将u8缓冲区转换为Rust中的结构

将u8缓冲区转换为Rust中的结构

本文介绍了将u8缓冲区转换为Rust中的结构的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我有一个未知大小的字节缓冲区,并且我想创建一个指向该缓冲区开头的内存的局部struct变量。遵循在C语言中所做的事情,我在Rust中尝试了很多不同的操作,并不断出错。这是我最近的尝试:

I have a byte buffer of unknown size, and I want to create a local struct variable pointing to the memory of the beginning of the buffer. Following what I'd do in C, I tried a lot of different things in Rust and kept getting errors. This is my latest attempt:

use std::mem::{size_of, transmute};

#[repr(C, packed)]
struct MyStruct {
    foo: u16,
    bar: u8,
}

fn main() {
    let v: Vec<u8> = vec![1, 2, 3];
    let buffer = v.as_slice();
    let s: MyStruct = unsafe { transmute(buffer[..size_of::<MyStruct>()]) };
}

我收到一个错误:

error[E0277]: the size for values of type `[u8]` cannot be known at compilation time
   --> src/main.rs:12:42
    |
12  |     let s: MyStruct = unsafe { transmute(buffer[..size_of::<MyStruct>()]) };
    |                                          ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ doesn't have a size known at compile-time
    |
    = help: the trait `std::marker::Sized` is not implemented for `[u8]`
    = note: to learn more, visit <https://doc.rust-lang.org/book/ch19-04-advanced-types.html#dynamically-sized-types-and-the-sized-trait>


推荐答案

如果您不想将数据复制到结构,但可以将其保留在适当的位置,可以使用。而是创建& MyStruct

If you don't want to copy the data to the struct but instead leave it in place, you can use slice::align_to. This creates a &MyStruct instead:

#[repr(C, packed)]
#[derive(Debug, Copy, Clone)]
struct MyStruct {
    foo: u16,
    bar: u8,
}

fn main() {
    let v = vec![1u8, 2, 3];

    // I copied this code from Stack Overflow
    // without understanding why this case is safe.
    let (head, body, _tail) = unsafe { v.align_to::<MyStruct>() };
    assert!(head.is_empty(), "Data was not aligned");
    let my_struct = &body[0];

    println!("{:?}", my_struct);
}

在这里,使用 align_to 将某些字节转换为 MyStruct ,因为我们已使用 repr(C,packed)和所有 MyStruct 中的类型可以是任意字节。

Here, it's safe to use align_to to transmute some bytes to MyStruct because we've used repr(C, packed) and all of the types in MyStruct can be any arbitrary bytes.

另请参见:




  • How to read a struct from a file in Rust?
  • Can I take a byte array and deserialize it into a struct?

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08-20 08:59