# Module std::pin 1.33.0[−][src]

Types that pin data to its location in memory.

It is sometimes useful to have objects that are guaranteed not to move, in the sense that their placement in memory does not change, and can thus be relied upon. A prime example of such a scenario would be building self-referential structs, as moving an object with pointers to itself will invalidate them, which could cause undefined behavior.

At a high level, a `Pin<P>` ensures that the pointee of any pointer type `P` has a stable location in memory, meaning it cannot be moved elsewhere and its memory cannot be deallocated until it gets dropped. We say that the pointee is "pinned". Things get more subtle when discussing types that combine pinned with non-pinned data; see below for more details.

By default, all types in Rust are movable. Rust allows passing all types by-value, and common smart-pointer types such as `Box<T>` and `&mut T` allow replacing and moving the values they contain: you can move out of a `Box<T>`, or you can use `mem::swap`. `Pin<P>` wraps a pointer type `P`, so `Pin``<``Box``<T>>` functions much like a regular `Box<T>`: when a `Pin``<``Box``<T>>` gets dropped, so do its contents, and the memory gets deallocated. Similarly, `Pin``<&mut T>` is a lot like `&mut T`. However, `Pin<P>` does not let clients actually obtain a `Box<T>` or `&mut T` to pinned data, which implies that you cannot use operations such as `mem::swap`:

    use std::pin::Pin;
    fn swap_pins<T>(x: Pin<&mut T>, y: Pin<&mut T>) {
        // `mem::swap` needs `&mut T`, but we cannot get it.
        // We are stuck, we cannot swap the contents of these references.
        // We could use `Pin::get_unchecked_mut`, but that is unsafe for a reason:
        // we are not allowed to use it for moving things out of the `Pin`.
    }

Run

It is worth reiterating that `Pin<P>` does not change the fact that a Rust compiler considers all types movable. `mem::swap` remains callable for any `T`. Instead, `Pin<P>` prevents certain values (pointed to by pointers wrapped in `Pin<P>`) from being moved by making it impossible to call methods that require `&mut T` on them (like `mem::swap`).

`Pin<P>` can be used to wrap any pointer type `P`, and as such it interacts with `Deref` and `DerefMut`. A `Pin<P>` where `P: Deref` should be considered as a "`P`-style pointer" to a pinned `P::Target` -- so, a `Pin``<``Box``<T>>` is an owned pointer to a pinned `T`, and a `Pin``<``Rc``<T>>` is a reference-counted pointer to a pinned `T`. For correctness, `Pin<P>` relies on the implementations of `Deref` and `DerefMut` not to move out of their `self` parameter, and only ever to return a pointer to pinned data when they are called on a pinned pointer.

# `Unpin`

Many types are always freely movable, even when pinned, because they do not rely on having a stable address. This includes all the basic types (like `bool`, `i32`, and references) as well as types consisting solely of these types. Types that do not care about pinning implement the `Unpin` auto-trait, which cancels the effect of `Pin<P>`. For `T: Unpin`, `Pin``<``Box``<T>>` and `Box<T>` function identically, as do `Pin``<&mut T>` and `&mut T`.

Note that pinning and `Unpin` only affect the pointed-to type `P::Target`, not the pointer type `P` itself that got wrapped in `Pin<P>`. For example, whether or not `Box<T>` is `Unpin` has no effect on the behavior of `Pin``<``Box``<T>>` (here, `T` is the pointed-to type).

# Example: self-referential struct

Before we go into more details to explain the guarantees and choices associated with `Pin<T>`, we discuss some examples for how it might be used. Feel free to skip to where the theoretical discussion continues.

    use std::pin::Pin;
    use std::marker::PhantomPinned;
    use std::ptr::NonNull;

    // This is a self-referential struct because the slice field points to the data field.
    // We cannot inform the compiler about that with a normal reference,
    // as this pattern cannot be described with the usual borrowing rules.
    // Instead we use a raw pointer, though one which is known not to be null,
    // as we know it's pointing at the string.
    struct Unmovable {
        data: String,
        slice: NonNull<String>,
        _pin: PhantomPinned,
    }

    impl Unmovable {
        // To ensure the data doesn't move when the function returns,
        // we place it in the heap where it will stay for the lifetime of the object,
        // and the only way to access it would be through a pointer to it.
        fn new(data: String) -> Pin<Box<Self>> {
            let res = Unmovable {
                data,
                // we only create the pointer once the data is in place
                // otherwise it will have already moved before we even started
                slice: NonNull::dangling(),
                _pin: PhantomPinned,
            };
            let mut boxed = Box::pin(res);

            let slice = NonNull::from(&boxed.data);
            // we know this is safe because modifying a field doesn't move the whole struct
            unsafe {
                let mut_ref: Pin<&mut Self> = Pin::as_mut(&mut boxed);
                Pin::get_unchecked_mut(mut_ref).slice = slice;
            }
            boxed
        }
    }

    let unmoved = Unmovable::new("hello".to_string());
    // The pointer should point to the correct location,
    // so long as the struct hasn't moved.
    // Meanwhile, we are free to move the pointer around.
    let mut still_unmoved = unmoved;
    assert_eq!(still_unmoved.slice, NonNull::from(&still_unmoved.data));

    // Since our type doesn't implement Unpin, this will fail to compile:
    // let mut new_unmoved = Unmovable::new("world".to_string());
    // std::mem::swap(&mut *still_unmoved, &mut *new_unmoved);

Run

# Example: intrusive doubly-linked list

In an intrusive doubly-linked list, the collection does not actually allocate the memory for the elements itself. Allocation is controlled by the clients, and elements can live on a stack frame that lives shorter than the collection does.

To make this work, every element has pointers to its predecessor and successor in the list. Elements can only be added when they are pinned, because moving the elements around would invalidate the pointers. Moreover, the `Drop` implementation of a linked list element will patch the pointers of its predecessor and successor to remove itself from the list.

Crucially, we have to be able to rely on `drop` being called. If an element could be deallocated or otherwise invalidated without calling `drop`, the pointers into it from its neighboring elements would become invalid, which would break the data structure.

Therefore, pinning also comes with a `drop`-related guarantee.

# `Drop` guarantee

The purpose of pinning is to be able to rely on the placement of some data in memory. To make this work, not just moving the data is restricted; deallocating, repurposing, or otherwise invalidating the memory used to store the data is restricted, too. Concretely, for pinned data you have to maintain the invariant that its memory will not get invalidated or repurposed from the moment it gets pinned until when `drop` is called. Only once `drop` returns or panics, the memory may be reused.

Memory can be "invalidated" by deallocation, but also by replacing a `Some(v)` by `None`, or calling `Vec::set_len` to "kill" some elements off of a vector. It can be repurposed by using `ptr::write` to overwrite it without calling the destructor first. None of this is allowed for pinned data without calling `drop`.

This is exactly the kind of guarantee that the intrusive linked list from the previous section needs to function correctly.

Notice that this guarantee does not mean that memory does not leak! It is still completely okay not ever to call `drop` on a pinned element (e.g., you can still call `mem::forget` on a `Pin``<``Box``<T>>`). In the example of the doubly-linked list, that element would just stay in the list. However you may not free or reuse the storage without calling `drop`.

# `Drop` implementation

If your type uses pinning (such as the two examples above), you have to be careful when implementing `Drop`. The `drop` function takes `&mut self`, but this is called even if your type was previously pinned! It is as if the compiler automatically called `Pin::get_unchecked_mut`.

This can never cause a problem in safe code because implementing a type that relies on pinning requires unsafe code, but be aware that deciding to make use of pinning in your type (for example by implementing some operation on `Pin``<&Self>` or `Pin``<&mut Self>`) has consequences for your `Drop` implementation as well: if an element of your type could have been pinned, you must treat `Drop` as implicitly taking `Pin``<&mut Self>`.

For example, you could implement `Drop` as follows:

    impl Drop for Type {
        fn drop(&mut self) {
            // `new_unchecked` is okay because we know this value is never used
            // again after being dropped.
            inner_drop(unsafe { Pin::new_unchecked(self)});
            fn inner_drop(this: Pin<&mut Type>) {
                // Actual drop code goes here.
            }
        }
    }

Run

The function `inner_drop` has the type that `drop` should have, so this makes sure that you do not accidentally use `self`/`this` in a way that is in conflict with pinning.

Moreover, if your type is `#[repr(packed)]`, the compiler will automatically move fields around to be able to drop them. It might even do that for fields that happen to be sufficiently aligned. As a consequence, you cannot use pinning with a `#[repr(packed)]` type.

# Projections and Structural Pinning

When working with pinned structs, the question arises how one can access the fields of that struct in a method that takes just `Pin``<&mut Struct>`. The usual approach is to write helper methods (so called projections) that turn `Pin``<&mut Struct>` into a reference to the field, but what type should that reference have? Is it `Pin``<&mut Field>` or `&mut Field`? The same question arises with the fields of an `enum`, and also when considering container/wrapper types such as `Vec<T>`, `Box<T>`, or `RefCell<T>`. (This question applies to both mutable and shared references, we just use the more common case of mutable references here for illustration.)

It turns out that it is actually up to the author of the data structure to decide whether the pinned projection for a particular field turns `Pin``<&mut Struct>` into `Pin``<&mut Field>` or `&mut Field`. There are some constraints though, and the most important constraint is consistency: every field can be either projected to a pinned reference, or have pinning removed as part of the projection. If both are done for the same field, that will likely be unsound!

As the author of a data structure you get to decide for each field whether pinning "propagates" to this field or not. Pinning that propagates is also called "structural", because it follows the structure of the type. In the following subsections, we describe the considerations that have to be made for either choice.

## Pinning is not structural for `field`

It may seem counter-intuitive that the field of a pinned struct might not be pinned, but that is actually the easiest choice: if a `Pin``<&mut Field>` is never created, nothing can go wrong! So, if you decide that some field does not have structural pinning, all you have to ensure is that you never create a pinned reference to that field.

Fields without structural pinning may have a projection method that turns `Pin``<&mut Struct>` into `&mut Field`:

    impl Struct {
        fn pin_get_field(self: Pin<&mut Self>) -> &mut Field {
            // This is okay because `field` is never considered pinned.
            unsafe { &mut self.get_unchecked_mut().field }
        }
    }

Run

You may also `impl Unpin for Struct` even if the type of `field` is not `Unpin`. What that type thinks about pinning is not relevant when no `Pin``<&mut Field>` is ever created.

## Pinning is structural for `field`

The other option is to decide that pinning is "structural" for `field`, meaning that if the struct is pinned then so is the field.

This allows writing a projection that creates a `Pin``<&mut Field>`, thus witnessing that the field is pinned:

    impl Struct {
        fn pin_get_field(self: Pin<&mut Self>) -> Pin<&mut Field> {
            // This is okay because `field` is pinned when `self` is.
            unsafe { self.map_unchecked_mut(|s| &mut s.field) }
        }
    }

Run

However, structural pinning comes with a few extra requirements:

1. The struct must only be `Unpin` if all the structural fields are `Unpin`. This is the default, but `Unpin` is a safe trait, so as the author of the struct it is your responsibility not to add something like `impl<T> Unpin for Struct<T>`. (Notice that adding a projection operation requires unsafe code, so the fact that `Unpin` is a safe trait does not break the principle that you only have to worry about any of this if you use `unsafe`.)

2. The destructor of the struct must not move structural fields out of its argument. This is the exact point that was raised in the previous section: `drop` takes `&mut self`, but the struct (and hence its fields) might have been pinned before. You have to guarantee that you do not move a field inside your `Drop` implementation. In particular, as explained previously, this means that your struct must not be `#[repr(packed)]`. See that section for how to write `drop` in a way that the compiler can help you not accidentally break pinning.

3. You must make sure that you uphold the `Drop` guarantee: once your struct is pinned, the memory that contains the content is not overwritten or deallocated without calling the content's destructors. This can be tricky, as witnessed by `VecDeque<T>`: the destructor of `VecDeque<T>` can fail to call `drop` on all elements if one of the destructors panics. This violates the `Drop` guarantee, because it can lead to elements being deallocated without their destructor being called. (`VecDeque<T>` has no pinning projections, so this does not cause unsoundness.)

4. You must not offer any other operations that could lead to data being moved out of the structural fields when your type is pinned. For example, if the struct contains an `Option<T>` and there is a `take`-like operation with type `fn(Pin<&mut Struct<T>>) -> Option<T>`, that operation can be used to move a `T` out of a pinned `Struct<T>` -- which means pinning cannot be structural for the field holding this data.

   For a more complex example of moving data out of a pinned type, imagine if `RefCell<T>` had a method `fn get_pin_mut(self: Pin<&mut Self>) -> Pin<&mut T>`. Then we could do the following:

   ⓘ

       fn exploit_ref_cell<T>(rc: Pin<&mut RefCell<T>>) {
           { let p = rc.as_mut().get_pin_mut(); } // Here we get pinned access to the `T`.
           let rc_shr: &RefCell<T> = rc.into_ref().get_ref();
           let b = rc_shr.borrow_mut();
           let content = &mut *b; // And here we have `&mut T` to the same data.
       }

   Run

   This is catastrophic, it means we can first pin the content of the `RefCell<T>` (using `RefCell::get_pin_mut`) and then move that content using the mutable reference we got later.

## Examples

For a type like `Vec<T>`, both possibilities (structural pinning or not) make sense. A `Vec<T>` with structural pinning could have `get_pin`/`get_pin_mut` methods to get pinned references to elements. However, it could not allow calling `pop` on a pinned `Vec<T>` because that would move the (structurally pinned) contents! Nor could it allow `push`, which might reallocate and thus also move the contents.

A `Vec<T>` without structural pinning could `impl<T> Unpin for Vec<T>`, because the contents are never pinned and the `Vec<T>` itself is fine with being moved as well. At that point pinning just has no effect on the vector at all.

In the standard library, pointer types generally do not have structural pinning, and thus they do not offer pinning projections. This is why `Box<T>: Unpin` holds for all `T`. It makes sense to do this for pointer types, because moving the `Box<T>` does not actually move the `T`: the `Box<T>` can be freely movable (aka `Unpin`) even if the `T` is not. In fact, even `Pin``<``Box``<T>>` and `Pin``<&mut T>` are always `Unpin` themselves, for the same reason: their contents (the `T`) are pinned, but the pointers themselves can be moved without moving the pinned data. For both `Box<T>` and `Pin``<``Box``<T>>`, whether the content is pinned is entirely independent of whether the pointer is pinned, meaning pinning is not structural.

When implementing a `Future` combinator, you will usually need structural pinning for the nested futures, as you need to get pinned references to them to call `poll`. But if your combinator contains any other data that does not need to be pinned, you can make those fields not structural and hence freely access them with a mutable reference even when you just have `Pin``<&mut Self>` (such as in your own `poll` implementation).

## Structs

Pin | A pinned pointer.