第五次上机作业

task 1_1

#include <stdio.h>
#include<stdlib.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");
system("pause");
return 0;
}

 

task 1_2

#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;
}

 

task 2_1

#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;
}

task 2_2

#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);
    
    system("pause");
    return 0;
}

 2_1

(1)s1的大小为24,sizeof(s1)计算的是字符串的内存大小包含'\0',而strlen(s1)统计的是s1的长度

(2)不能,数组不能直接进行赋值,要借助strcpy;

(3)交换了

2_2

(1)s1中存放的是字符串。

    sizeof(s1)计算的是指针所占的地址空间。

    strlen(s1)统计的是字符串的长度。

(2)可以替换成该写法。

(3)交换的是指针的指向,没有交换。

 

task 3

#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++);
}

task 4

#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;
}

task 5

#include <stdio.h>
#include<stdlib.h>
#define N 80

void func(char *);

int main()
{
    char s[N];
     
    while (scanf("%s", s) != EOF)
    {
        func(s);
        puts(s);
    }
    
    system("pause");
    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';
}

 

task 6_1

#include<stdio.h>
#include<stdlib.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]);
    
    system("pause");
    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;
            }
}

 task 6_2

#include <stdio.h>
#include <string.h>
#include<stdlib.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]);
    system("pause");  
    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;
        }
    }
}

 回答:交换的是指针变量的值。

task 7

#include <stdio.h>
#include <string.h>
#include<stdlib.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]);
    
    system("pause");
    return 0;
}

int check_id(char *str)
{
    char *p;
    int i;
 
    p = str;
 
    while((*p >= '0' && *p <= '9')||*p=='X')
            p++;
 
    if(*p == '\0' && strlen(str) == 18)
        return 1;
    else
        return 0;
}

task 8

#include <stdio.h>
#include<stdlib.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);
    
    system("pause");
    return 0;
}

void encoder (char *s)
{
    int i;
    for(i=0;s[i]!='\0';i++)
    {
        if(s[i]=='Z')
            s[i]='A';
        if(s[i]=='z')
            s[i]='a';    
        if(('A'<=s[i]&&s[i]<'Z')||('a'<=s[i]&&s[i]<'z'))
            s[i]=s[i]+1;
    }
}

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

 

posted @ 2023-05-10 23:44  chen,,  阅读(15)  评论(0编辑  收藏  举报