rust - 为什么取消引用从函数返回的未处理的原始指针会导致未定义的行为？

Question

我发现取消引用一个未绑定的原始指针会导致未定义的行为，但我不知道为什么。我发现这Pin解决了我的情况。为什么它不起作用？

操场

fn main() {
    assert_not_eq(&Ref::pinned(1), &Ref::pinned(2));
    assert_not_eq(&Ref::raw(1), &Ref::raw(2)); // test failed
}

struct Ref<T> {
    value: T,
    reference: *const T,
}

use std::pin::Pin;
impl<T> Ref<T> {
    fn pinned(value: T) -> Pin<Box<Ref<T>>>
    where
        T: std::marker::Unpin,
    {
        let mut instance = Pin::new(Box::new(Self::dangling(value)));
        instance.reference = &mut instance.value;
        return instance;
    }

    fn raw(value: T) -> Ref<T> {
        let mut instance = Self::dangling(value);
        instance.reference = &mut instance.value;
        return instance;
    }

    fn dangling(value: T) -> Ref<T> {
        Ref {
            value,
            reference: 0 as *const T,
        }
    }
}

fn assert_not_eq<T: std::cmp::PartialEq + std::fmt::Debug + Copy>(first: &Ref<T>, second: &Ref<T>) {
    assert_eq!(unsafe { *first.reference }, first.value);
    assert_eq!(unsafe { *second.reference }, second.value);
    assert_ne!(first.value, second.value);
}