实验五

task1-1

typedef struct
{
char name[M]; // 书名
char author[M]; // 作者
}Book;
int main()
{
Book x[N] = { {"一九八四", "乔治.奥威尔"},
              {"美丽新世界", "赫胥黎"},
              {"昨日的世界", "斯蒂芬.茨威格"},
              {"万历十五年", "黄仁宇"},
              {"一只特立独行的猪", "王小波"}
            };
int i;
FILE *fp;
// 以写的方式打开文本文件data1.txt
fp = fopen("data1.txt", "w");
// 如果打开文件失败，输出提示信息并返回
if(fp == NULL)
{
printf("fail to open file\n");
return 1;
}
// 将结构体数组x中的图书信息写到fp指向的文件data1.txt
// 同时也输出到屏幕上
for(i=0; i<N; ++i)
{
fprintf(fp, "%-20s %-20s\n", x[i].name, x[i].author);
printf("%-20s %-20s\n", x[i].name, x[i].author);
}
fclose(fp);

return 0;
}

#include <stdio.h>
#define N 5
#define M 80
typedef struct
{
char name[M]; // 书名
char author[M]; // 作者
}Book;
int main()
{
Book x[N];
int i;
FILE *fp;
// 以读的方式打开文本文件data1.txt
fp = fopen("data1.txt", "r");
// 如果打开文件失败，输出提示信息并返回
if(fp == NULL)
{
printf("fail to open file\n");
return 1;
}
// 从fp指向的文件data1.txt中读取信息到结构体数组x
// 同时，把x的内容输出到屏幕上
for(i=0; i<N; ++i)
{
fscanf(fp, "%s %s\n", x[i].name, x[i].author);
printf("%-20s %-20s\n", x[i].name, x[i].author);
}
fclose(fp);
return 0;
}

问题：因为在字符串的数组中，数组名就代表数组的地址

 

task2-1

#include <stdio.h>
#define N 5
#define M 80
typedef struct
{
char name[M]; // 书名
char author[M]; // 作者
}Book;
int main()
{
Book x[N] = { {"一九八四", "乔治.奥威尔"},
            {"美丽新世界", "赫胥黎"},
            {"昨日的世界", "斯蒂芬.茨威格"},
            {"万历十五年", "黄仁宇"},
            {"一只特立独行的猪", "王小波"}
            };
int i;
FILE *fp;
// 以写的方式打开二进制文件data2.dat
fp = fopen("data2.dat", "wb");
// 如果打开文件失败，输出提示信息并返回
if(fp == NULL)
{
printf("fail to open file\n");
return 1;
}
// 将结构体数组x中的图书信息写以数据块方式写入文件
// 把从地址x处开始sizeof(Book)×N个字节大小的数据块写入fp指向的文件
fwrite(x, sizeof(Book), N, fp);
fclose(fp);
return 0;
}

 

#include <stdio.h>
#define N 5
#define M 80
typedef struct
{
char name[M]; // 书名
char author[M]; // 作者
}Book;
int main()
{
Book x[N];
int i;
FILE *fp;
// 以读的方式打开二进制文件data2.dat
fp = fopen("data2.dat", "rb");
// 如果打开文件失败，输出提示信息并返回
if(fp == NULL)
{
printf("fail to open file\n");
return 1;
}
// 从fp指向的文件中读取数据块到x对应的地址单元
// 数据块大小为sizeof(Book)×N
fread(x, sizeof(Book), N, fp);
// 在屏幕上输出结构体数组x中保存的数据
for(i=0; i<N; ++i)
printf("%-20s%-20s\n", x[i].name, x[i].author);
fclose(fp);
return 0;
}

task3-2

#include <stdio.h>
int main()
{
FILE *fin, *fout;
char ch;
// 以只读、文本方式打开文件data3_1.txt
fin = fopen("data3_1.txt", "r");
// 如果打开失败，输出提示信息并返回
if(fin == NULL)
{
printf("fail to open data3_1.txt\n");
return 1;
}

if(fout == NULL)
{
printf("fail to open data3_2.txt\n");
return 1;
}
int a=0;
// 当fin指向的文件data1_txt没有结束时
while( !feof(fin) )
{
// 从fin指向的文件data1_txt读取单个字符
ch = fgetc(fin);
// 如果ch是小写字母，转换成大写
if((ch >= 'a' && ch <= 'z')||(ch>=48&&ch<=57)||ch==45)
a++;
}
fclose(fin);
printf("%d",a);
return 0;
}

 task5

#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

#define N 10

typedef struct
{
    long int id;
    char name[20];
    float objective;
    float subjective;
    float sum;
    char level[10];
}STU;

void input(STU s[], int n);
void output(STU s[], int n);
void process(STU s[], int n);

int main()
{
    STU stu[N];

    printf("从文件读入%d个考生信息：准考证号，姓名，客观题得分（<=40），操作题得分（<=60）\n", N);
    input(stu, N);

    printf("\n对考生信息进行处理：计算总分，确定等级\n");
    process(stu, N);

    printf("\n打印考生完整信息，并保存到文件中");
    output(stu, N);

    return 0;
}

void input(STU s[], int n)
{
    int i;
    FILE* fin;

    fin = fopen("examinee.txt", "r");
    if (fin == NULL)
    {
        printf("fail to open file\n");
        exit(0);
    }

    while (!feof(fin))
    {
        for (i = 0; i < n; i++)
            fscanf(fin, "%ld %s %f %f", &s[i].id, s[i].name, &s[i].objective, &s[i].subjective);
    }

    fclose(fin);
}

void output(STU s[], int n)
{
    FILE* fout;
    int i;

    printf("\n");
    printf("准考证号\t姓名\t客观题得分\t操作题得分\t总分\t\t等级\n");
    for (i = 0; i < n; i++)
        printf("%ld\t\t%s\t%.2f\t\t%.2f\t\t%.2f\t\t%s\n", s[i].id, s[i].name, s[i].objective, s[i].subjective, s[i].sum, s[i].level);

    fout = fopen("result.txt", "w");
    if (!fout)
    {
        printf("fail to open or create result.txt\n");
        exit(0);
    }
    fprintf(fout, "准考证号\t\t姓名\t客观题得分\t操作题得分\t总分\t\t等级\n");
    for (i = 0; i < n; i++)
        fprintf(fout, "%ld\t\t%s\t%.2f\t\t%.2f\t\t%.2f\t\t%s\n", s[i].id,
            s[i].name, s[i].objective, s[i].subjective, s[i].sum, s[i].level);


    fclose(fout);
}

void process(STU s[], int n)
{
    int i, j;
    for (i = 0; i < n; i++)
        s[i].sum = s[i].objective  + s[i].subjective ;

    STU temp;
    for (i = 0; i < n; i++)
        for (j = 0; j < n - 1 - i; j++)
            if (s[j].sum < s[j + 1].sum)
            {
                temp = s[j];
                s[j] = s[j + 1];
                s[j + 1] = temp;
            }
    for (i = 0; i < n; i++)
    {
        if ((i + 1) <= n * 0.1)
            strcpy(s[i].level, "优秀");
        else if ((i + 1) > n * 0.1 && (i + 1) <= n * 0.5)
            strcpy(s[i].level, "合格");
        else
            strcpy(s[i].level, "不合格");
    }
}

task6

int main()
{
    int i,j;
    STU x[N], lucky[M];
    FILE* fin,* fout;
    fin = fopen("list.txt", "r");
    if (fin == NULL)
    {
        printf("fail to open file\n");
        exit(0);
    }
    for (i = 0; i < N; i++)
    {
        fscanf(fin, "%ld %s %s", &x[i].no, x[i].name, x[i].bj);
    }
    fclose(fin);
    srand(time(0));
    for (i = 0; i < M; i++)
    {
        j = rand() % 80;
        lucky[i] = x[j];
    }
    fout = fopen("lucky.txt", "w");
    if (fout == NULL)
    {
        printf("fail to open file\n");
        return 1;
    }
    for (i = 0; i < M; i++)
    {
        fprintf(fout, "%ld %s %s\n", lucky[i].no, lucky[i].name, lucky[i].bj);
        printf("%ld %s %s\n", lucky[i].no, lucky[i].name, lucky[i].bj);
    }
    fclose(fout);
    return 0;
}