当前位置:网站首页>How much disk space does a file of 1 byte actually occupy
How much disk space does a file of 1 byte actually occupy
2020-11-06 21:04:14 【Zhang Yanfei Allen】
In the foreword 《 Whether a new empty file takes up disk space ? How much does it take 》 We learned about the disk overhead of an empty file . Today, let's think about another question , If we only write 1 Bytes , So the actual disk usage of this file is also 1 Bytes ?
see 1 Byte file
As before , Let's not talk about the principle , Just do it yourself .
# mkdir tempDir
# cd tempDir
# du -h
0 .
# touch test
# du -h
0 .
After creating an empty file in a directory , adopt du
The footprint of the folder as seen by the command has not changed . This is in line with our previous understanding , Because empty files only occupy inode. good , Let's modify the file , Add a letter
echo "a" > test
# du -h
4.0K .
After saving, check the space usage of the directory again . We found that the original 0 Increased to 4K. So , No matter how small the contents of the document , Even a byte , In fact, the operating system will also assign you 4K Of . Oh , Of course, we have to calculate the above mentioned inode And file names stored in the folder data structure . therefore , Don't maintain a lot of broken files in your system . File size , It takes up a lot of disks !
Notice that my experimental environment is in ext Under the file system . If it is xfs There may be some discrepancy in performance .
Keep talking about this 4K
And then linux Source code file fs/ext2/ext2.h About inode The definition of , We find the data node defined in the structure block Array :
struct ext2_inode {
......
__le32 i_block[EXT2_N_BLOCKS]; # An array that points to blocks that store file data
......
When a file has no data to store , This array is empty . And when we write 1 A few bytes later , The file system needs to apply for block To store , After the application , The pointer is in this array . Even if the file contains only one byte , It will still be assigned a whole Block, Because this is the smallest working unit of a file system . So this block How big is it ,ext You can go through dumpe2fs
see .
#dumpe2fs -h /dev/mapper/vgroot-lvroot
......
Block size: 4096
It's on my machine , One Block yes 4KB.
What to do if the content of the document is too large
I don't know if you pay attention to ,inode As defined in block How about the array size , Only EXT2_N_BLOCKS
individual . Let's look at the definition of this constant again , Find out it's 15, The definition in the relevant kernel is as follows :
#define EXT2_NDIR_BLOCKS 12
#define EXT2_IND_BLOCK EXT2_NDIR_BLOCKS
#define EXT2_DIND_BLOCK (EXT2_IND_BLOCK + 1)
#define EXT2_TIND_BLOCK (EXT2_DIND_BLOCK + 1)
#define EXT2_N_BLOCKS (EXT2_TIND_BLOCK + 1)
Just press the 4K Of block size Look at ,15 individual block Just enough to save 15*4=60K The file of . I believe you are not satisfied with the size of this file , You save one avi Big movies have to be on G 了 . that Linux How to achieve large file storage ? Um. , In fact, the definition process of macro above has already told you , It's just 12 Arrays are stored directly block The pointer , The rest is used for indirect indexing (EXT2_IND_BLOCK), Secondary indirect index (EXT2_DIND_BLOCK) And the third level index (EXT2_TIND_BLOCK).
such , The space that a file can use expands exponentially . When the files were small , They all use direct index , disk IO Less , Good performance . When the files are big , Visit one block It might have to be done three times first IO, Performance is a little bit slow , However, there are OS Page caching at level 、 Directory item cache bonus , It's OK .
Conclusion
File systems are managed in blocks , So no matter how small your file is , Even if it's only one byte , Will consume a whole block . This block size can be passed through dumpe2fs
Wait for the order to see . What if you want to change the size of this block ? I'm sorry , You can only reformat .
Development of hard disk album of internal training :
- 1. Disk opening : Take off the hard coat of the mechanical hard disk !
- 2. Disk partitioning also implies technical skills
- 3. How can we solve the problem that mechanical hard disks are slow and easy to break down ?
- 4. Disassemble the SSD structure
- 5. How much disk space does a new empty file take ?
- 6. Only 1 How much disk space does a byte file actually take up
- 7. When there are too many documents ls Why is the command stuck ?
- 8. Understand the principle of formatting
- 9.read How much disk does a byte of file actually take place on IO?
- 10.write When to write to disk after one byte of file IO?
- 11. Mechanical hard disk random IO Slower than you think
- 12. How much faster is a server equipped with a SSD than a mechanical hard disk ?
My official account is 「 Develop internal skill and practice 」, I'm not just talking about technical theory here , It's not just about practical experience . It's about combining theory with practice , Deepen the understanding of theory with practice 、 Use theory to improve your technical practice ability . Welcome to my official account , Please also share with your friends ~~~
版权声明
本文为[Zhang Yanfei Allen]所创,转载请带上原文链接,感谢
边栏推荐
- C++ 数字、string和char*的转换
- C++学习——centos7上部署C++开发环境
- C++学习——一步步学会写Makefile
- C++学习——临时对象的产生与优化
- C++学习——对象的引用的用法
- C++编程经验(6):使用C++风格的类型转换
- Won the CKA + CKS certificate with the highest gold content in kubernetes in 31 days!
- C + + number, string and char * conversion
- C + + Learning -- capacity() and resize() in C + +
- C + + Learning -- about code performance optimization
猜你喜欢
-
C + + programming experience (6): using C + + style type conversion
-
Latest party and government work report ppt - Park ppt
-
在线身份证号码提取生日工具
-
Online ID number extraction birthday tool
-
️野指针?悬空指针?️ 一文带你搞懂!
-
Field pointer? Dangling pointer? This article will help you understand!
-
HCNA Routing&Switching之GVRP
-
GVRP of hcna Routing & Switching
-
Seq2Seq实现闲聊机器人
-
【闲聊机器人】seq2seq模型的原理
随机推荐
- LeetCode 91. 解码方法
- Seq2seq implements chat robot
- [chat robot] principle of seq2seq model
- Leetcode 91. Decoding method
- HCNA Routing&Switching之GVRP
- GVRP of hcna Routing & Switching
- HDU7016 Random Walk 2
- [Code+#1]Yazid 的新生舞会
- CF1548C The Three Little Pigs
- HDU7033 Typing Contest
- HDU7016 Random Walk 2
- [code + 1] Yazid's freshman ball
- CF1548C The Three Little Pigs
- HDU7033 Typing Contest
- Qt Creator 自动补齐变慢的解决
- HALCON 20.11:如何处理标定助手品质问题
- HALCON 20.11:标定助手使用注意事项
- Solution of QT creator's automatic replenishment slowing down
- Halcon 20.11: how to deal with the quality problem of calibration assistant
- Halcon 20.11: precautions for use of calibration assistant
- “十大科学技术问题”揭晓!|青年科学家50²论坛
- "Top ten scientific and technological issues" announced| Young scientists 50 ² forum
- 求反转链表
- Reverse linked list
- js的数据类型
- JS data type
- 记一次文件读写遇到的bug
- Remember the bug encountered in reading and writing a file
- 单例模式
- Singleton mode
- 在这个 N 多编程语言争霸的世界,C++ 究竟还有没有未来?
- In this world of N programming languages, is there a future for C + +?
- es6模板字符
- js Promise
- js 数组方法 回顾
- ES6 template characters
- js Promise
- JS array method review
- 【Golang】️走进 Go 语言️ 第一课 Hello World
- [golang] go into go language lesson 1 Hello World