将* mut T强制解除引用到* mut ManuallyDrop <T>是不确定的行为？

Question

According to the docs, ManuallyDrop<T> is a zero-cost wrapper. Does that mean I can dereference a raw pointer to ManuallyDrop<T> casted from a raw pointer to T ?

Answer 1

ManuallyDrop is declared as #[repr(transparent)] :

#[stable(feature = "manually_drop", since = "1.20.0")]
#[lang = "manually_drop"]
#[derive(Copy, Clone, Debug, Default, PartialEq, Eq, PartialOrd, Ord, Hash)]
#[repr(transparent)]
pub struct ManuallyDrop<T: ?Sized> {
    value: T,
}

#[repr(transparent)] is described as :

The attribute can be applied to a newtype-like structs that contains a single field. It indicates that the newtype should be represented exactly like that field's type, ie, the newtype should be ignored for ABI purpopses [sic]: not only is it laid out the same in memory, it is also passed identically in function calls.

[...]

PtrWithCustomZst is also represented exactly like *const Foo

I believe that it is safe to perform this transformation.

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The real question is why would you want to do this? Having a pointer to a ManuallyDrop structure seems rather pointless. If you have a pointer to a T , the underlying value won't be dropped to start with. If you convert the pointer to a reference (while ensuring you uphold the rules of references), the reference won't drop the underlying value either.