实验5
实验1
源代码1
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 27 for(i = 0; i < n; ++i) 28 printf("%d ", x[i]); 29 printf("\n"); 30 } 31 void find_min_max(int x[], int n, int *pmin, int *pmax) { 32 int i; 33 34 *pmin = *pmax = x[0]; 35 for(i = 0; i < n; ++i) 36 if(x[i] < *pmin) 37 *pmin = x[i]; 38 else if(x[i] > *pmax) 39 *pmax = x[i]; 40 }
问题1：用指针指向最大数和最小数的地址 实现输出
问题2：x[0]
源代码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 }
问题1：返回最大数的索引
问题2：可以
实验2
源代码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 }
运行结果
问题1：80，数组字符数，实际字符数
问题2：不能
问题3：可以
源代码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 }
运行结果
问题1：Learning makes me happy的地址 地址所占字符数 表示的字符串的字符数
问题2：可以
问题3：地址，没有
实验3
源代码
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 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 }
运行结果
问题1：二维数组的第一行即二维数组x的x[0]元素的地址
问题2：int *ptr[4];表示一个含有四个元素的指针数组 存放的都是地址
实验4
源代码
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] = "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 }
运行结果
问题1：replace的功能是将句子中的i换成*
问题2：可以
实验5
源代码
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 while(*p!='\0'){ 27 if(*p==x){ 28 *p='\0'; 29 break; 30 } 31 p++; 32 } 33 return str; 34 }
运行结果
问题：第二次的字符串自动输入为换行符，吸收输入一个字符后的换行符
实验6
源代码
1 #include <stdio.h> 2 #include <string.h> 3 #define N 5 4 5 int check_id(char *str); 6 7 int main() 8 { 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])) 18 printf("%s\tTrue\n", pid[i]); 19 else 20 printf("%s\tFalse\n", pid[i]); 21 22 return 0; 23 } 24 25 int check_id(char *str) { 26 int i; 27 28 for(i=0;i<17;++i){ 29 if(*(str+i)>'9'||*(str+i)<'0') 30 return 0; 31 } 32 if(*(str+17)>'9'||*(str+17)<'0') 33 if(*(str+17)!='X') 34 return 0; 35 if(*(str+18)!='\0') 36 return 0; 37 return 1; 38 }
运行结果
实验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 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 return 0; 24 } 25 26 void encoder(char *str,int n){ 27 while(*str){ 28 if(*str>='a'&&*str<='z'){ 29 *str='a'+((*str-'a')+n)%26; 30 } 31 else if(*str>='A'&&*str<='Z'){ 32 *str='A'+((*str-'A')+n)%26; 33 } 34 str++; 35 } 36 } 37 void decoder(char *str,int n){ 38 while(*str){ 39 if(*str>='a'&&*str<='z'){ 40 *str='a'+((*str-'a')-n+26)%26; 41 } 42 else if(*str>='A'&&*str<='Z'){ 43 *str='A'+((*str-'A')-n+26)%26; 44 } 45 str++; 46 } 47 }
运行结果
实验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 *tmp; 7 8 for(i = 1; i < argc; ++i){ 9 for(j = 1; j < argc-i; ++j){ 10 if(strcmp(argv[j],argv[j+1])>0){ 11 tmp=argv[j]; 12 argv[j]=argv[j+1]; 13 argv[j+1]=tmp; 14 } 15 } 16 } 17 18 for(i = 1; i < argc; ++i) 19 printf("hello, %s\n", argv[i]); 20 21 return 0; 22 }
运行结果
