实验五

1.1

#include <stdio.h>
#define N 4
int main()
{
int x[N] = {1, 9, 8, 4};
int i;
int *p;
// 方式1：通过数组名和下标遍历输出数组元素
for (i = 0; i < N; ++i)
printf("%d", x[i]);
printf("\n");
// 方式2：通过指针变量遍历输出数组元素 （写法1）
for (p = x; p < x + N; ++p)
printf("%d", *p);
printf("\n");
// 方式2：通过指针变量遍历输出数组元素（写法2）
p = x;
for (i = 0; i < N; ++i)
printf("%d", *(p + i));
printf("\n");
// 方式2：通过指针变量遍历输出数组元素（写法3）
p = x;
for (i = 0; i < N; ++i)
printf("%d", p[i]);
printf("\n");
return 0;
}

1.2

#include <stdio.h>
int main()
{
int x[2][4] = {{1, 9, 8, 4}, {2, 0, 4, 9}};
int i, j;
int *p; // 指针变量，存放int类型数据的地址
int(*q)[4]; // 指针变量，指向包含4个int型元素的一维数组
// 使用数组名、下标访问二维数组元素
for (i = 0; i < 2; ++i)
{
for (j = 0; j < 4; ++j)
printf("%d", x[i][j]);
printf("\n");
} 
for (p = &x[0][0], i = 0; p < &x[0][0] + 8; ++p, ++i)
{
printf("%d", *p);
if ((i + 1) % 4 == 0)
printf("\n");
}

for (q = x; q < x + 2; ++q)
{
for (j = 0; j < 4; ++j)
printf("%d", *(*q + j));
printf("\n");
} 
return 0;
}

2.1

#include <stdio.h>
#include <string.h>
#define N 80
int main()
{
char s1[] = "Learning makes me happy";
char s2[] = "Learning makes me sleepy";
char tmp[N];
printf("sizeof(s1) vs. strlen(s1): \n");
printf("sizeof(s1) = %d\n", sizeof(s1));
printf("strlen(s1) = %d\n", strlen(s1));
printf("\nbefore swap: \n");
printf("s1: %s\n", s1);
printf("s2: %s\n", s2);
printf("\nswapping...\n");
strcpy(tmp, s1);
strcpy(s1, s2);
strcpy(s2, tmp);
printf("\nafter swap: \n");
printf("s1: %s\n", s1);
printf("s2: %s\n", s2);

return 0;
}

问题1：24

问题2：不能

问题3：s1与s2互换

2.2

#include <stdio.h>
#include <string.h>
#define N 80
int main()
{
char *s1 = "Learning makes me happy";
char *s2 = "Learning makes me sleepy";
char *tmp;
printf("sizeof(s1) vs. strlen(s1): \n");
printf("sizeof(s1) = %d\n", sizeof(s1));
printf("strlen(s1) = %d\n", strlen(s1));
printf("\nbefore swap: \n");
printf("s1: %s\n", s1);
printf("s2: %s\n", s2);
printf("\nswapping...\n");
tmp = s1;
s1 = s2;
s2 = tmp;
printf("\nafter swap: \n");
printf("s1: %s\n", s1);
printf("s2: %s\n", s2);
return 0;
}

问题1：字符串的起始地址，第一个单词的长度，字符串的字符数

问题2：可以

问题3：交换的是地址

 

3

#include <stdio.h>
void str_cpy(char *target, const char *source);
void str_cat(char *str1, char *str2);
int main()
{
char s1[80], s2[20] = "1984";
str_cpy(s1, s2);
puts(s1);
str_cat(s1, " Animal Farm");
puts(s1);
return 0;
} 
void str_cpy(char *target, const char *source)
{
while (*target++ = *source++)
;
} 
void str_cat(char *str1, char *str2)
{
while (*str1)
str1++;
while (*str1++ = *str2++)
;
}

4.1

#include <stdio.h>
#define N 80
int func(char *);
int main()
{
char str[80];
while (gets(str) != NULL)
{
if (func(str))
printf("yes\n");
else
printf("no\n");
} 
return 0;
} 
int func(char *str)
{
char *begin, *end;
begin = end = str;
while (*end)
end++;
end--;
while (begin < end)
{
if (*begin != *end)
return 0;
else
{
begin++;
end--;
}
} 
return 1;
}

5.1

#include <stdio.h>

#define N 80
void func(char *);
int main()
{
char s[N];
while (scanf("%s", s) != EOF)
{
func(s);
puts(s);
} 
return 0;
} 
void func(char *str)
{
int i;
char *p1, *p2, *p;
p1 = str;
while (*p1 == '*')
p1++;
p2 = str;
while (*p2)
p2++;
p2--;
while (*p2 == '*')
p2--;
p = str;
i = 0;
while (p < p1)
{
str[i] = *p;
p++;
i++;
} 
while (p <= p2)
{
if (*p != '*')
{
str[i] = *p;
i++;
} p
++;
} 
while (*p != '\0')
{
str[i] = *p;
p++;
i++;
} 
str[i] = '\0';
}

6.1

#include <stdio.h>
#include <string.h>
void sort(char *name[], int n);
int main()
{
char *course[4] = {"C Program",
"C++ Object Oriented Program",
"Operating System",
"Data Structure and Algorithms"};
int i;
sort(course, 4);
for (i = 0; i < 4; i++)
printf("%s\n", course[i]);
return 0;
} 
void sort(char *name[], int n)
{
int i, j;
char *tmp;
for (i = 0; i < n - 1; ++i)
for (j = 0; j < n - 1 - i; ++j)
if (strcmp(name[j], name[j + 1]) > 0)
{
tmp = name[j];
name[j] = name[j + 1];
name[j + 1] = tmp;
}
}

6.2

#include <stdio.h>
#include <string.h>
void sort(char *name[], int n);
int main()
{
char *course[4] = {"C Program",
"C++ Object Oriented Program",
"Operating System",
"Data Structure and Algorithms"};
int i;
sort(course, 4);
for (i = 0; i < 4; i++)
printf("%s\n", course[i]);
return 0;
} 
void sort(char *name[], int n)
{
int i, j, k;
char *tmp;
for (i = 0; i < n - 1; i++)
{
k = i;
for (j = i + 1; j < n; j++)
if (strcmp(name[j], name[k]) < 0)
k = j;
if (k != i)
{
tmp = name[i];
name[i] = name[k];
name[k] = tmp;
}
}
}

7.1

 

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define N 5

int check_id(char *str);

int main(){
    char *pid[N]={"31010120000721656X",
                  "330106199609203301",
                  "53010220051126571",
                  "510104199211197977",
                  "53010220051126133Y"};

    int i;

    for(i=0;i<N;++i)
        if(check_id(pid[i]))
            printf("%s\tTure\n",pid[i]);
        else
            printf("%s\tFalse\n",pid[i]);

    system("pause");
    return 0;

}
        

int check_id(char *str){
int n,j,k;
char bj[]="1234567890X";
for(n=0;(str[n])!='\0';n++)
    ;
j=0;
for(k=0;k<=10;k++){
if(str[n-1]==bj[k])
    j=1;
}
if((j==1)&&(n==18))
    return 1;
else 
    return 0;
}

8

#include <stdio.h>
#define N 80
void encoder(char *s);
void decoder(char *s);

int main()
{
    char words[N];
    
    printf("输入英文文本：");
    gets(words);
    
    printf("编码后的英文文本: ");
    encoder(words);
    printf("%s\n", words);
    
    printf("对编码后的英文文本解码：");
    decoder(words);
    printf("%s\n", words);
    
    return 0;
    }
void encoder(char *s)
{   while(*s){
       if(*s>=65&&*s<=90||*s>=97&&*s<=122){
              *s=*s+1;
              s++;continue;
       }
       else if(*s==90||*s==122){
              *s=*s-25;
              s++;continue;
       }
       s++;
}
 return;
    
}

void decoder(char *s)
{ while(*s){
       if(*s>=66&&*s<=90||*s>=98&&*s<=122){
              *s=*s-1;
              s++;continue;
       }
       else if(*s==65||*s==97){
              *s=*s+25;
              s++;continue;
       }
       s++;
}    return;
}