实验五

task1.1

#include <stdio.h>
#define N 5

void input(int x[], int n);
void output(int x[], int n);
void find_min_max(int x[], int n, int *pmin, int *pmax);

int main() {
    int a[N];
    int min, max;

    printf("录入%d个数据:\n", N);
    input(a, N);

    printf("数据是: \n");
    output(a, N);

    printf("数据处理...\n");
    find_min_max(a, N, &min, &max);

    printf("输出结果:\n");
    printf("min = %d, max = %d\n", min, max);

    return 0;
}

void input(int x[], int n) {
    int i;

    for(i = 0; i < n; ++i)
        scanf("%d", &x[i]);
}

void output(int x[], int n) {
    int i;
    
    for(i = 0; i < n; ++i)
        printf("%d ", x[i]);
    printf("\n");
}

void find_min_max(int x[], int n, int *pmin, int *pmax) {
    int i;
    
    *pmin = *pmax = x[0];

    for(i = 1; i < n; ++i)
        if(x[i] < *pmin)
            *pmin = x[i];
        else if(x[i] > *pmax)
            *pmax = x[i];
}

Q1:找出输入数据中的最大值和最小值

Q2:指针变量pmin,pmax都指向x[0]

task1.2

#include <stdio.h>
#define N 5

void input(int x[], int n);
void output(int x[], int n);
int *find_max(int x[], int n);

int main() {
    int a[N];
    int *pmax;

    printf("录入%d个数据:\n", N);
    input(a, N);

    printf("数据是: \n");
    output(a, N);

    printf("数据处理...\n");
    pmax = find_max(a, N);

    printf("输出结果:\n");
    printf("max = %d\n", *pmax);

    return 0;
}

void input(int x[], int n) {
    int i;

    for(i = 0; i < n; ++i)
        scanf("%d", &x[i]);
}

void output(int x[], int n) {
    int i;
    
    for(i = 0; i < n; ++i)
        printf("%d ", x[i]);
    printf("\n");
}

int *find_max(int x[], int n) {
    int max_index = 0;
    int i;

    for(i = 1; i < n; ++i)
        if(x[i] > x[max_index])
            max_index = i;
    
    return &x[max_index];
}

Q:返回最大数据的地址

Q:ok

task2.1

#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[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("\nswaping...\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.数组的大小是24字节,计算s1的字节长度,统计s1去掉空字符后的字节长度

2.不能,不可以对数组名进行赋值操作

3.交换

task2.2

#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);

    return 0;
}

1.s1中存放字符串的初始地址,计算指针变量s1占用的字节数,统计s1指向字符串的实际长度

2.s1是指针变量,可以进行赋值,该语义是将字符串的初始地址复制给s1

3.没有交换

task3

#include<stdio.h>
int main()
{
    int x[2][4]={{1,9,8,4},{2,0,4,9}};
    int i,j;
    int *ptr1;
    int(*ptr2)[4];//指向int型一维数组

    printf("输出1:使用数组名、下标直接访问二维数组元素\n");
    for(i=0;i<2;++i)
    {
        for(j=0;j<4;++j)
            printf("%d",x[i][j]);
        printf("\n");
    }

    printf("\n输出2:使用指向元素的指针变量ptr1间接访问二维数组元素\n");
    for(ptr1=&x[0][0],i=0;ptr1<&x[0][0]+8;++ptr1,++i)
    {
        printf("%d",*ptr1);

        if((i+1)%4==0)
            printf("\n");
    }

    printf("\n输出3:使用指向一维数组的指针变量ptr2间接访问二维数组元素\n");
    for(i=0,ptr2=x;i<2;++i,++ptr2)
    {
        for(j=0;j<4;++j)
                printf("%d",*(*ptr2+j));
        printf("\n");
    }

    return 0;
}

一个是指针变量一个是数组

task4.1

#include<stdio.h>
#define N 80

void replace(char *str,char old_char,char new_char);

int main()
{
    char text[N]="c programming is difficult or not,it is a question.";


    printf("原始文本:\n");
    printf("%s",text);

    replace(text,'i','*');//函数调用,注意字符形参写法,单引号不能少

    printf("处理后文本:\n");
    printf("%s\n",text);

    return 0;
}
void replace(char *str,char old_char,char new_char)
{
    while(*str)
    {
        if(*str==old_char)
            *str=new_char;
        *str++;
    }
}

1.新字符替换旧字符

2.可以改写成*str!='\0'

task4.2

#include <stdio.h>
#define N 80

void str_trunc(char *str, char x);

int main() {
    char str[N];
    char ch;

    printf("输入字符串: ");
    gets(str);

    printf("输入一个字符: ");
    ch = getchar();

    printf("截断处理...\n");
    str_trunc(str, ch);

    printf("截断处理后的字符串: %s\n", str);

}

void str_trunc(char *str, char x) {
    while(*str) {
        if(*str == x)
            *str='\0';     // blank1
             str++;   // blank2
    }
        // blank3
}

task5.1

#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;
            }
}

task5.2

#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;
        }
    }
}

内存中存储字符串的位置发生了交换

task6

#include<stdio.h>
#include<string.h>
#define N 5

int check_id(char *str);

int main()
{
    char *pid[N]={"31010120000721656X",
                    "330106199609203301",
                    "53010220051126571",
                    "510104199X11197977",
                    "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 k=0;

    while(*str)
        {
            str++;
            k++;
    }

    if(k!=18)
        return 0;

    str=str-k;

    for(k=0;k<18;k++)
    {
        if(k==17)
        {
            if(*(str+k)!='X'&&(*(str+k)<'0'||*(str+k)>'9'))
                 return 0;
        }
        else if(*(str+k)<'0'||*(str+k)>'9')
                return 0;
    }

}

task7

#include <stdio.h>
#define N 80
void encoder(char *str); // 函数声明
void decoder(char *str); // 函数声明

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); // 函数调用
    printf("%s\n", words);

    printf("对编码后的英文文本解码: ");
    decoder(words); // 函数调用
    printf("%s\n", words);

    return 0;
}

/*函数定义
功能:对s指向的字符串进行编码处理
编码规则:
对于a~z或A~Z之间的字母字符,用其后的字符替换; 其中,z用a替换,Z用A替换
其它非字母字符,保持不变
*/
void encoder(char *str) {
      int i=0;
      int len;
      len=strlen(str);
      for(;i<len;++i){
      if(str[i]>='a'&&str[i]<='z')
      str[i]=str[i]+1; 
      else if(str[i]>='A'&&str[i]<='Z')
      str[i]=str[i]+1;
      }
}

/*函数定义
功能:对s指向的字符串进行解码处理
解码规则:
对于a~z或A~Z之间的字母字符,用其前面的字符替换; 其中,a用z替换,A用Z替换
其它非字母字符,保持不变
*/
void decoder(char *str) {
    int i=0;
    int len;
      len=strlen(str);
      for(;i<len;++i){
      if(str[i]>='a'&&str[i]<='z')
      str[i]=str[i]-1; 
      else if(str[i]>='A'&&str[i]<='Z')
      str[i]=str[i]-1;
      }
    
}

 

posted @ 2024-05-27 12:58  中年男士  阅读(10)  评论(0)    收藏  举报