实验6

#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");

    return 0;
}

 

 

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

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

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

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

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

    return 0;
}

 

 1.task1_1执行++p后，p中存放地址为2004

2.task1_2执行++p之后，p中存放地址为2001

3.因为存放int型数据占用4个字节，char型占用一个字节

 

#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];

    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;
}

 

 

#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]; 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;
}

 

 task2_1

1.2004

2.2016

task2_2

1.2001

2.2004

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

 

#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;
}

 

 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

 

#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;
}

 

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

3.没有

#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,flag=1;
    if(strlen(str)==18)
    {
        for(i=0;i<18;i++)
        {
            if((str[i]=='X')||(str[i]>='0'&&str[i]<='9'))
                flag;
            else
                flag=0;
        }
    }
    else
        flag=0;
    return flag;
}

 

 

#define _CRT_SECURE_NO_WARNINGS
#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_s(str);
    flag = is_palindrome(str);
    if (flag)
        printf("YES\n");
    else
        printf("NO\n");
    return 0;
}
int is_palindrome(char* s)
{
    int i, j = 0;
    char a[N];
    for (i = strlen(s); i >0; i--)
    {
        a[j++] = s[i-1];
    }
    a[j] = '\0';
    if (strcmp(a, s) == 0)
        return 1;
    else
        return 0;
}

 

 

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