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如何使用ioctl+nix宏获取可变大小的缓冲区

usb-hid ioctl usb rust

Juan Leni · 技术社区 · 6 年前

我想检索一个可变大小的缓冲区。还有一个 ioctl

#define HID_MAX_DESCRIPTOR_SIZE     4096
#define HIDIOCGRDESC        _IOR('H', 0x02, struct hidraw_report_descriptor)

struct hidraw_report_descriptor {
    __u32 size;
    __u8 value[HID_MAX_DESCRIPTOR_SIZE];
};

我按以下方式定义宏:

ioctl_read_buf!(hid_read_descr, b'H', 0x02, u8);

稍后打电话给:

let mut desc_raw = [0u8; 4 + 4096];
let err = unsafe { hid_read_descr(file.as_raw_fd(), &mut desc_raw); };

做这个的时候, desc_raw 满是零。我本以为前4个字节包含 size 基于结构定义。

另一种方法,似乎也行不通

ioctl_read!(hid_read_descr2, b'H', 0x02, [u8; 4+4096]);
// ...
let mut desc_raw = [0xFFu8; 4 + 4096];
let err = unsafe { hid_read_descr2(file.as_raw_fd(), &mut desc_raw); };

在这两种情况下,我都试图初始化使用0xFF并在调用之后,它似乎没有受到影响。

ioctl_read_buf 宏错误?

2 回复 | 直到 6 年前

Shepmaster Tim Diekmann 5 年前

既然 Digikata has thoughtfully provided enough code to drive the program ...

ioctl_read_buf 宏错误?

我可以这么说这里不正确。不想读取数据数组,只想读取特定类型的单个实例。那就是 ioctl_read! 是为了。

repr(C) 模仿C定义的结构。这确保了诸如对齐、填充、字段排序等重要细节都与我们调用的代码一一匹配。

然后,我们可以构造此结构的未初始化实例,并将其传递给新定义的函数。

use libc; // 0.2.66
use nix::ioctl_read; // 0.16.1
use std::{
    fs::OpenOptions,
    mem::MaybeUninit,
    os::unix::{fs::OpenOptionsExt, io::AsRawFd},
};

const HID_MAX_DESCRIPTOR_SIZE: usize = 4096;

#[repr(C)]
pub struct hidraw_report_descriptor {
    size: u32,
    value: [u8; HID_MAX_DESCRIPTOR_SIZE],
}

ioctl_read!(hid_read_sz, b'H', 0x01, libc::c_int);
ioctl_read!(hid_read_descr, b'H', 0x02, hidraw_report_descriptor);

fn main() -> Result<(), Box<dyn std::error::Error>> {
    let file = OpenOptions::new()
        .read(true)
        .write(true)
        .custom_flags(libc::O_NONBLOCK)
        .open("/dev/hidraw0")?;

    unsafe {
        let fd = file.as_raw_fd();

        let mut size = 0;
        hid_read_sz(fd, &mut size)?;
        println!("{}", size);

        let mut desc_raw = MaybeUninit::<hidraw_report_descriptor>::uninit();
        (*desc_raw.as_mut_ptr()).size = size as u32;
        hid_read_descr(file.as_raw_fd(), desc_raw.as_mut_ptr())?;
        let desc_raw = desc_raw.assume_init();
        let data = &desc_raw.value[..desc_raw.size as usize];
        println!("{:02x?}", data);
    }

    Ok(())
}

Digikata 6 年前

我想你有几个问题。有些在生锈的一面,有些在使用 HIDIOCGRDESC

struct hidraw_report_descriptor rpt_desc;

memset(&rpt_desc, 0x0, sizeof(rpt_desc));

/* Get Report Descriptor */
rpt_desc.size = desc_size;
res = ioctl(fd, HIDIOCGRDESC, &rpt_desc);

desc_size HIDIOCGRDESCSIZE ioctl呼叫。除非填写正确的大小参数,否则ioctl将返回一个错误( ENOTTY EINVAL ).

O_NONBLOCK 在不使用的情况下打开HID设备的标志 libc::open . 我的结局是:

#[macro_use]
extern crate nix;

extern crate libc;

ioctl_read!(hid_read_sz, b'H', 0x01, i32);
ioctl_read_buf!(hid_read_descr, b'H', 0x02, u8);

fn main() {
    // see /usr/include/linux/hidraw.h
    // and hid-example.c
    extern crate ffi;
    use std::ffi::CString;
    let fname = CString::new("/dev/hidraw0").unwrap();
    let fd = unsafe { libc::open(fname.as_ptr(), libc::O_NONBLOCK | libc::O_RDWR) };

    let mut sz = 0i32;
    let err = unsafe { hid_read_sz(fd, &mut sz) };
    println!("{:?} size is {:?}", err, sz);

    let mut desc_raw = [0x0u8; 4 + 4096];

    // sz on my system ended up as 52 - this handjams in the value
    // w/ a little endian swizzle into the C struct .size field, but
    // really we should properly define the struct
    desc_raw[0] = sz as u8;

    let err = unsafe { hid_read_descr(fd, &mut desc_raw) };
    println!("{:?}", err);

    for (i, &b) in desc_raw.iter().enumerate() {
        if b != 0 {
            println!("{:4} {:?}", i, b);
        }
    }
}