为 C/FFI 库调用分配对象 [英] Allocating an object for C / FFI library calls

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问题描述

我有一个 C 库,它具有 gpio 实现.有特定于目标的 gpio_type,每个 MCU 对 gpio_type 有不同的定义.库中的函数之一:

I have a C library, which has gpio implementation. There's gpio_type which is target specific, each MCU has different definition for gpio_type. One of the functions in the library:

void gpio_init(gpio_type *object, int32_t pin);

我想使用 C 库函数在 Rust 中编写 Gpio 对象的抽象.因此需要类似不透明指针类型的东西(在 C++ 中,我只需创建一个类型为:gpio_type 的成员变量).我想我会创建一个空的枚举(或结构),分配对象所需的空间并将其转换为匹配 C 层中的类型.

I want to write abstraction of Gpio object in Rust, using C library functions. Therefore need something like opaque pointer type (in C++ I would just create a member variable with type: gpio_type). I figured I would create an empty enum (or struct), allocate a space needed for the object and transmute it to match the type in C layer.

pub enum gpio_type {}

#[link(name = "gpio_lib", kind = "static")]
extern {
    pub fn gpio_init(obj: *mut gpio_type, value: i32);
}

pub struct Gpio {
    gpio : *mut gpio_type,
}

impl Gpio {
    pub fn new(pin: u32) -> Gpio {
        unsafe {
            let mut gpio_ptr : &'static [u8; 4] = init(); // size of gpio in C is 4 bytes for one target, will be changed later to obtain it dynamically
            let gpio_out = Gpio { gpio: transmute(gpio_ptr)};
            gpio_init(gpio_out.gpio, pin);
            gpio_out
        }
    }
}

这针对嵌入式设备,因此没有 std,没有 libc.我不想为 rust 中的每个目标重新定义 gpio_type(复制每个目标的 C 声明),寻找一些东西来为 C 将处理的对象分配内存.

This targets embedded devices, therefore no std, no libc. I don't want to redefine gpio_type for each target in rust (copy the C declaration for each target), looking for something to just allocate memory for the object which C will handle.

下面的代码片段根据反汇编生成指向地址 0 的指针.Gpio新方法的反汇编:

The following snippet below produces pointer to address 0 according to disassembly. Disassembly for Gpio new method:

 45c:   b580        push    {r7, lr}
 45e:   466f        mov r7, sp
 460:   4601        mov r1, r0
 462:   2000        movs    r0, #0
 464:   f000 fae6   bl  a34 <gpio_init>
 468:   2000        movs    r0, #0
 46a:   bd80        pop {r7, pc}

知道为什么 462 是 0 吗?

Any ideas why 462 is 0 ?

推荐答案

寻找一些东西来为 C 将处理的对象分配内存

looking for something to just allocate memory for the object which C will handle

这样的事情呢?给结构一个实际大小(在这种情况下,给它一个字节大小的固定大小数组),在堆上分配那个空间,然后把它当作一个原始指针.

What about something like this? Give the struct an actual size (in this case by giving it a fixed-size array of byte-sized items), allocate that space on the heap, then treat that as a raw pointer.

use std::mem;

#[allow(missing_copy_implementations)]
pub struct Gpio([u8; 4]);

impl Gpio {
    fn new() -> Gpio { Gpio([0,0,0,0]) }
}

fn main() {
    // Allocate some bytes and get a raw pointer
    let a: *mut u8 = unsafe { mem::transmute(Box::new(Gpio::new())) };

    // Use it here!

    // When done... back to a box
    let b: Box<Gpio> = unsafe { mem::transmute(a) };

    // Now it will be dropped automatically (and free the allocated memory)

    // Or you can be explicit
    drop(b);
}

但是,我建议做这样的事情;这更明显,不需要堆分配:

However, I'd suggest doing something like this; it's a lot more obvious and doesn't need a heap allocation:

#[allow(missing_copy_implementations)]
pub struct Gpio([u8; 4]);

impl Gpio {
    fn new() -> Gpio { Gpio([0,0,0,0]) }

    fn as_mut_ptr(&mut self) -> *mut u8 {
        self.0.as_mut_ptr()
    }
}

fn main() {
    let mut g = Gpio::new();
    let b = g.as_mut_ptr();
}

作为奖励,你有一个不错的地方来挂一些方法.as_mut_ptr 可能不需要是公共的,并且可以隐藏在 Gpio 结构上的公共方法后面.

As a bonus, you get a nice place to hang some methods on. Potentially as_mut_ptr wouldn't need to be public, and could be hidden behind public methods on the Gpio struct.

(也可以使用 uninitialized 而不是 [0,0,0,0])

// This depends on your library, check the FFI guide for details
extern {
    fn gpio_init(gpio: *mut u8, pin: u8);
    fn gpio_pin_on(gpio: *mut u8);
    fn gpio_pin_off(gpio: *mut u8);
}

#[allow(missing_copy_implementations)]
pub struct Gpio([u8; 4]);

impl Gpio {
    fn new(pin: u8) -> Gpio {
        let mut g = Gpio([0,0,0,0]);
        g.init(pin);
        g
    }

    fn as_mut_ptr(&mut self) -> *mut u8 {
        self.0.as_mut_ptr()
    }

    fn init(&mut self, pin: u8) { unsafe { gpio_init(self.as_mut_ptr(), pin) } }
    pub fn on(&mut self) { unsafe { gpio_pin_on(self.as_mut_ptr()) } }
    pub fn off(&mut self) { unsafe { gpio_pin_off(self.as_mut_ptr()) } }
}

static BLUE_LED_PIN: u8 = 0x4;

fn main() {
    let mut g = Gpio::new(BLUE_LED_PIN);
    g.on();
    g.off();
}

这篇关于为 C/FFI 库调用分配对象的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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