gueal

 

实验5

实验任务1

代码:

#include<stdio.h>
#include<stdlib.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");

    //方式2:通过指针变量遍历输出数组元素(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(2)
    p=x;
    for(i=0;i<N;++i)
        printf("%d",*(p+i));
    printf("\n");

    system("pause");

    return 0;
}
#include<stdio.h>
#include<stdlib.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类型元素的一维数组

    //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");
    }

    system("pause");

    return 0;
}

 

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实验任务2

代码:

#include<stdio.h>
#include<stdlib.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);

    system("pause");

    return 0;
}
#include<stdio.h>
#include<stdlib.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);

    system("pause");

    return 0;
}

 

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答案:

1.数组s1的大小是24,sizeof计算的是大小包括'0',strlen计算的是长度
2.不能,不能像这样直接声明一个没有确定大小的数组。字符串赋值只能一位一位的。
3.交换了
task2_2:
1.指针s1存放的是字符数组的首地址,sizeof计算的一个指针的大小,strlen计算的是长度
2.能
3.交换的是指针,储存单元中的内容没有交换

实验任务3

代码:

#include<stdio.h>
#include<stdlib.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);

    system("pause");

    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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实验任务4

代码:

#include<stdio.h>
#include<stdlib.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");
    }

    system("pause");

    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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实验任务5

代码:

#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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实验任务6

代码:

#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;
            }
}
#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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答案:交换的是内存中字符串的储存位置

实验任务7

代码:

#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 len = strlen(str);
    int i;
    
    if(len != 18) return 0;
    
    for(i = 0;i < len; i++) {
        char x = *(str + i);
        if(i == 17) {
            if((x >= '0' && x <= '9') || x =='X') continue ;
            else return 0;
        }
        else {
            if(x >= '0' && x <= '9') continue ;
            else return 0;
        }
    }
    return 1;
}

 

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实验任务8

代码:

#include <stdio.h>
#include <string.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, len;
    
    len = strlen(s);
    
    for(i = 0;i < len; i++) {
        char x;
        x = *(s + i);
        if((x >= 'a' && x <= 'z') || (x >= 'A' && x <= 'Z')) {
            if(x == 'z') *(s + i) = 'a';
            else if(x == 'Z') *(s + i) = 'A';
            else *(s + i) = x + 1;
        }
    }
}

void decoder(char *s) {
    int i, len;
    
    len = strlen(s);
    
    for(i = 0;i < len; i++) {
        char x;
        x = *(s + i);
        if((x >= 'a' && x <= 'z') || (x >= 'A' && x <= 'Z')) {
            if(x == 'a') *(s + i) = 'z';
            else if(x == 'A') *(s + i) = 'Z';
            else *(s + i) = x - 1;
        }
    }
}

 

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posted on 2023-05-11 09:08  gueal  阅读(11)  评论(0编辑  收藏  举报

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