实验5
源代码1.1;
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 }
截图1.1;
问题;
找出五个数中的最大值和最小值;
min,max
源代码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 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 }
截图1.2;
问题;
最大值;
不可以,返回的指针指向无效的地址
源代码2.1;
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 }
截图2.1;
问题;
1:80字节
2:s1所占字节数
3:s1中字符的个数
不能,不能对已定义的数组整体赋值
交换了
源代码2.2;
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 }
截图2.2;
问题;
存放的是learning makes me happy的首地址
s1所占内存大小
s1中字符个数
可以;第一种是在声明时赋值,第二种是先声明后赋值
交换的是s1,s2的地址,没有交换
源代码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 }
截图3;
问题;
1.ptr是指针变量,指向包含4个int元素的一维数组
2.ptr是一个数组的名称
源代码4;
1 #include <stdio.h> 2 #define N 80 3 4 void replace(char *str, char old_char, char new_char); // 函数声明 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', '*'); // 函数调用 注意字符形参写法,单引号不能少 13 14 printf("处理后文本: \n"); 15 printf("%s\n", text); 16 17 return 0; 18 } 19 20 // 函数定义 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 }
截图4;
问题;将传入的字符串 str 中指定的字符 old_char 替换为字符 new_char
可以
源代码5;
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); // 函数调用 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 char *str_trunc(char *str,char x){ 24 int i=0; 25 while(str[i]!='\0'){ 26 if(str[i]==x){ 27 str[i]='\0'; 28 break; 29 } 30 i++; 31 } 32 return str; 33 34 }
截图5;
问题;getchar会读取换行符
源代码6;
1 #include <stdio.h> 2 #include <string.h> 3 #define N 5 4 5 int check_id(char *str); // 函数声明 6 7 int main() 8 { 9 char *pid[N] = {"31010120000721656X", 10 "3301061996X0203301", 11 "53010220051126571", 12 "510104199211197977", 13 "53010220051126133Y"}; 14 int i; 15 16 for (i = 0; i < N; ++i) 17 if (check_id(pid[i])) // 函数调用 18 printf("%s\tTrue\n", pid[i]); 19 else 20 printf("%s\tFalse\n", pid[i]); 21 22 return 0; 23 } 24 25 // 函数定义 26 // 功能: 检查指针str指向的身份证号码串形式上是否合法 27 // 形式合法,返回1,否则,返回0 28 int check_id(char *str) { 29 int len = strlen(str); 30 31 if(len!=18){ 32 return 0; 33 } 34 35 int i; 36 37 for(i=0;i<len-1;i++){ 38 if(!('0'<=str[i]&&str[i]<='9')){ 39 return 0; 40 } 41 } 42 43 char lastchar = str[len-1]; 44 if(lastchar == 'X'){ 45 return 1; 46 } 47 else if('0'<=lastchar&&lastchar<='9'){ 48 return 1; 49 } 50 else{ 51 return 0; 52 } 53 }
截图6;
源代码7;
1 #include <stdio.h> 2 #define N 80 3 void encoder(char *str, int n); 4 void decoder(char *str, int n); 5 6 int main() { 7 char words[N]; 8 int n; 9 10 printf("输入英文文本: "); 11 gets(words); 12 13 printf("输入n: "); 14 scanf("%d", &n); 15 16 printf("编码后的英文文本: "); 17 encoder(words, n); 18 printf("%s\n", words); 19 20 printf("对编码后的英文文本解码: "); 21 decoder(words, n); 22 printf("%s\n", words); 23 24 return 0; 25 } 26 27 void encoder(char *str, int n) { 28 while (*str!= '\0') { 29 if ((*str >= 'a' && *str <= 'z') || (*str >= 'A' && *str <= 'Z')) { 30 if (*str >= 'a' && *str <= 'z') { 31 *str = ((*str - 'a') + n) % 26 + 'a'; 32 } else if (*str >= 'A' && *str <= 'Z') { 33 *str = ((*str - 'A') + n) % 26 + 'A'; 34 } 35 } 36 str++; 37 } 38 } 39 40 void decoder(char *str, int n) { 41 while (*str!= '\0') { 42 if ((*str >= 'a' && *str <= 'z') || (*str >= 'A' && *str <= 'Z')) { 43 if (*str >= 'a' && *str <= 'z') { 44 *str = ((*str - 'a') - n + 26) % 26 + 'a'; 45 } else if (*str >= 'A' && *str <= 'Z') { 46 *str = ((*str - 'A') - n + 26) % 26 + 'A'; 47 } 48 } 49 str++; 50 } 51 }
截图7;
源代码8;
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; 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 } 16 17 for(i = 1; i < argc; ++i) 18 printf("hello, %s\n", argv[i]); 19 20 return 0; 21 }
截图8;