为什么当只需要一个字节时，Rust为什么使用两个字节来表示这个枚举? [英] Why does Rust use two bytes to represent this enum when only one is necessary?

问题描述

即使只有8 * 8 = 64的可能性，似乎仅对A使用一个字节足够聪明，但对B仅使用一个字节不够聪明.有什么办法哄Rust解决这个问题，还是我必须手动实现更紧凑的布局?

It appears to be smart enough to only use one byte for A, but not smart enough to use one byte for B, even though there are only 8*8=64 possibilities. Is there any way to coax Rust to figure this out or do I have to manually implement a more compact layout?

游乐场链接.

#![allow(dead_code)]

enum A {
    L,
    UL,
    U,
    UR,
    R,
    DR,
    D,
    DL,
}

enum B {
    C(A, A),
}

fn main() {
    println!("{:?}", std::mem::size_of::<A>()); // prints 1
    println!("{:?}", std::mem::size_of::<B>()); // prints 2
}

推荐答案

两个字节对于保持借用结构成员的能力都是必需的.

Both bytes are necessary to preserve the ability to borrow struct members.

Rust中的类型不是理想的值集:它具有数据布局，该布局描述了如何存储值.控制该语言的规则"之一是将类型放入struct或enum不会改变其数据布局:它在另一种类型内的布局与独立类型的布局相同，这使您可以引用构造成员并与其他任何参考文献互换使用.*

A type in Rust is not an ideal set of values: it has a data layout, which describe how the values are stored. One of the "rules" governing the language is that putting a type inside a struct or enum doesn't change its data layout: it has the same layout inside another type as it does standalone, which allows you to take references to struct members and use them interchangeably with any other reference.*

在满足此约束的同时，无法将两个A装入一个字节，因为A的大小是一个完整的字节-即使使用repr(packed)，也无法寻址一个字节的一部分.未使用的位只是保持未使用(除非可以通过利基填充将它们重新用于存储枚举标签).

There's no way to fit two As into one byte while satisfying this constraint, because the size of A is one whole byte -- you can't address a part of a byte, even with repr(packed). The unused bits just remain unused (unless they can be repurposed to store the enum tag by niche-filling).

*好吧，repr(packed)实际上可以使此错误. 引用压缩字段可能导致不确定的行为，即使在安全代码中！

*Well, repr(packed) can actually make this untrue. Taking a reference to a packed field can cause undefined behavior, even in safe code!

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