macro_rules! module_fs {
(type: $type:ty, $($f:tt)*) => { ... };
}
Expand description
Declares a kernel module that exposes a single file system.
The type
argument must be a type which implements the Type
trait. Also accepts various
forms of kernel metadata.
Examples
ⓘ
use kernel::prelude::*;
use kernel::{c_str, fs};
module_fs! {
type: MyFs,
name: "my_fs_kernel_module",
author: "Rust for Linux Contributors",
description: "My very own file system kernel module!",
license: "GPL",
}
struct MyFs;
#[vtable]
impl fs::Context<Self> for MyFs {
type Data = ();
fn try_new() -> Result {
Ok(())
}
}
impl fs::Type for MyFs {
type Context = Self;
const SUPER_TYPE: fs::Super = fs::Super::Independent;
const NAME: &'static CStr = c_str!("example");
const FLAGS: i32 = 0;
fn fill_super(_data: (), sb: fs::NewSuperBlock<'_, Self>) -> Result<&fs::SuperBlock<Self>> {
let sb = sb.init(
(),
&fs::SuperParams {
magic: 0x6578616d,
..fs::SuperParams::DEFAULT
},
)?;
let sb = sb.init_root()?;
Ok(sb)
}
}