task 1.1
源代码
#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;
}
运行结果
task 1.2
源代码
# 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 (q = x; q<x+2; ++q){
for (j=0;j<4;++j)
printf("%d",*(*q+j));
printf("\n");
}
return 0;
}
运行结果
task 2
源代码
# 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("\nebefore 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;
}
运行结果
问题1：数组s1的大小为24;sizeo计算的是s1的数组长度包含结束字符而strlen不包含，strlen指的该字符的实际长度
问题2：不能 s1[]表示的是这串字符串数组的内容而s1是变量名称
问题3：交换了
task 2.2
源代码
# 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("\nebefore 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;
}
运行结果
问题1：s1存放的是字符串；sizeof计算的都是存放的所需字节，strlen统计的是该字符串实际的字符长度
问题2：不能；s1是指针变量的名称不能存放内容
问题3：交换的是数组内容；没有交换
task 3
源代码
# 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++);
}
实验结果
task 4
源代码
# 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;
}
运行结果
task 5
源代码
#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';
}
运行结果
task 6.1
源代码
#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",
"Date 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; ++j)
if(strcmp(name[j], name[j + 1]) > 0)
{
tmp = name[j];
name[j] = name[j + 1];
name[j + 1] = tmp;
}
}
运行结果
task 6.2
源代码
#include<stdio.h>
void sort(char *name[], int n);
int main()
{
char *course[4] = {"C Program",
"C++ Object Oriented Program",
"Operating System",
"Date 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;
}
}
}
运行结果
task 7
源代码
#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]);
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 flag = 0;
char *tmp;
tmp = str;
while((*tmp >= '0' && *tmp <= '9') || *tmp == 'X')
tmp++;
if(*tmp == '\0' && strlen(str) == 18)
flag = 1;
return flag;
}
运行结果
task 8
源代码
#include <stdio.h>
#define N 80
void encoder(char *s); // 函数声明
void decoder(char *s); // 函数声明
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); // 函数调用
printf("%s\n", words);
printf("对编码后的英文文本解码: ");
decoder(words); // 函数调用
printf("%s\n", words);
return 0;
}
/*函数定义
功能：对s指向的字符串进行编码处理
编码规则：
对于a~z或A~Z之间的字母字符，用其后的字符替换; 其中，z用a替换，Z用A替换
其它非字母字符，保持不变
*/
void encoder(char *s)
{
while(*s)
{
if((*s >= 'a' && *s < 'z') || (*s >= 'A' && *s < 'Z'))
*s = *s+1;
else if(*s == 'z' || *s == 'Z')
*s = *s-25;
s++;
}
}
/*函数定义
功能：对s指向的字符串进行解码处理
解码规则：
对于a~z或A~Z之间的字母字符，用其前面的字符替换; 其中，a用z替换，A用Z替换
其它非字母字符，保持不变
*/
void decoder(char *s)
{
while(*s)
{
if((*s > 'a' && *s <= 'z') || (*s > 'A' && *s <= 'Z'))
*s = *s-1;
else if(*s == 'a' || *s == 'A')
*s = *s+25;
s++;
}
}
运行结果
标签：int,s1,5yuan,char,++,实验,str,printf From： https://www.cnblogs.com/chenyihan/p/17372016.html