338 lines
10 KiB
Markdown
338 lines
10 KiB
Markdown
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# 实验01 Linux I/O 编程
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## 实验目的
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1. 练习 UNIX I/O 函数（`open`、`close`、`read`、`write`、`lseek`）的使用
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2. 掌握标准 I/O 函数（`fgets`、`fread`、`fwrite`）的操作方式
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3. 建立 API 开销的概念，理解系统调用与库函数的性能差异
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4. 熟悉结构体的二进制 I/O 读写方法
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5. 综合运用文件 I/O 完成文本处理任务
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## 涉及知识点
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- 文件描述符与 `open`/`close`/`read`/`write` 系统调用
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- 标准 I/O：`fopen`/`fclose`/`fgets`/`fprintf`/`fread`/`fwrite`
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- 文件打开模式：`O_RDONLY`、`O_WRONLY`、`O_CREAT`、`O_TRUNC`、`O_APPEND`
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- 结构体与文件 I/O 结合（二进制序列化）
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- `gettimeofday` 高精度计时
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- 字符串处理：`strtok`、`strcmp`、`strstr`、`sscanf`、`%[^:]`
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- 排序算法（词频统计中的字典序排列）
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---
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## 任务一：task41.c —— 学生信息文件字段处理
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### 任务要求
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1. 创建文件 `student.txt`，写入若干学生记录，每行格式为 `姓名:学号:学院:年龄:性别`
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2. 从 `student.txt` 中查找所有属于"计算机与网络安全学院"的记录
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3. 将找到的记录字段顺序调整为 `学号:姓名:性别:年龄:学院`
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4. 将调整后的记录写入 `csStudent.txt`
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### 关键代码提示
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```c
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#include <fcntl.h>
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#include <unistd.h>
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#include <string.h>
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#include <stdio.h>
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#include <stdlib.h>
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int main() {
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// ---- 第一步：创建并写入 student.txt ----
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int fd = open("student.txt", O_WRONLY | O_CREAT | O_TRUNC, 0644);
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if (fd < 0) { perror("open student.txt"); exit(1); }
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const char *records[] = {
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"张三:2023001:计算机与网络安全学院:20:男\n",
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"李四:2023002:电子信息学院:21:女\n",
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"王五:2023003:计算机与网络安全学院:22:男\n",
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"赵六:2023004:数学学院:19:女\n",
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"钱七:2023005:计算机与网络安全学院:20:男\n",
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};
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for (int i = 0; i < 5; i++)
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write(fd, records[i], strlen(records[i]));
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close(fd);
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// ---- 第二步：读取、筛选、重组字段 ----
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FILE *fin = fopen("student.txt", "r");
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FILE *fout = fopen("csStudent.txt", "w");
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char line[256];
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while (fgets(line, sizeof(line), fin) != NULL) {
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if (strstr(line, "计算机与网络安全学院") != NULL) {
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char name[64], id[64], college[64], age[16], gender[16];
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sscanf(line, "%[^:]:%[^:]:%[^:]:%[^:]:%s",
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name, id, college, age, gender);
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// 调整字段顺序：学号:姓名:性别:年龄:学院
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fprintf(fout, "%s:%s:%s:%s:%s\n", id, name, gender, age, college);
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}
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}
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fclose(fin);
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fclose(fout);
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return 0;
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}
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```
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### 常见问题
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| 问题 | 原因 | 解决方法 |
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|------|------|----------|
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| `write` 后文件内容为空 | 忘记 `close`，数据还在内核缓冲区 | 写完后务必 `close(fd)` |
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| 读取中文出现乱码 | 编码不匹配 | 确保源文件为 UTF-8 编码，终端 locale 一致 |
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| `strtok` 分割结果不对 | 行末换行符干扰 | 分割前先去除 `\n` |
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| `sscanf` 读取不完整 | 格式字符串匹配错误 | 使用 `%[^:]` 匹配非冒号字符序列 |
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---
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## 任务二：task42.c —— 结构体二进制文件读写
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### 任务要求
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1. 从键盘读入 5 个学生的信息（学号、姓名、语文、数学、英语成绩），存入结构体数组
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2. 将结构体数组以二进制方式写入文件 `score.dat`（使用 `write` 写入原始字节）
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3. 从文件中读取第 1、3、5 条记录并显示
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### 关键代码提示
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```c
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#include <fcntl.h>
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#include <unistd.h>
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#include <stdio.h>
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typedef struct {
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int id;
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char name[32];
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float chinese;
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float math;
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float english;
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} Student;
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int main() {
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Student stu[5];
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// 从键盘读入
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for (int i = 0; i < 5; i++) {
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printf("请输入第%d个学生(学号 姓名 语文 数学 英语): ", i + 1);
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scanf("%d %s %f %f %f", &stu[i].id, stu[i].name,
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&stu[i].chinese, &stu[i].math, &stu[i].english);
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}
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// 二进制写入
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int fd = open("score.dat", O_WRONLY | O_CREAT | O_TRUNC, 0644);
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write(fd, stu, sizeof(Student) * 5);
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close(fd);
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// 读取第 1、3、5 条（下标 0、2、4）
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fd = open("score.dat", O_RDONLY);
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Student temp;
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for (int i = 0; i < 5; i++) {
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read(fd, &temp, sizeof(Student));
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if (i == 0 || i == 2 || i == 4) {
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printf("学号:%d 姓名:%s 语文:%.1f 数学:%.1f 英语:%.1f\n",
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temp.id, temp.name, temp.chinese, temp.math, temp.english);
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}
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}
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// 也可用 lseek 精确定位到第 3 条
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lseek(fd, sizeof(Student) * 2, SEEK_SET);
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read(fd, &temp, sizeof(Student));
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printf("第3条: %s\n", temp.name);
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close(fd);
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return 0;
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}
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```
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### 常见问题
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| 问题 | 原因 | 解决方法 |
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| 读出的数值不对 | 结构体内存对齐（padding） | `sizeof(Student)` 可能大于各字段大小之和，属正常现象 |
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| `lseek` 定位不准 | 偏移量计算错误 | 偏移量 = `sizeof(Student) * (n - 1)` |
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| 中文姓名存储异常 | `char name[32]` 对 UTF-8 中文不够 | 增大缓冲区（一个汉字占 3 字节） |
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---
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## 任务三：task43.c —— API 执行时间测量（选做）
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### 任务要求
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1. 分别测量 `read`/`write` 和 `fread`/`fwrite` 在不同数据量下的执行时间
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2. 对比系统调用与库函数的性能差异
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3. 绘制或输出性能对比表
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### 关键代码提示
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```c
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#include <sys/time.h>
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#include <fcntl.h>
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#include <unistd.h>
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#include <stdio.h>
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#include <stdlib.h>
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long time_diff(struct timeval *start, struct timeval *end) {
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return (end->tv_sec - start->tv_sec) * 1000000L
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+ (end->tv_usec - start->tv_usec);
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}
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int main() {
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struct timeval start, end;
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int N = 1000000; // 循环次数
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char buf[1];
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// 测量 write（逐字节）
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int fd = open("test.dat", O_WRONLY | O_CREAT | O_TRUNC, 0644);
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gettimeofday(&start, NULL);
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for (int i = 0; i < N; i++)
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write(fd, buf, 1);
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gettimeofday(&end, NULL);
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close(fd);
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printf("write 逐字节: %ld 微秒\n", time_diff(&start, &end));
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// 测量 fwrite（逐字节，带用户缓冲）
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FILE *fp = fopen("test2.dat", "w");
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gettimeofday(&start, NULL);
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for (int i = 0; i < N; i++)
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fwrite(buf, 1, 1, fp);
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gettimeofday(&end, NULL);
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fclose(fp);
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printf("fwrite 逐字节: %ld 微秒\n", time_diff(&start, &end));
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return 0;
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}
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```
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### 测量方案
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| 测量项 | 操作 | 说明 |
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|--------|------|------|
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| `write` | 逐字节写 1MB | 基准：每次陷入内核 |
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| `read` | 逐字节读 1MB | 基准：每次陷入内核 |
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| `fwrite` | 逐字节写 1MB | 带用户空间缓冲 |
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| `fread` | 逐字节读 1MB | 带用户空间缓冲 |
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| `write` | 块写入（4KB） | 对比块大小影响 |
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### 常见问题
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| 问题 | 原因 | 解决方法 |
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| 计时结果为 0 | 操作太快，微秒级精度不够 | 增加循环次数到百万级 |
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| 系统调用比库函数慢很多 | 每次 `read`/`write` 都陷入内核 | 正常现象，体现用户缓冲的价值 |
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| 结果波动大 | 系统调度干扰 | 多次测量取平均值 |
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---
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## 任务四：task44.c —— 英文文章词频统计
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### 任务要求
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1. 读取一篇英文文章（从文件或标准输入）
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2. 统计每个单词出现的次数
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3. 输出格式：`单词:次数`
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4. 按字典序排列所有单词
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5. 额外输出出现频度最高的 10 个单词
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### 关键代码提示
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```c
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#include <stdio.h>
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#include <string.h>
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#include <stdlib.h>
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#include <ctype.h>
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#define MAX_WORDS 10000
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typedef struct {
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char word[64];
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int count;
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} WordEntry;
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WordEntry dict[MAX_WORDS];
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int dict_size = 0;
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// 查找已有单词或插入新单词
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int find_or_insert(const char *word) {
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for (int i = 0; i < dict_size; i++) {
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if (strcmp(dict[i].word, word) == 0) {
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dict[i].count++;
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return i;
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}
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}
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strcpy(dict[dict_size].word, word);
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dict[dict_size].count = 1;
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return dict_size++;
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}
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// qsort 比较函数：字典序
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int cmp_alpha(const void *a, const void *b) {
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return strcmp(((WordEntry *)a)->word, ((WordEntry *)b)->word);
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}
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// qsort 比较函数：频度降序
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int cmp_freq(const void *a, const void *b) {
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return ((WordEntry *)b)->count - ((WordEntry *)a)->count;
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}
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int main() {
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FILE *fp = fopen("article.txt", "r");
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if (!fp) { perror("fopen"); return 1; }
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char word[64];
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while (fscanf(fp, "%63s", word) == 1) {
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// 去除标点，统一小写
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char clean[64];
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int j = 0;
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for (int i = 0; word[i]; i++) {
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if (isalpha(word[i]))
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clean[j++] = tolower(word[i]);
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}
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clean[j] = '\0';
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if (j > 0)
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find_or_insert(clean);
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}
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fclose(fp);
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// 按字典序输出
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qsort(dict, dict_size, sizeof(WordEntry), cmp_alpha);
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for (int i = 0; i < dict_size; i++)
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printf("%s:%d\n", dict[i].word, dict[i].count);
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// 按频度降序输出前 10 个
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qsort(dict, dict_size, sizeof(WordEntry), cmp_freq);
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printf("\n频度最高的10个单词:\n");
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for (int i = 0; i < 10 && i < dict_size; i++)
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printf("%s:%d\n", dict[i].word, dict[i].count);
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|||
|
|
return 0;
|
|||
|
|
}
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
### 注意事项
|
|||
|
|
|
|||
|
|
- 单词提取时需过滤标点符号（逗号、句号、引号等）
|
|||
|
|
- 不区分大小写（统一转为小写）
|
|||
|
|
- 连字符（如 "well-known"）可按需决定是否拆分
|
|||
|
|
- 文件较大时注意 `MAX_WORDS` 的上限，可改用动态分配
|
|||
|
|
|
|||
|
|
### 常见问题
|
|||
|
|
|
|||
|
|
| 问题 | 原因 | 解决方法 |
|
|||
|
|
|------|------|----------|
|
|||
|
|
| 单词带着标点 | 没有清理非字母字符 | 用 `isalpha` 逐字符过滤 |
|
|||
|
|
| 大小写被当成不同单词 | 未统一大小写 | 提取前用 `tolower` 转换 |
|
|||
|
|
| 排序结果不对 | `qsort` 比较函数写错 | 注意比较函数的参数类型转换 |
|
|||
|
|
| 数组越界 | 单词数超过 `MAX_WORDS` | 动态扩容（`realloc`）或增大数组 |
|
|||
|
|
|
|||
|
|
---
|
|||
|
|
|
|||
|
|
## 实验总结
|
|||
|
|
|
|||
|
|
通过本实验，应掌握以下能力：
|
|||
|
|
|
|||
|
|
1. 熟练使用底层 I/O（`open`/`read`/`write`）和标准 I/O（`fopen`/`fgets`/`fprintf`）
|
|||
|
|
2. 理解文件描述符与 `FILE *` 的区别
|
|||
|
|
3. 能用结构体进行二进制文件读写
|
|||
|
|
4. 了解系统调用与库函数的性能差异
|
|||
|
|
5. 综合运用字符串处理和文件 I/O 解决实际问题
|