if some files are pseudo, why open() function still can access it?

Question

We know there are many files that are pseudo files, i.e. not a real file.

ex:

/sys/xxx
/proc/xxx
/dev/xxxx

In my understanding, the open() will call x86 ASM code, the ASM code will do hardware interrupt to access to the disk.

The problem is if the open() will eventually access to the disk, how pseudo file still get access by open()?

But you don't always need to access the disk… if you're accessing a pure memory based file — 炸鱼薯条德里克, Mar 15 '19 at 08:43

score 4 · Answer 1 · answered Mar 15 '19 at 08:38

4

The open() system call does not work as you describe. Instead, it asks the kernel to open a file. The kernel knows which filesystem that file is on and what device it is associated with. This may be a physical hard drive, a block of memory, etc. If the associated device is just a block of memory, no disk access is performed.

answered Mar 15 '19 at 08:38

Fox

8,193

2

Also, the hardware architecture may not even be x86. – Kusalananda Mar 15 '19 at 08:40

score 2 · Accepted Answer · edited Mar 15 '19 at 15:04

As Fox noted in their answer, the kernel handles the open() syscall, and meanwhile, filesystems implement file operations in their own way. Filesystems, on the other hand, implement their version of syscalls, and that's what kernel should be using.

Consider, for instance, ext4 call to open directory or file operations in procfs (which notably has no .open mentioned anywhere), and pipefs which handles named and unnamed pipes. But the general idea is that the open() syscall is not going to be necessarily assembly, nor it is guaranteed to be specific to a particular architecture.

And to quote an answer by the user Warren Young who noted that way before this question appeared,

there is no single file where mkdir() exists. Linux supports many different file systems and each one has its own implementation of the "mkdir" operation. The abstraction layer that lets the kernel hide all that behind a single system call is called the VFS. So, you probably want to start digging in fs/namei.c, with vfs_mkdir(). The actual implementations of the low-level file system modifying code are elsewhere. For instance, the ext4 implementation is called ext4_mkdir(), defined in fs/ext4/namei.c.

As for why open() works this way, this is also due to Unix design philosophy that everything is a file. If you're using an API, you want to deal with consistent interface and not reinvent the wheel for every filesystem in existence (unless a person is a kernel developer, in which case they have our gratitude and respect).

if some files are pseudo, why open() function still can access it?

2 Answers2