data-structures - 将 u8 缓冲区转换为 Rust 中的结构

Question

我有一个未知大小的字节缓冲区，我想创建一个指向缓冲区开头内存的本地结构变量。按照我在 C 中所做的，我在 Rust 中尝试了很多不同的东西，但不断出错。这是我最近的尝试:

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>()]) };
}

我得到一个错误:

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>

Answer 1

如果您不想将数据复制到结构中，而是将其保留在原地，您可以使用 slice::align_to .这将创建一个 &MyStruct:

#[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 将一些字节转换为 MyStruct 是安全的，因为我们已经使用了 repr(C, packed) 和所有MyStruct 中的类型可以是任意字节。

另见:

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