ccs0216

实验五 
#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 = 0; i < n; ++i)
        if(x[i] < *pmin)
            *pmin = x[i];
        else if(x[i] > *pmax)
            *pmax = x[i];
}
task1.1

image

 

1.功能是找到一列数组中最大和最小值

2.x[0]的地址

#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 = 0; i < n; ++i)
        if(x[i] > x[max_index])
            max_index = i;
    
    return &x[max_index];
}
task1.2

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1.找到最大值的地址;返回最大值的地址

2.可以

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

int main() {
    char s1[N] = "Learning makes me happy";
    char s2[N] = "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);

    return 0;
}
task2.1

image

1.80;内存大小;s1长度

2.不能，s1是起始地址不能赋值

3.交换

#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;
}
task2.2

image

1.L的地址，str1占用的字节数，有效字符数

2.可以，2.1中是将地址赋值给了指针，2.2中是字符串赋值数组

3.交换地址；不交换

#include <stdio.h>

int main() {
    int x[2][4] = {{1, 9, 8, 4}, {2, 0, 4, 9}};
    int i, j;
    int *ptr1;     // 指针变量，存放int类型数据的地址
    int(*ptr2)[4]; // 指针变量，指向包含4个int元素的一维数组

    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: 使用指针变量ptr1(指向元素)间接访问\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: 使用指针变量ptr2(指向一维数组)间接访问\n");
    for (ptr2 = x; ptr2 < x + 2; ++ptr2) {
        for (j = 0; j < 4; ++j)
            printf("%d ", *(*ptr2 + j));
        printf("\n");
    }

    return 0;
}
task3

image

 1.指向数组指针

2.指向储存指针的数组

#include <stdio.h>
#define N 80

void replace(char *str, char old_char, char new_char); // 函数声明

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

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

// 函数定义
void replace(char *str, char old_char, char new_char) {
    int i;

    while(*str) {
        if(*str == old_char)
            *str = new_char;
        str++;
    }
}
task4

image

1.将old_char替换成new_char

2.可以

#include<stdio.h>
#define N 80

char*str_trunc(char *str,char x);

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

    while(printf("输入字符串:"),gets(str)!=NULL){
        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\n",str);
        getchar();
    }
    return 0;
}
char*str_trunc(char *str,char x){
    int i=0;
    while(*str!='\0'){
        if(*str==x){
            *str='\0';
        }
        str++;
        i++;
    }
}
task.5

image

 处理换行符

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

int check_id(char *str);

int main()
{
    char *pid[N] = {"31010120000721656X",
                    "3301061996X0203301",
                    "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 cnt=0;
    if(strlen(str)!=18)
        return 0;
    while(cnt<17){
        if(*str>='0'&&*str<='9'){
            cnt++;
            str++;
        }
        else
            return 0;
    }
    while(cnt=18){
        if(*str=='X'||(*str>='0'&&*str<='9'))
            return 1;
        else
            return 0;
    }
}
task.6

image

 

#include<stdio.h>
#define N 80
void encoder(char *str,int n);
void decoder(char *str,int n);

int main(){
    char words[N];
    int n;

    printf("请输入文本：");
    gets(words);

    printf("输入n:");
    scanf("%d",&n);

    printf("编码后的英文文本：");
    encoder(words,n);
    printf("%s\n",words);

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

    return 0;
}

void encoder(char *str,int n){
    while(*str!='\0'){
        if(*str>='a'&&*str<='z'){
            *str='a'+(*str+n-97)%26;
        }
        else if(*str>='A'&&*str<='Z'){
            *str='A'+(*str+n-65)%26;
        }
        str++;
    }
}

void decoder(char *str,int n){
    while(*str!='\0'){
        if(*str>='a'&&*str<='z'){
            if(*str-97-n>=0)
                *str='a'+*str-97-n;
            else
                *str='z'-(97+n-*str)%26+1;
        }
        else if(*str>='A'&&*str<='Z'){
            if(*str-65-n>=0)
                *str='A'+*str-65-n;
            else
                *str='Z'-(65+n-*str)%26+1;
        }
        str++;
    }
}
task.7

image

#include<stdio.h>
#include<string.h>
void com(int n,char *s[]);
int main(int argc,char *argv[]){
    int i;
    com(argc-1,argv+1);
    for(i=1;i<argc;++i)
        printf("hello,%s\n",argv[i]);

    return 0;
}

void com(int n,char *s[]){
    int i,j;
    char *temp;
    for(i=0;i<n-1;i++){
        for(j=0;j<n-1-i;j++){
            if(strcmp(s[j],s[j+1])>0){
                temp=s[j];
                s[j]=s[j+1];
                s[j+1]=temp;
            }
        }
    }
}
task.8

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posted on 2025-12-12 13:51  cc_sx  阅读(0)  评论(0)    收藏  举报