# The Dot Operator The dot operator will perform a lot of magic to convert types. It will perform auto-referencing, auto-dereferencing, and coercion until types match. The detailed mechanics of method lookup are defined [here][method_lookup], but here is a brief overview that outlines the main steps. Suppose we have a function `foo` that has a receiver (a `self`, `&self` or `&mut self` parameter). If we call `value.foo()`, the compiler needs to determine what type `Self` is before it can call the correct implementation of the function. For this example, we will say that `value` has type `T`. We will use [fully-qualified syntax][fqs] to be more clear about exactly which type we are calling a function on. - First, the compiler checks if it can call `T::foo(value)` directly. This is called a "by value" method call. - If it can't call this function (for example, if the function has the wrong type or a trait isn't implemented for `Self`), then the compiler tries to add in an automatic reference. This means that the compiler tries `<&T>::foo(value)` and `<&mut T>::foo(value)`. This is called an "autoref" method call. - If none of these candidates worked, it dereferences `T` and tries again. This uses the `Deref` trait - if `T: Deref` then it tries again with type `U` instead of `T`. If it can't dereference `T`, it can also try _unsizing_ `T`. This just means that if `T` has a size parameter known at compile time, it "forgets" it for the purpose of resolving methods. For instance, this unsizing step can convert `[i32; 2]` into `[i32]` by "forgetting" the size of the array. Here is an example of the method lookup algorithm: ```rust,ignore let array: Rc> = ...; let first_entry = array[0]; ``` How does the compiler actually compute `array[0]` when the array is behind so many indirections? First, `array[0]` is really just syntax sugar for the [`Index`][index] trait - the compiler will convert `array[0]` into `array.index(0)`. Now, the compiler checks to see if `array` implements `Index`, so that it can call the function. Then, the compiler checks if `Rc>` implements `Index`, but it does not, and neither do `&Rc>` or `&mut Rc>`. Since none of these worked, the compiler dereferences the `Rc>` into `Box<[T; 3]>` and tries again. `Box<[T; 3]>`, `&Box<[T; 3]>`, and `&mut Box<[T; 3]>` do not implement `Index`, so it dereferences again. `[T; 3]` and its autorefs also do not implement `Index`. It can't dereference `[T; 3]`, so the compiler unsizes it, giving `[T]`. Finally, `[T]` implements `Index`, so it can now call the actual `index` function. Consider the following more complicated example of the dot operator at work: ```rust fn do_stuff(value: &T) { let cloned = value.clone(); } ``` What type is `cloned`? First, the compiler checks if it can call by value. The type of `value` is `&T`, and so the `clone` function has signature `fn clone(&T) -> T`. It knows that `T: Clone`, so the compiler finds that `cloned: T`. What would happen if the `T: Clone` restriction was removed? It would not be able to call by value, since there is no implementation of `Clone` for `T`. So the compiler tries to call by autoref. In this case, the function has the signature `fn clone(&&T) -> &T` since `Self = &T`. The compiler sees that `&T: Clone`, and then deduces that `cloned: &T`. Here is another example where the autoref behavior is used to create some subtle effects: ```rust # use std::sync::Arc; # #[derive(Clone)] struct Container(Arc); fn clone_containers(foo: &Container, bar: &Container) { let foo_cloned = foo.clone(); let bar_cloned = bar.clone(); } ``` What types are `foo_cloned` and `bar_cloned`? We know that `Container: Clone`, so the compiler calls `clone` by value to give `foo_cloned: Container`. However, `bar_cloned` actually has type `&Container`. Surely this doesn't make sense - we added `#[derive(Clone)]` to `Container`, so it must implement `Clone`! Looking closer, the code generated by the `derive` macro is (roughly): ```rust,ignore impl Clone for Container where T: Clone { fn clone(&self) -> Self { Self(Arc::clone(&self.0)) } } ``` The derived `Clone` implementation is [only defined where `T: Clone`][clone], so there is no implementation for `Container: Clone` for a generic `T`. The compiler then looks to see if `&Container` implements `Clone`, which it does. So it deduces that `clone` is called by autoref, and so `bar_cloned` has type `&Container`. We can fix this by implementing `Clone` manually without requiring `T: Clone`: ```rust,ignore impl Clone for Container { fn clone(&self) -> Self { Self(Arc::clone(&self.0)) } } ``` Now, the type checker deduces that `bar_cloned: Container`. [fqs]: ../book/ch19-03-advanced-traits.html#fully-qualified-syntax-for-disambiguation-calling-methods-with-the-same-name [method_lookup]: https://rustc-dev-guide.rust-lang.org/method-lookup.html [index]: ../std/ops/trait.Index.html [clone]: ../std/clone/trait.Clone.html#derivable