实验6 指针

TASK 1

2004

2001

存放的数据类型不同

#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;
}

 

TASK 2

2004，2006

2001，2004

因为p存储的是每一个元素的地址，而q所指向的是每一行元素的地址

#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;
} 

 

 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

 不能

 已交换

#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;
}

 

 

TASK 3

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号的个数，strlen统计的是字符串长度

不能

没有

#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;
}

 

 

#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;
}

 

 

TASK 4

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

 

 

TASK 5

#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 n,i,flag=1;
    
    n=strlen(s);
    for(i=0;i<n/2;i++)
    {    
        if(s[i]!=s[n-1-i])
            flag=0;
            break;
    }
    
    if(flag==1)return 1;
    if(flag==0)return 0;
} 

 

 

 

 

TASK 6

//task 6 
#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]-=25;
    }
}

void decoder(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]=='A'||s[i]=='a')
            s[i]+=25;
    }
}