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 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 = 0; i < n; ++i) 34 if(x[i] < *pmin) 35 *pmin = x[i]; 36 else if(x[i] > *pmax) 37 *pmax = x[i]; 38 }
运行截图
问题一：找出五个数据中的最小值和最大值
问题二：指向函数中的min和max变量的地址
1.2
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 for(i = 0; i < n; ++i) 30 scanf("%d", &x[i]); 31 } 32 33 void output(int x[], int n) { 34 int i; 35 for(i = 0; i < n; ++i) 36 printf("%d ", x[i]); 37 printf("\n"); 38 } 39 40 int *find_max(int x[], int n) { 41 int max_index = 0; 42 int i; 43 44 for(i = 0; i < n; ++i) 45 if(x[i] > x[max_index]) 46 max_index = i; 47 48 return &x[max_index]; 49 }
问题一：找出数组中的最大值，返回该值的地址
问题二：可以
实验任务二
源代码
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 }
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问题一：大小80字节，数组在内存占的字节数，字符串的长度
问题二：不能，数组名是指针，不能进行赋值运算
问题三：进行交换
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 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 tmp = s1; 16 s1 = s2; 17 s2 = tmp; 18 printf("\nafter swap: \n"); 19 printf("s1: %s\n", s1); 20 printf("s2: %s\n", s2); 21 return 0; 22 }
运行截图
问题一:字符串的地址，指针本身的大小，指针指向的字符串长度
问题二：可以，2.1是字符数组初始化，2.2是指针指向字符串常量
问题三：指针指向，没有交换
试验任务三
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 if((i + 1) % 4 == 0) 20 printf("\n"); 21 } 22 23 printf("\n输出3：使用指针变量ptr2(指向一维数组)间接访问\n"); 24 for(ptr2 = x; ptr2 < x + 2; ++ptr2) { 25 for(j = 0; j < 4; ++j) 26 printf("%d ", *(*ptr2 + j)); 27 printf("\n"); 28 } 29 30 return 0; 31 }
实验任务四
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 void replace(char *str, char old_char, char new_char) { 21 while(*str) { 22 if(*str == old_char) 23 *str = new_char; 24 str++; 25 } 26 }
问题一：将字符串中的old_char字符替换为new_char字符
问题二：可以
实验任务五
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 char *str_trunc(char *str, char x) { 25 char *p = str; 26 27 while(*p && *p != x) 28 p++; 29 30 if(*p == x) 31 *p = '\0'; 32 33 return str; 34 }
问题：回车键会被读取，作用是吸收换行符
实验任务六
1 #include <stdio.h> 2 #include <string.h> 3 #include <ctype.h> 4 #define N 5 5 6 int check_id(char *str); 7 8 int main() { 9 char *pid[N] = { 10 "31010120000721656X", 11 "3301061996X0203301", 12 "53010220051126571", 13 "510104199211197977", 14 "53010220051126133Y" 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 25 return 0; 26 } 27 28 int check_id(char *str) { 29 int i; 30 31 if(strlen(str) != 18) 32 return 0; 33 34 for(i = 0; i < 17; i++) { 35 if(!isdigit(str[i])) 36 return 0; 37 } 38 39 if((isdigit(str[17])) || (str[17] == 'X')) 40 return 1; 41 42 return 0; 43 }
实验任务七
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("%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 void encoder(char *str, int n) { 29 while(*str) { 30 if(*str >= 'a' && *str <= 'z') { 31 *str = (*str - 'a' + n) % 26 + 'a'; 32 } 33 else if(*str >= 'A' && *str <= 'Z') { 34 *str = (*str - 'A' + n) % 26 + 'A'; 35 } 36 str++; 37 } 38 } 39 40 void decoder(char *str, int n) { 41 while(*str) { 42 if(*str >= 'a' && *str <= 'z') { 43 *str = (*str - 'a' - n + 26) % 26 + 'a'; 44 } 45 else if(*str >= 'A' && *str <= 'Z') { 46 *str = (*str - 'A' - n + 26) % 26 + 'A'; 47 } 48 str++; 49 } 50 }
实验任务八
1 #include <stdio.h> 2 #include <string.h> 3 int main(int argc, char *argv[]) { 4 int i,j; 5 for (i=1;i<argc;i++){ 6 for(j=1;j<argc-i;j++){ 7 8 if(strcmp(argv[j],argv[j+1])>0){ 9 char *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 }
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