实验5
task1_1
代码
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 = 1; i < n; ++i) 48 if(x[i] < *pmin) 49 *pmin = x[i]; 50 else if(x[i] > *pmax) 51 *pmax = x[i]; 52 }
结果
回答
功能是找出最大和最小值
分别指向数组a的第一个元素
task1_2
代码
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 = 1; i < n; ++i) 47 if(x[i] > x[max_index]) 48 max_index = i; 49 50 return &x[max_index]; 51 }
结果
回答
返回的是最大值的地址
可以
task2_1
代码
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[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 }
结果
回答
1：s1大小是24，sizeof计算的是数组大小及末尾的空字符，strlen统计的是字符数，不包括末尾空字符
2：不能，因为没有初始化且不能直接对数组进行赋值
3：已经交换
task2_2
代码
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 }
结果
回答
1：s1存放这段字符串的地址，sizeof计算的是指针变量s1的大小，strlen统计的是字符数
2：能够替换，区别是下面的代码未初始化s1，原代码直接初始化
3：交换的是指针的指向，两段字符串在单元中没有交换
task3
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 printf("输出1: 使用数组名、下标直接访问二维数组元素\n"); 9 for (i = 0; i < 2; ++i) { 10 for (j = 0; j < 4; ++j) 11 printf("%d ", x[i][j]); 12 printf("\n"); 13 } 14 15 printf("\n输出2: 使用指向元素的指针变量p间接访问二维数组元素\n"); 16 for (ptr1 = &x[0][0], i = 0; ptr1 < &x[0][0] + 8; ++ptr1, ++i) { 17 printf("%d ", *ptr1); 18 19 if ((i + 1) % 4 == 0) 20 printf("\n"); 21 } 22 23 printf("\n输出3: 使用指向一维数组的指针变量q间接访问二维数组元素\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 }
结果
task4_1
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] = "c 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 int i; 22 23 while(*str) { 24 if(*str == old_char) 25 *str = new_char; 26 str++; 27 } 28 }
结果
回答
功能是将原始文本中的旧字符改为新字符，新旧为自己指定。
可以。
task4_2
代码
1 #include <stdio.h> 2 #define N 80 3 4 void str_trunc(char *str, char x); 5 6 int main() { 7 char str[N]; 8 char ch; 9 10 printf("输入字符串: "); 11 gets(str); 12 13 printf("输入一个字符: "); 14 ch = getchar(); 15 16 printf("截断处理...\n"); 17 str_trunc(str, ch); 18 19 printf("截断处理后的字符串: %s\n", str); 20 21 } 22 23 void str_trunc(char *str, char x) { 24 while(*str) { 25 if(*str == x) 26 break; 27 28 str++; 29 } 30 31 *str='\0'; 32 }
结果
task5_1
代码
1 #include <stdio.h> 2 #include <string.h> 3 void sort(char *name[], int n); 4 5 int main() { 6 char *course[4] = {"C Program", 7 "C++ Object Oriented Program", 8 "Operating System", 9 "Data Structure and Algorithms"}; 10 int i; 11 12 sort(course, 4); 13 14 for (i = 0; i < 4; i++) 15 printf("%s\n", course[i]); 16 17 return 0; 18 } 19 20 void sort(char *name[], int n) { 21 int i, j; 22 char *tmp; 23 24 for (i = 0; i < n - 1; ++i) 25 for (j = 0; j < n - 1 - i; ++j) 26 if (strcmp(name[j], name[j + 1]) > 0) { 27 tmp = name[j]; 28 name[j] = name[j + 1]; 29 name[j + 1] = tmp; 30 } 31 }
结果
task5_2
代码
1 #include <stdio.h> 2 #include <string.h> 3 void sort(char *name[], int n); 4 5 int main() { 6 char *course[4] = {"C Program", 7 "C++ Object Oriented Program", 8 "Operating System", 9 "Data Structure and Algorithms"}; 10 int i; 11 12 sort(course, 4); 13 for (i = 0; i < 4; i++) 14 printf("%s\n", course[i]); 15 16 return 0; 17 } 18 19 void sort(char *name[], int n) { 20 int i, j, k; 21 char *tmp; 22 23 for (i = 0; i < n - 1; i++) { 24 k = i; 25 for (j = i + 1; j < n; j++) 26 if (strcmp(name[j], name[k]) < 0) 27 k = j; 28 29 if (k != i) { 30 tmp = name[i]; 31 name[i] = name[k]; 32 name[k] = tmp; 33 } 34 } 35 }
结果
回答
交换的是指针的指向。
task6
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] = {"31010120000721656X", 9 "330106199609203301", 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 char *i=str; 26 if(strlen(str)!=18){ 27 return 0; 28 } 29 while(*i){ 30 if(!(*i>='0'&&*i<='9')&&*i!='X'){ 31 return 0; 32 } 33 i++; 34 } 35 return 1; 36 }
结果
task7
代码
1 #include <stdio.h> 2 #define N 80 3 void encoder(char *str); 4 void decoder(char *str); 5 6 int main() { 7 char words[N]; 8 9 printf("输入英文文本: "); 10 gets(words); 11 12 printf("编码后的英文文本: "); 13 encoder(words); 14 printf("%s\n", words); 15 16 printf("对编码后的英文文本解码: "); 17 decoder(words); 18 printf("%s\n", words); 19 20 return 0; 21 } 22 23 24 void encoder(char *str) { 25 char *i=str; 26 while(*i){ 27 if(*i=='z'||*i=='Z'){ 28 *i=*i-26; 29 }else if((*i>='a'&&*i<'z')||(*i>='A'&&*i<'Z')){ 30 *i=*i+1; 31 } 32 i++; 33 } 34 } 35 36 37 void decoder(char *str) { 38 char *i=str; 39 while(*i){ 40 if(*i=='a'||*i=='A'){ 41 *i=*i+26; 42 }else if((*i>'a'&&*i<='z')||(*i>'A'&&*i<='Z')){ 43 *i=*i-1; 44 } 45 i++; 46 } 47 }
结果
task8
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;++i) { 8 for(j=i+1;j<argc;++j) { 9 if(strcmp(argv[i],argv[j])>0) { 10 temp = argv[i]; 11 argv[i] = argv[j]; 12 argv[j] = temp; 13 } 14 } 15 } 16 for(i = 1; i < argc; ++i) 17 printf("hello, %s\n", argv[i]); 18 19 return 0; 20 }
结果
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