如何将结构与指定的字节边界对齐? [英] How can I align a struct to a specifed byte boundary?

问题描述

我需要将一个结构与Rust中的16个字节边界对齐.似乎可以通过 repr属性给出有关对齐的提示，但是它可以不支持这种确切的用例.

I need to align a struct to a 16 byte boundary in Rust. It seems possible to give hints about alignment thorough the repr attribute, but it doesn't support this exact use case.

对我要实现的功能的测试是类型Foo这样的

A functional test of what I'm trying to achieve is a type Foo such that

assert_eq!(mem::align_of::<Foo>(), 16);

或具有字段baz的结构Bar

println!("{:p}", Bar::new().baz);

总是打印一个可被16整除的数字.

always prints a number divisible by 16.

Rust目前有可能吗?有什么解决方法吗?

Is this currently possible in Rust? Are there any work-arounds?

推荐答案

huon的回答很好，但是已经过时了.

huon's answer is good, but it's out of date.

从Rust 1.25.0开始，您现在可以将类型与N字节对齐使用属性#[repr(align(N))] .它记录在参考的类型布局"部分下.请注意，对齐方式必须是2的幂，您不能混合使用align和packed表示形式，并且对齐类型可能会为该类型添加额外的填充. 这是如何使用此功能的示例:

As of Rust 1.25.0, you may now align a type to N bytes using the attribute #[repr(align(N))]. It is documented under the reference's Type Layout section. Note that the alignment must be a power of 2, you may not mix align and packed representations, and aligning a type may add extra padding to the type. Here's an example of how to use the feature:

#[repr(align(64))]
struct S(u8);

fn main() {
    println!("size of S: {}", std::mem::size_of::<S>());
    println!("align of S: {}", std::mem::align_of::<S>());
}

这篇关于如何将结构与指定的字节边界对齐?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！