实验5
实验任务1
源代码：
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 for (int i = 0; i < n; i++) 21 scanf_s("%d", &x[i]); 22 23 } 24 void output(int x[], int n) { 25 for (int i = 0; i < n; i++) 26 printf("%d", x[i]); 27 printf("\n"); 28 } 29 int* find_max(int x[], int n) { 30 int i = 0; 31 int max = 0; 32 for (; i < n; i++) 33 if (x[i] > x[max]) 34 max = i; 35 return &x[max]; 36 37 }
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问题1：找到数组最大数的地址，最大数的地址
问题2：可以
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("max=%d,min=%d\n", max,min); 17 return 0; 18 } 19 void input(int x[], int n) { 20 for (int i = 0; i < n; i++) 21 scanf_s("%d", &x[i]); 22 23 } 24 void output(int x[], int n) { 25 for (int i = 0; i < n; i++) 26 printf("%d", x[i]); 27 printf("\n"); 28 } 29 void find_min_max(int x[], int n,int *pmin,int *pmax) { 30 int i = 0; 31 *pmax =*pmin=x[0]; 32 for (; i < n; i++) 33 if (x[i] > *pmax) 34 *pmax = x[i]; 35 else if (x[i] < *pmin) 36 *pmin = x[i]; 37 38 }
问题1：找数组最大值与最小值
问题2：min,max
实验任务2：
源代码：
1 #include<stdio.h> 2 #include<string.h> 3 #define N 80 4 int main() { 5 char s1[N] = "Learning makes me happy"; 6 char s2[N] = "Learning makes me sleepy"; 7 char temp[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 printf("\nbefore swap:\n"); 12 printf("s1:%s\n", s1); 13 printf("s2:%s\n", s2); 14 printf("\nswapping...\n"); 15 strcpy(temp, s1); 16 strcpy(s1, s2); 17 strcpy(s2, temp); 18 printf("\nafter swap:\n"); 19 printf("s1:%s\n", s1); 20 printf("s2:%s\n", s2); 21 return 0;}
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问题1：80，数组的大小，字符串的大小
问题2：不可以，s1表示的是地址
问题3：是
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 *temp; 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 printf("\nbefore swap:\n"); 12 printf("s1:%s\n", s1); 13 printf("s2:%s\n", s2); 14 printf("\nswapping...\n"); 15 temp = s1; 16 s1 = s2; 17 s2 = temp; 18 printf("\nafter swap:\n"); 19 printf("s1:%s\n", s1); 20 printf("s2:%s\n", s2); 21 return 0; 22 23 }
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1.字符串第一个字符的地址；指针变量s1的大小；字符串长度
2.可以，一个储存字符串，一个是地址
3.地址，没有
实验任务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; // 指针变量，存放int类型数据的地址 7 int(*ptr2)[4]; // 指针变量，指向包含4个int元素的一维数组 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 }
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试验任务4
源代码：
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] = "Programming is difficult or not ,it is a question."; 6 printf("原始文本\n"); 7 printf("%s\n", text); 8 replace(text, 'i', '*'); 9 printf("处理后的文本\n"); 10 printf("%s\n", text); 11 return 0; 12 } 13 void replace(char* str, char old_char, char new_char) { 14 int i; 15 while (*str) { 16 if (*str == old_char) 17 *str = new_char; 18 str++; 19 } 20 }
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将字符替换
可以
实验任务5
源代码
1 #include<stdio.h> 2 #define N 80 3 char *str_trunc(char* str, char x); 4 char* sgets(char* m, int n); 5 int main() { 6 char str[N]; 7 char ch; 8 while (printf("输入字符串："), sgets(str, N) != NULL) { 9 printf("输入一个字符："); 10 ch = getchar(); 11 printf("截断处理...\n"); 12 str_trunc(str, ch); 13 printf("截断处理后的字符串：%s\n\n", str); 14 getchar(); 15 }return 0; 16 } 17 char* sgets(char* m, int n) { 18 char* real; 19 int i = 0; 20 real = fgets(m, n, stdin); 21 if (real) { 22 while (real[i] != '\n' && real[i] != '\0') 23 i++; 24 if (real[i] == '\n') 25 real[i] = '\0'; 26 else 27 while (getchar() != '\n') 28 continue; 29 } 30 return real; 31 } 32 char* str_trunc(char* str, char x) { 33 while (*str != x && *str != '\0') 34 str++; 35 if (*str == x) 36 *str = '\0'; 37 return str; 38 39 }
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第二组开始输出都是空字符串；line18是在吞掉输入输入字符时多输入的'\n'，
实验任务6：
源代码：
1 #include<stdio.h> 2 #include<string.h> 3 #define N 5 4 int check_id(char* str); 5 int main() { 6 char *str[N] = { 7 "31010120000721656X", 8 "3301061996x0203301", 9 "53010220051126571", 10 "510104199211197977", 11 "53010220051126133Y"}; 12 int i; 13 for (i = 0; i < N; i++) { 14 if (check_id(str[i])) 15 printf("%s\tTure\n", str[i]); 16 else 17 printf("%s\tFalse\n", str[i]); 18 } 19 20 21 return 0; 22 } 23 int check_id(char* str) { 24 25 if (strlen(str) != 18) 26 return 0; 27 int i = 0; 28 29 while (*str) 30 { 31 if (*str < '0') 32 33 return 0; 34 35 36 37 else if (*str > '9' && *str != 'X') 38 39 return 0; 40 41 42 else if (*str == 'X' && i != 17) 43 return 0; 44 i++; 45 46 str++; 47 } 48 49 return 1; 50 51 }
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实验任务7
源代码：
1 #include<stdio.h> 2 #define N 80 3 void encoder(char* str, int n); 4 void decoder(char* str, int n); 5 char* sgets(char* str, int n); 6 int main() { 7 char word[N]; 8 int n; 9 printf("输入英文文本\n"); 10 sgets(word, 80); 11 printf("输入n:\n"); 12 scanf_s("%d", &n); 13 printf("输入编码后的英文文本"); 14 encoder(word, n); 15 puts(word); 16 printf("对编码后的英文文本解码："); 17 decoder(word, n); 18 puts(word); 19 return 0; 20 }char* sgets(char* str, int n) { 21 int i = 0; 22 char* real; 23 real = fgets(str, n, stdin); 24 if (real) { 25 while (str[i] != '\n' && str[i] != '\0') 26 i++; 27 if (str[i] == '\n') 28 str[i] = '\0'; 29 else 30 while (getchar() != '\n') 31 continue; 32 33 } 34 return real; 35 } 36 void encoder(char* str, int n) { 37 while (*str) { 38 if ((*str >= 'a' && *str <= 'z') || (*str >= 'A' && *str <= 'Z')) { 39 if (*str <= 'Z' && (*str + n) > 'Z') 40 *str = *str + n - 26; 41 else if (*str + n > 'z') 42 *str = *str + n - 26; 43 else 44 *str = *str + n; 45 } 46 47 str++; 48 } 49 } 50 void decoder(char* str, int n) { 51 while (*str) { 52 if ((*str >= 'a' && *str <= 'z') || (*str >= 'A' && *str <= 'Z')) { 53 if (*str >= 'a' && (*str - n)<'a') 54 *str = *str -n + 26; 55 else if (*str + n < 'A') 56 *str = *str - n + 26; 57 else 58 *str = *str - n; 59 } 60 61 str++; 62 } 63 }
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实验任务8：
源代码
1 #include <stdio.h> 2 #include <string.h> 3 4 int main(int argc, char* argv[]) { 5 int i, j; 6 char* temp; 7 for (i = 1; i < argc - 1; ++i) { 8 for (j = 1; j < argc - 1; ++j) { 9 if (strcmp(argv[j], argv[j + 1])>0) { 10 temp = argv[j]; 11 argv[j] = argv[j + 1]; 12 argv[j + 1] = temp; 13 } 14 } 15 } 16 for (i = 1; i < argc; ++i) 17 printf("hello, %s\n", argv[i]); 18 19 return 0; 20 }
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