如何将引用元组转换为元组引用?

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我想将引用元组(都是对同一结构成员的引用)转换为元组的引用.

I'd like to convert a tuple of references (which are all references to members of the same struct) to a reference of a tuple.

我试图以各种方式胁迫他们，但如果不进行克隆我就无法做到.

I've tried to coerce them in various ways, however I wasn't able to do it without cloning.

struct Bar(); struct Foo(Bar, Bar, Bar); fn main() { let a: &Foo = &Foo(Bar(), Bar(), Bar()); let b: &(Bar, Bar) = &(a.0, a.1); }

error[E0507]: cannot move out of borrowed content --> src/main.rs:7:28 | 7 | let b: &(Bar, Bar) = &(a.0, a.1); | ^^^ cannot move out of borrowed content error[E0507]: cannot move out of borrowed content --> src/main.rs:7:33 | 7 | let b: &(Bar, Bar) = &(a.0, a.1); | ^^^ cannot move out of borrowed content

我希望 b 的类型为 &(Bar, Bar) 给定 a 的类型为 &Foo.

I expect b to be of type &(Bar, Bar) given a is of type &Foo.

推荐答案

这不可能.

引用引用到一个值.你希望有一个 &(Bar, Bar) 但内存中没有任何地方有 (Bar, Bar) 的 2 元组.你不能引用不存在的东西.

A reference refers to a value. You wish to have a &(Bar, Bar) but there is nowhere in memory that has a 2-tuple of (Bar, Bar). You cannot refer to something that does not exist.

&(A, B) 和 (&A, &B) 的内存布局根本不兼容，所以你也不能使用不安全的 Rust 技术.

The memory layouts of &(A, B) and (&A, &B) are fundamentally incompatible, so you cannot use unsafe Rust techniques either.

在这种特殊情况中，您可能能够使用不安全的 Rust 将您的 &Foo 直接转换为 &(Bar, Bar)，但是...

In this particular case, you might be able to use unsafe Rust to convert your &Foo directly to a &(Bar, Bar), but...

• 它要求元组结构和元组的布局相同；我不知道这是保证1
• 它要求元组结构的布局紧密排列，以便您可以通过成员大小进行偏移以到达下一个；我不知道这是保证1
• 它要求元组结构的布局按照它们定义的相同顺序放置成员；我不知道这是保证1
• 你只能对连续的片段进行；没有得到第一和第三项

// I copied this unsafe block from Stack Overflow // without properly documenting why I think this code is safe. let b: &(Bar, Bar) = unsafe { &*(a as *const Foo as *const (Bar, Bar)) }; println!("{:?}", b);

// I copied this unsafe block from Stack Overflow // without properly documenting why I think this code is safe. let c: &(Bar, Bar) = unsafe { let p = a as *const Foo as *const Bar; let p = p.offset(1); &*(p as *const (Bar, Bar)) }; println!("{:?}", c);

1 — 事实上，引用明确指出:

元组对其布局没有任何保证.

Tuples do not have any guarantees about their layout.

唯一的例外是单元元组 (())，它保证为零大小类型，大小为 0，对齐方式为 1.

The exception to this is the unit tuple (()) which is guaranteed as a zero-sized type to have a size of 0 and an alignment of 1.

这意味着虽然此代码可能会打印出您期望的内容并且 Miri 没有抱怨，但这是未定义的行为.

This means that while this code may print out what you expect and Miri does not complain, it's undefined behavior.