任务一
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;
int(*q)[4];
for (i = 0;i<N;++i)
{
for (j = 0;j<4;++i)
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;
}
任务二
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;
}
S1大小是24，sizeof统计的是占用内存大小，包含休止符，strlen统计的是字符串长度，不包含休止符。
不能，等号两边类型不同
对
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;
}
sizeof计算的是内存大小，strlen计算的是字符串长度。
可以，1是地址常量，2是指针变量。
交换s1和s2的指向关系，储存内容不变。
任务三
#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++)
;
}
任务四
#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;
}
任务五
#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';
}
任务六
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;
char *tmp;
for (i = 0; i < n - 1; ++i)
{
k=i;
for (j = i +1; j < n;j++)
if (strcmp(name[j], name[j + 1]) > 0)
k=j;
if (k!=i)
{
tmp = name[j];
name[j] = name[j + 1];
name[j + 1] = tmp;
}
}
}
变换的是指针变量，内存不变
任务七
#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 i, j, k;
char* tmp;
tmp=str;
while ((*tmp <='9' && *tmp >= '0') || *tmp == 'X')
tmp++;
if (*tmp == '\0' && strlen(str) == 18)
return 1;
else
return 0;
}
任务八
#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)
{
char* tmp;
tmp = s;
int i = 0;
while (*tmp) {
if (*tmp == 'z')
s[i++] = 'a', tmp++;
else if (*tmp == 'Z')
s[i++] = 'A', tmp++;
else if (*tmp < 'A' || *tmp>'z')
s[i] = s[i++], tmp++;
else
s[i++] = *tmp + 1,tmp++;
}
}
void decoder(char* s) {
char* tmp;
tmp = s;
int i = 0;
while (*tmp) {
if (*tmp == 'a')
s[i++] = 'z', tmp++;
else if (*tmp == 'A')
s[i++] = 'Z', tmp++;
else if (*tmp < 'A' || *tmp>'z')
s[i] = s[i++], tmp++;
else
s[i++] = *tmp - 1, tmp++;
}
}
标签：tmp,int,s1,char,++,实验,printf From： https://www.cnblogs.com/xsh123456/p/17371914.html