改变不可变的局部变量是未定义的行为吗？
Rust Reference似乎说改变不可变的本地数据（不在 内UnsafeCell）是未定义的行为：
行为被视为未定义
- 改变不可变数据。const 项中的所有数据都是不可变的。此外，所有通过共享引用访问的数据或由不可变绑定拥有的数据都是不可变的，除非该数据包含在
UnsafeCell<U>.
以下代码通过将不可变局部变量重新解释为AtomicU32. 目前代码运行得很好并打印了预期的结果，但它的行为实际上是未定义的吗？
use std::sync::atomic::{AtomicU32, Ordering};
#[repr(C, align(4))]
struct Bytes([u8; 4]);
fn main() {
let bytes = Bytes([11; 4]);
let x = unsafe { &*(&bytes as *const Bytes as *const AtomicU32) };
x.store(12345, Ordering::SeqCst);
println!("{:?}", bytes.0); // [57, 48, 0, 0]
}
Miri 不会抱怨下面的代码示例，其中字节是可变的。由于这些字节是通过共享引用 ( &AtomicU32)进行变异的，在我看来，根据The Rust Reference，下面的代码也应该具有未定义的行为 - 鉴于“通过共享引用 [..] 到达的所有数据都是不可变的”和“改变不可变数据[被认为是未定义的行为]”。
use std::sync::atomic::{AtomicU32, Ordering};
#[repr(C, align(4))]
struct Bytes([u8; 4]);
fn main() {
let mut bytes = Bytes([11; 4]);
let x = unsafe { &*(&mut bytes as *mut Bytes as *const AtomicU32) };
x.store(12345, Ordering::SeqCst);
println!("{:?}", bytes.0); // [57, 48, 0, 0]
}
回答
是的，根据美里的说法：
error: Undefined Behavior: trying to reborrow for SharedReadWrite at alloc1377, but parent tag <untagged> does not have an appropriate item in the borrow stack
--> src/main.rs:8:22
|
8 | let x = unsafe { &*(&bytes as *const Bytes as *const AtomicU32) };
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ trying to reborrow for SharedReadWrite at alloc1377, but parent tag <untagged> does not have an appropriate item in the borrow stack
|
= help: this indicates a potential bug in the program: it performed an invalid operation, but the rules it violated are still experimental
= help: see https://github.com/rust-lang/unsafe-code-guidelines/blob/master/wip/stacked-borrows.md for further information
= note: inside `main` at src/main.rs:8:22
= note: inside `<fn() as std::ops::FnOnce<()>>::call_once - shim(fn())` at /playground/.rustup/toolchains/nightly-x86_64-unknown-linux-gnu/lib/rustlib/src/rust/library/core/src/ops/function.rs:227:5
= note: inside `std::sys_common::backtrace::__rust_begin_short_backtrace::<fn(), ()>` at /playground/.rustup/toolchains/nightly-x86_64-unknown-linux-gnu/lib/rustlib/src/rust/library/std/src/sys_common/backtrace.rs:125:18
= note: inside closure at /playground/.rustup/toolchains/nightly-x86_64-unknown-linux-gnu/lib/rustlib/src/rust/library/std/src/rt.rs:66:18
= note: inside `std::ops::function::impls::<impl std::ops::FnOnce<()> for &dyn std::ops::Fn() -> i32 + std::marker::Sync + std::panic::RefUnwindSafe>::call_once` at /playground/.rustup/toolchains/nightly-x86_64-unknown-linux-gnu/lib/rustlib/src/rust/library/core/src/ops/function.rs:259:13
= note: inside `std::panicking::r#try::do_call::<&dyn std::ops::Fn() -> i32 + std::marker::Sync + std::panic::RefUnwindSafe, i32>` at /playground/.rustup/toolchains/nightly-x86_64-unknown-linux-gnu/lib/rustlib/src/rust/library/std/src/panicking.rs:379:40
= note: inside `std::panicking::r#try::<i32, &dyn std::ops::Fn() -> i32 + std::marker::Sync + std::panic::RefUnwindSafe>` at /playground/.rustup/toolchains/nightly-x86_64-unknown-linux-gnu/lib/rustlib/src/rust/library/std/src/panicking.rs:343:19
= note: inside `std::panic::catch_unwind::<&dyn std::ops::Fn() -> i32 + std::marker::Sync + std::panic::RefUnwindSafe, i32>` at /playground/.rustup/toolchains/nightly-x86_64-unknown-linux-gnu/lib/rustlib/src/rust/library/std/src/panic.rs:396:14
= note: inside `std::rt::lang_start_internal` at /playground/.rustup/toolchains/nightly-x86_64-unknown-linux-gnu/lib/rustlib/src/rust/library/std/src/rt.rs:51:25
= note: inside `std::rt::lang_start::<()>` at /playground/.rustup/toolchains/nightly-x86_64-unknown-linux-gnu/lib/rustlib/src/rust/library/std/src/rt.rs:65:5
也可以看看：
- FFI 函数可以修改未声明为可变的变量吗？
- 别名可变原始指针 (*mut T) 会导致未定义的行为吗？
因为这些字节是通过共享引用 (
&AtomicU32)
AtomicU32包含UnsafeCell，因此它符合您引用的豁免标准：
pub struct $atomic_type {
v: UnsafeCell<$int_type>,
}
这是 API as 的一部分AtomicU32::from_mut，不需要unsafe.