试验任务1
task1.1
源码
#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;
}
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task1.2
源码
#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");
}
// 使用指针变量p间接访问二维数组元素
for (p = &x[0][0], i = 0; p < &x[0][0] + 8; ++p, ++i)
{
printf("%d", *p);
if ((i + 1) % 4 == 0)
printf("\n");
}
// 使用指针变量q间接访问二维数组元素
for (q = x; q < x + 2; ++q)
{
for (j = 0; j < 4; ++j)
printf("%d", *(*q + j));
printf("\n");
}
return 0;
}
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试验任务2
task2.1
源码
#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("\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;
}
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问题1：数组s1的大小是24；sizeof(s1)计算的是数组s1的大小；返回s1计数器值而不包含\0
问题2：不可以，
问题3：已经交换
task2.2
源码
#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("\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;
}
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试验任务3
源码
#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++)
;
}
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试验任务4
源码
#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;
}
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试验任务5
源码
#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';
}
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试验任务6
task6.1
源码
#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",
"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;
}
}
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task6.2
源码
#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",
"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;
}
}
}
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试验任务7
源码
#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]);
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;
};
}
}
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试验任务8
源码
#include <stdio.h>
#define N 80
#include<stdlib.h>
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);
system("pause");
return 0;
}
void encoder(char* s)
{
int i;
for (i = 0;*(s + i) != 0;i++)
{
if ((*(s + i) >= 65 && *(s + i) < 90) || (*(s + i) >= 97 && *(s + i) < 122))
{
*(s + i) += 1;continue;
}
if (*(s + i) == 90) { *(s + i) = 65;continue; }
if (*(s + i) == 122) { *(s + i) = 97;continue; }
}
}
void decoder(char* s)
{
int i;
for (i = 0;*(s + i) != 0;i++)
{
if (*(s + i) > 65 && *(s + i) <= 90)
{
*(s + i) -= 1;continue;
}
if (*(s + i) > 97 && *(s + i) <= 122)
{
*(s + i) -= 1;continue;
}
if (*(s + i) == 65) { *(s + i) = 90;continue; }
if (*(s + i) == 97)
{
*(s + i) = 122;continue;
}
}
}
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