取消引用 *mut T 转换为 *mut ManuallyDrop<T> 是未定义的行为吗？
Is it undefined behavior to dereference a *mut T cast to *mut ManuallyDrop<T>?
根据文档，ManuallyDrop<T> 是一个零成本包装器。这是否意味着我可以取消引用从指向 T 的原始指针转换为 ManuallyDrop<T> 的原始指针？
ManuallyDrop 是 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)] 是 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, i.e., 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
我相信执行此转换是安全的。
真正的问题是为什么你想这样做？拥有指向 ManuallyDrop 结构的指针似乎毫无意义。如果您有一个指向 T 的指针，则不会删除基础值。如果将指针转换为引用（同时确保遵守引用规则），引用也不会删除基础值。
根据文档，ManuallyDrop<T> 是一个零成本包装器。这是否意味着我可以取消引用从指向 T 的原始指针转换为 ManuallyDrop<T> 的原始指针？
ManuallyDrop 是 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)] 是 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, i.e., 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.
[...]
PtrWithCustomZstis also represented exactly like*const Foo
我相信执行此转换是安全的。
真正的问题是为什么你想这样做？拥有指向 ManuallyDrop 结构的指针似乎毫无意义。如果您有一个指向 T 的指针，则不会删除基础值。如果将指针转换为引用（同时确保遵守引用规则），引用也不会删除基础值。