Rust ‘&T as *const T as *mut T’ 在 ‘static mut’ 项目中合适吗

想象一下，您正试图编写一个围绕一些可变静态变量的安全包装器:

#![feature(strict_provenance)]
#![deny(fuzzy_provenance_casts)]
#![deny(lossy_provenance_casts)]

struct Wrapper<T>(*mut T);
struct Guard<T>(*mut T);

impl<T> Wrapper<T> {
    fn guard(&self) -> Guard<T> {
        Guard(self.0)
    }
}

// imagine there's synchronization going on that makes this all safe
unsafe impl<T: Send + Sync> Sync for Wrapper<T> {}
impl<T> std::ops::Deref for Guard<T> {
    type Target = T;
    fn deref(&self) -> &Self::Target {
        unsafe { &*self.0 }
    }
}
impl<T> std::ops::DerefMut for Guard<T> {
    fn deref_mut(&mut self) -> &mut Self::Target {
        unsafe { &mut *self.0 }
    }
}

static mut GLOBAL_SINGLETON: isize = 0; // Imagine this couldn't be trivially replaced by an atomic type
static WRAPPER: Wrapper<isize> = Wrapper(unsafe { &GLOBAL_SINGLETON as *const _ as *mut _ });
//                                                ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
//                                             Is this sound in this sepcific instance?

// Actually dereference the raw pointer to provoke UB
fn main() {
    assert_eq!(0, *WRAPPER.guard());
    *WRAPPER.guard() += 1;
    assert_eq!(1, *WRAPPER.guard());
}

现在，通常情况下，强制转换&T as *const T as *mut T并写入结果原始指针是不合理的，但严格来源和MIRI都不会抱怨上面的代码.这是由于堆叠借款和/或MILI的限制而导致的假阴性吗？还是会有static mut件商品在全局被贴上SharedRW件的标签，并在这里获得免费通行证？Ctrl-F代表Stacked Borrows paper中的"静态"没有显示任何相关信息.

我也考虑过这样的事情，我更喜欢这样，因为它迫使T比Wrapper活得更长.

struct Wrapper<'a, T>(&'a std::cell::UnsafeCell<T>);
impl<'a, T> Wrapper<'a, T> {
    const fn new(t: &'a T) -> Self {
        Self(unsafe { std::mem::transmute(t) }
    }
}
// ...

static mut GLOBAL_SINGLETON: isize = 0;
static WRAPPER: Wrapper<'_, isize> = Wrapper::new(&GLOBAL_SINGLETON);
// ...

但这显然是不合理的(Miri会告诉你)，因为它需要&GLOBAL_SINGLETON个才能活着，而&UnsafeCell { GLOBAL_SINGLETON }个人才能活着.