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Add cloning example for dot operator behaviour (#292)
Co-authored-by: Yuki Okushi <yuki.okushi@huawei.com>
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src/dot-operator.md

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# The Dot Operator
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The dot operator will perform a lot of magic to convert types. It will perform
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auto-referencing, auto-dereferencing, and coercion until types match.
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The dot operator will perform a lot of magic to convert types.
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It will perform auto-referencing, auto-dereferencing, and coercion until types
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match.
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The detailed mechanics of method lookup are defined [here][method_lookup],
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but here is a brief overview that outlines the main steps.
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TODO: steal information from http://stackoverflow.com/questions/28519997/what-are-rusts-exact-auto-dereferencing-rules/28552082#28552082
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Suppose we have a function `foo` that has a receiver (a `self`, `&self` or
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`&mut self` parameter).
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If we call `value.foo()`, the compiler needs to determine what type `Self` is before
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it can call the correct implementation of the function.
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For this example, we will say that `value` has type `T`.
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We will use [fully-qualified syntax][fqs] to be more clear about exactly which
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type we are calling a function on.
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- First, the compiler checks if it can call `T::foo(value)` directly.
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This is called a "by value" method call.
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- If it can't call this function (for example, if the function has the wrong type
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or a trait isn't implemented for `Self`), then the compiler tries to add in an
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automatic reference.
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This means that the compiler tries `<&T>::foo(value)` and `<&mut T>::foo(value)`.
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This is called an "autoref" method call.
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- If none of these candidates worked, it dereferences `T` and tries again.
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This uses the `Deref` trait - if `T: Deref<Target = U>` then it tries again with
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type `U` instead of `T`.
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If it can't dereference `T`, it can also try _unsizing_ `T`.
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This just means that if `T` has a size parameter known at compile time, it "forgets"
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it for the purpose of resolving methods.
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For instance, this unsizing step can convert `[i32; 2]` into `[i32]` by "forgetting"
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the size of the array.
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Here is an example of the method lookup algorithm:
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```rust,ignore
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let array: Rc<Box<[T; 3]>> = ...;
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let first_entry = array[0];
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```
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How does the compiler actually compute `array[0]` when the array is behind so
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many indirections?
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First, `array[0]` is really just syntax sugar for the [`Index`][index] trait -
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the compiler will convert `array[0]` into `array.index(0)`.
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Now, the compiler checks to see if `array` implements `Index`, so that it can call
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the function.
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Then, the compiler checks if `Rc<Box<[T; 3]>>` implements `Index`, but it
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does not, and neither do `&Rc<Box<[T; 3]>>` or `&mut Rc<Box<[T; 3]>>`.
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Since none of these worked, the compiler dereferences the `Rc<Box<[T; 3]>>` into
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`Box<[T; 3]>` and tries again.
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`Box<[T; 3]>`, `&Box<[T; 3]>`, and `&mut Box<[T; 3]>` do not implement `Index`,
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so it dereferences again.
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`[T; 3]` and its autorefs also do not implement `Index`.
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It can't dereference `[T; 3]`, so the compiler unsizes it, giving `[T]`.
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Finally, `[T]` implements `Index`, so it can now call the actual `index` function.
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Consider the following more complicated example of the dot operator at work:
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```rust
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fn do_stuff<T: Clone>(value: &T) {
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let cloned = value.clone();
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}
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```
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What type is `cloned`?
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First, the compiler checks if it can call by value.
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The type of `value` is `&T`, and so the `clone` function has signature
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`fn clone(&T) -> T`.
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It knows that `T: Clone`, so the compiler finds that `cloned: T`.
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What would happen if the `T: Clone` restriction was removed? It would not be able
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to call by value, since there is no implementation of `Clone` for `T`.
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So the compiler tries to call by autoref.
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In this case, the function has the signature `fn clone(&&T) -> &T` since
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`Self = &T`.
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The compiler sees that `&T: Clone`, and then deduces that `cloned: &T`.
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Here is another example where the autoref behavior is used to create some subtle
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effects:
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```rust
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# use std::sync::Arc;
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#
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#[derive(Clone)]
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struct Container<T>(Arc<T>);
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fn clone_containers<T>(foo: &Container<i32>, bar: &Container<T>) {
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let foo_cloned = foo.clone();
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let bar_cloned = bar.clone();
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}
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```
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What types are `foo_cloned` and `bar_cloned`?
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We know that `Container<i32>: Clone`, so the compiler calls `clone` by value to give
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`foo_cloned: Container<i32>`.
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However, `bar_cloned` actually has type `&Container<T>`.
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Surely this doesn't make sense - we added `#[derive(Clone)]` to `Container`, so it
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must implement `Clone`!
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Looking closer, the code generated by the `derive` macro is (roughly):
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```rust,ignore
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impl<T> Clone for Container<T> where T: Clone {
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fn clone(&self) -> Self {
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Self(Arc::clone(&self.0))
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}
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}
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```
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The derived `Clone` implementation is [only defined where `T: Clone`][clone],
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so there is no implementation for `Container<T>: Clone` for a generic `T`.
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The compiler then looks to see if `&Container<T>` implements `Clone`, which it does.
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So it deduces that `clone` is called by autoref, and so `bar_cloned` has type
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`&Container<T>`.
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We can fix this by implementing `Clone` manually without requiring `T: Clone`:
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```rust,ignore
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impl<T> Clone for Container<T> {
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fn clone(&self) -> Self {
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Self(Arc::clone(&self.0))
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}
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}
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```
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Now, the type checker deduces that `bar_cloned: Container<T>`.
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[fqs]: ../book/ch19-03-advanced-traits.html#fully-qualified-syntax-for-disambiguation-calling-methods-with-the-same-name
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[method_lookup]: https://rustc-dev-guide.rust-lang.org/method-lookup.html
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[index]: ../std/ops/trait.Index.html
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[clone]: ../std/clone/trait.Clone.html#derivable

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