实验5

实验任务1

#include <stdio.h>
#define N 4
int main()
{
int x[N] = {1, 9, 8, 4};
int i;
int *p;

for (i = 0; i < N; ++i)
printf("%d", x[i]);
printf("\n");


for (p = x; p < x + N; ++p)
printf("%d", *p);
printf("\n");


p = x;
for (i = 0; i < N; ++i)
printf("%d", *(p + i));
printf("\n");


p = x;
for (i = 0; i < N; ++i)
printf("%d", p[i]);
printf("\n");

system("pause");
return 0;
}

#include <stdio.h>

int main()
{

int x[2][4] = {{1, 9, 8, 4}, {2, 0, 4, 9}};
int i, j;
int *p; 
int(*q)[4]; 

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 (j = 0; j < 4; ++j)
printf("%d", *(*q + j));
printf("\n");
}

system("pause");
return 0;
}

 

 

实验任务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[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;
}

#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);

system("pause");
return 0;
}

 

问题：

task2_1
1.s1的大小是 24 ，sizeof 计算的包括'0'，strlen计算的是长度。
2.不能。因为没有确定大小。字符串赋值只能一位一位赋。
3.交换了。
task2_2
1.指针s1存放的是首地址，sizeof计算的一个指针的大小，strlen计算的是长度。
2.能。
3.交换的是指针，内容没有交换。

实验任务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);

system("pause");
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++);

}

 

实验任务4

#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;
}

 

实验任务5

#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';
}

 

实验任务6

#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]);
system("pause");
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;
}
}

#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]);
system("pause");
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;
}
}
}

 交换的是字符串的储存位置。

实验任务7

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

int check_id(char *str); // 函数声明

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

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

    system("pause");
    return 0;
}

// 函数定义
// 功能: 检查指针str指向的身份证号码串形式上是否合法。
// 形式合法，返回1，否则，返回0
int check_id(char *str)
{
    int j;
    char x;

    if( strlen(str) != 18)
        return 0;
    else
        {
            for(j=0;j<18;j++)
        {
            x = *(str+j);
            if((x >= '0'&& x<='9')|| x == 'X')
                continue;
            else
                return 0;
            
    };
}
}

 

 

实验任务8

#include <stdio.h>
#include <string.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);
    system("pause");
    printf("%s\n", words);
    
    return 0;
}

void encoder(char *s) {
    int i, len;
    
    len = strlen(s);
    
    for(i = 0;i < len; i++) {
        char x;
        x = *(s + i);
        if((x >= 'a' && x <= 'z') || (x >= 'A' && x <= 'Z')) {
            if(x == 'z') *(s + i) = 'a';
            else if(x == 'Z') *(s + i) = 'A';
            else *(s + i) = x + 1;
        }
    }
}

void decoder(char *s) {
    int i, len;
    
    len = strlen(s);
    
    for(i = 0;i < len; i++) {
        char x;
        x = *(s + i);
        if((x >= 'a' && x <= 'z') || (x >= 'A' && x <= 'Z')) {
            if(x == 'a') *(s + i) = 'z';
            else if(x == 'A') *(s + i) = 'Z';
            else *(s + i) = x - 1;
        }
    }
}