对齐结构成员，但不结构

发布于 2025-01-23 11:08:55 字数 675 浏览 3 评论 0原文

我正在编写一些与C接口的生锈代码。C定义了一些结构，例如

struct foo {
    // other fields
    uint32_t count;
    const struct bar* array;
};

我想通过此布局来定义锈结构，但是为了安全，我想放置count 和array < /code>在自己作为私人成员的结构中，就像这样：

#[repr(C)]
pub struct Foo {
    // other fields
    pub bars: BarSlice,
}
#[repr(C)]
pub struct BarSlice {
    count: u32,
    ptr: *const Bar,
}

这是错误的，因为在64位系统上，barslice将与8个字节保持一致，并且在count> Count之前的字段可能是8个字节对齐和4个字节长。我如何告诉Rust正确对齐结构的成员，但不必担心结构本身？

我也宁愿避免在可能的情况下

为每个对齐情况创建不同的切片类型，
其中foo

原文

I'm writing some Rust code that interfaces with C. The C defines some structures like

struct foo {
    // other fields
    uint32_t count;
    const struct bar* array;
};

I want to define a Rust structure with this layout, but for safety I want to put count and array in their own structure as private members, like so:

#[repr(C)]
pub struct Foo {
    // other fields
    pub bars: BarSlice,
}
#[repr(C)]
pub struct BarSlice {
    count: u32,
    ptr: *const Bar,
}

This is wrong though, since on 64-bit systems BarSlice will be aligned to 8 bytes, and the field immediately before count might be 8 byte aligned and 4 bytes long. How can I tell Rust to align the members of the struct correctly, but not worry about the struct itself?

I would also rather avoid if possible