实验六

task1_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");
	return 0;
}

问题回答：
1.执行++p后，指针变量p中存放的地址值是2004；
2.执行++p后，指针变量p中存放的地址值是2001；
3.指针变量存放的数据类型不同

task1_2

#include <stdio.h>
#define N 4
int main()
{
	char x[N] = { '1', '9', '8', '4' };
	int i;
	char* p;
	for (p = x; p < x + N; ++p)
		printf("%c", *p);
	printf("\n");
	return 0;
}

task2_1

#include <stdio.h>
int main()
{
	int x[2][4] = { {1,9,8,4}, {2,0,2,2}} ;
	int i, j;
	int *p;
	int (*q)[4];
// 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");
}

task2_2

#include<stdio.h>
int main()
{
	char x[2][4] = { {'1','9','8','4'},{'2','0','2','2'} };
	int i, j;
	char* p;
	char(*q)[4];
	for (i = 0;i<2;++i)
	{
		for (j = 0;j < 4;++j)
			printf("%c", x[i][j]);
		printf("\n");
}
	for (p = &x[0][0], i = 0;p < &x[0][0] + 8;++p, ++i)
	{
		printf("%c", *p);
		if ((i + 1) % 4 == 0)
			printf("\n");
	}
	for (q = x; q < x + 2; ++q)
	{
		for (j = 0; j < 4; ++j)
			printf("%c", *(*q + j));
		printf("\n");
	}
	return 0;
}
	for(q=x; q<x+2; ++q)
	{
		for(j=0; j<4; ++j)
			printf("%d", *(*q+j));
		printf("\n");
	}
	return 0;
}

问题回答：
1.2004；
2016；
2.2001；
2004
因为p存储的是每一个元素的地址，q所指向的是每一行元素的地址

task3_1

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <string.h>
#define N 80
int main()
{
	char s1[] = "C,I love u.";
	char s2[] = "C, I hate u.";
	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;
}

问题回答：
1.数组s1的大小是13，sizeof(s1)计算的是s1占用空间的大小，strlen(s1)统计的是s1字符串长度
2.不能替换
3.交换了

task3_2

#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<string.h>
#define N 80
int main()
{
	char *s1 = "C, I love u.";
	char *s2 = "C, I hate u.";
	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;
}

问题回答：
1.s1存放的是“C，I love you.”的首地址，sizeof计算的是所有字母以及结束符的个数，strlen统计的是字符串长度；
2.不能替换；
3.没有交换

task4

#include <stdio.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\tTrue\n", pid[i]);
		else
			printf("%s\tFalse\n", pid[i]);

	return 0;
}
int check_id(char* str)
{
	int i, j;
	char* temp;
	for (i = 1;i <= N;++i)
	{
		if (strlen(str) != 18)
		{
        		return 0;
        		break;
		}
    		else
    		{
        		temp = str;
        		for (;temp < str + 18;temp++)
        		{
            			if (*temp >= '0' && *temp <= '9' || *temp == 'X');
            			else
            			{
                			return 0;
                			break;
            			}
        		}
        		if (temp = str + 18)
            			return 1;
    		}
	}
}

task5

#include <stdio.h>
#include <string.h>
#define N 80
int is_palindrome(char* s); 
int main()
{
	char str[N];
	int flag;
	printf("Enter a string:\n");
	gets(str);
	flag = is_palindrome(str); 
	if (flag)
		printf("YES\n");
	else
		printf("NO\n");
	return 0;
}
int is_palindrome(char* s)
{
		int i, j;
		for (j = strlen(s) - 1, i = 0;i < j;i++, j--)
			if (s[i] == s[j])
				return 1;
			else
				return 0;
}

task6

#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)
{
	int i;
	for (i = 0;i < N;i++)
	{
		if (s[i] >= 'a' && s[i] < 'z' || s[i] >= 'A' && s[i] < 'Z')
			s[i] += 1;
		else if (s[i] == 'z' || s[i] == 'Z')
			s[i] = s[i] - 25;
	}
}
void decoder(char* s)
{
	int i;
	for (i = 0;i < N;i++)
	{
		if (s[i] == 'A' || s[i] == 'a')
			s[i] = s[i] + 25;
		else if (s[i] > 'A' && s[i] <= 'Z' || s[i] > 'a' && s[i] <= 'z')
			s[i] = s[i] - 1;
	}

}