实验5
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("¼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1：找出数组中的最大，最小值。
问2：min，max变量。
—_2
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1：找出数组的最大值
问2：可以。
task2.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 27 return 0; 28 }
问1：80 ， 数组s1所占用的字节数。字符串中有效字符的长度。
问2：不能，s1是常量，不可以被赋值。
——2
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1：字符串首个地址；指针本身大小；统计字符串有效长度。
问2：不能，原写法是指针变量存字符串地址且可赋值，而替换的不可被赋值。
task3.c
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("\n"); 10 for (i = 0; i < 2; ++i) { 11 for (j = 0; j < 4; ++j) 12 printf("%d ", x[i][j]); 13 printf("输出1：使用数组名,下标直接访问二维数组元素\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 }
task4.c
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 printf("处理后文本: \n"); 14 printf("%s\n", text); 15 16 return 0; 17 } 18 19 void replace(char *str, char old_char, char new_char) { 20 int i; 21 22 while(*str) { 23 if(*str == old_char) 24 *str = new_char; 25 str++; 26 } 27 }
问1：将字符串中所有指定的就字符替换为新字符。
问2：可以。
task5.c
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 printf("截断处理后的字符串：%s\n\n", str); 17 getchar(); 18 } 19 20 return 0; 21 }
问：输入字符串时会被跳过，是吸收输入时的换行符。
task6.c
1 #include <stdio.h> 2 #include <string.h> 3 #define N 5 4 5 int check_id(char *str); 6 int main() 7 { 8 char *pid[N] = {"31010120000721656X", 9 "3301061996X0203301", 10 "53010220051126571", 11 "510104199211197977", 12 "53010220051126133Y"}; 13 int i; 14 15 for (i = 0; i < N; ++i) 16 if (check_id(pid[i])) 17 printf("%s\tTrue\n", pid[i]); 18 else 19 printf("%s\tFalse\n", pid[i]); 20 21 return 0; 22 } 23 24 int check_id(char *str) { 25 if (strlen(str) != 18) { 26 return 0; 27 } 28 for (int i = 0; i < 17; ++i) { 29 if (str[i] < '0' || str[i] > '9') { 30 return 0; 31 } 32 } 33 char last_char = str[17]; 34 if (!((last_char >= '0' && last_char <= '9') || last_char == 'X')) { 35 return 0; 36 } 37 return 1; 38 39 }
task7.c
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 printf("输入英文文本："); 9 gets(words); 10 printf("输入n："); 11 scanf("%d", &n); 12 printf("编码后的英文文本："); 13 encoder(words, n); 14 printf("%s\n", words); 15 printf("对编码后的英文文本解码："); 16 decoder(words, n); 17 printf("%s\n", words); 18 return 0; 19 } 20 21 void encoder(char *str, int n) { 22 while (*str != '\0') { 23 if (*str >= 'a' && *str <= 'z') { 24 *str = 'a' + ((*str - 'a' + n) % 26); 25 } 26 else if (*str >= 'A' && *str <= 'Z') { 27 *str = 'A' + ((*str - 'A' + n) % 26); 28 } 29 30 str++; 31 } 32 } 33 34 void decoder(char *str, int n) { 35 36 while (*str != '\0') { 37 38 if (*str >= 'a' && *str <= 'z') { 39 40 *str = 'a' + ((*str - 'a' - n + 26) % 26); 41 } 42 43 else if (*str >= 'A' && *str <= 'Z') { 44 *str = 'A' + ((*str - 'A' - n + 26) % 26); 45 } 46 47 str++; 48 } 49 }
task8.c
1 #include <stdio.h> 2 #include <string.h> 3 int main(int argc, char *argv[]) { 4 int i, j; 5 char *temp; 6 for (i = 1; i < argc - 1; i++) { 7 for (j = 1; j < argc - i; j++) { 8 if (strcmp(argv[j],argv[j+1]) > 0) { 9 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 }
问了ai也不会.....
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