55 } 56 }
运行结果
实验任务8
源代码1
1 #include <stdio.h> 2 3 int main(int argc, char *argv[]) { 4 int i; 5 6 for(i = 1; i < argc; ++i) 7 printf("hello, %s\n", argv[i]); 8 9 return 0; 10 }
运行结果
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