% Closures

Rust not only has named functions, but anonymous functions as well. Anonymous
functions that have an associated environment are called 'closures', because they
close over an environment. Rust has a really great implementation of them, as
we'll see.

# Syntax

Closures look like this:

```rust
let plus_one = |x: i32| x + 1;

assert_eq!(2, plus_one(1));
```

We create a binding, `plus_one`, and assign it to a closure. The closure's
arguments go between the pipes (`|`), and the body is an expression, in this
case, `x + 1`. Remember that `{ }` is an expression, so we can have multi-line
closures too:

```rust
let plus_two = |x| {
    let mut result: i32 = x;

    result += 1;
    result += 1;

    result
};

assert_eq!(4, plus_two(2));
```

You'll notice a few things about closures that are a bit different than regular
functions defined with `fn`. The first of which is that we did not need to
annotate the types of arguments the closure takes or the values it returns. We
can:

```rust
let plus_one = |x: i32| -> i32 { x + 1 };

assert_eq!(2, plus_one(1));
```

But we don't have to. Why is this? Basically, it was chosen for ergonomic reasons.
While specifying the full type for named functions is helpful with things like
documentation and type inference, the types of closures are rarely documented
since they’re anonymous, and they don’t cause the kinds of error-at-a-distance
that inferring named function types can.

The second is that the syntax is similar, but a bit different. I've added spaces
here to make them look a little closer:

```rust
fn  plus_one_v1 (  x: i32 ) -> i32 { x + 1 }
let plus_one_v2 = |x: i32 | -> i32 { x + 1 };
let plus_one_v3 = |x: i32 |          x + 1  ;
```

Small differences, but they're similar in ways.

# Closures and their environment

Closures are called such because they 'close over their environment.' It
looks like this:

```rust
let num = 5;
let plus_num = |x: i32| x + num;

assert_eq!(10, plus_num(5));
```

This closure, `plus_num`, refers to a `let` binding in its scope: `num`. More
specifically, it borrows the binding. If we do something that would conflict
with that binding, we get an error. Like this one:

```rust,ignore
let mut num = 5;
let plus_num = |x: i32| x + num;

let y = &mut num;
```

Which errors with:

```text
error: cannot borrow `num` as mutable because it is also borrowed as immutable
    let y = &mut num;
                 ^~~
note: previous borrow of `num` occurs here due to use in closure; the immutable
  borrow prevents subsequent moves or mutable borrows of `num` until the borrow
  ends
    let plus_num = |x| x + num;
                   ^~~~~~~~~~~
note: previous borrow ends here
fn main() {
    let mut num = 5;
    let plus_num = |x| x + num;
    
    let y = &mut num;
}
^
```

A verbose yet helpful error message! As it says, we can't take a mutable borrow
on `num` because the closure is already borrowing it. If we let the closure go
out of scope, we can:

```rust
let mut num = 5;
{
    let plus_num = |x: i32| x + num;

} // plus_num goes out of scope, borrow of num ends

let y = &mut num;
```

If your closure requires it, however, Rust will take ownership and move
the environment instead:

```rust,ignore
let nums = vec![1, 2, 3];

let takes_nums = || nums;

println!("{:?}", nums);
```

This gives us:

```text
note: `nums` moved into closure environment here because it has type
  `[closure(()) -> collections::vec::Vec<i32>]`, which is non-copyable
let takes_nums = || nums;
                    ^~~~~~~
```

`Vec<T>` has ownership over its contents, and therefore, when we refer to it
in our closure, we have to take ownership of `nums`. It's the same as if we'd
passed `nums` to a function that took ownership of it.

## `move` closures

We can force our closure to take ownership of its environment with the `move`
keyword:

```rust
let num = 5;

let owns_num = move |x: i32| x + num;
```

Now, even though the keyword is `move`, the variables follow normal move semantics.
In this case, `5` implements `Copy`, and so `owns_num` takes ownership of a copy
of `num`. So what's the difference?

```rust
let mut num = 5;

{ 
    let mut add_num = |x: i32| num += x;

    add_num(5);
}

assert_eq!(10, num);
```

So in this case, our closure took a mutable reference to `num`, and then when
we called `add_num`, it mutated the underlying value, as we'd expect. We also
needed to declare `add_num` as `mut` too, because we’re mutating its
environment.

We also had to declare `add_num` as mut, since we will be modifying its
environment.

If we change to a `move` closure, it's different:

```rust
let mut num = 5;

{ 
    let mut add_num = move |x: i32| num += x;

    add_num(5);
}

assert_eq!(5, num);
```

We only get `5`. Rather than taking a mutable borrow out on our `num`, we took
ownership of a copy.

Another way to think about `move` closures: they give a closure its own stack
frame.  Without `move`, a closure may be tied to the stack frame that created
it, while a `move` closure is self-contained. This means that you cannot
generally return a non-`move` closure from a function, for example.

But before we talk about taking and returning closures, we should talk some more
about the way that closures are implemented. As a systems language, Rust gives
you tons of control over what your code does, and closures are no different.

# Closure implementation

Rust's implementation of closures is a bit different than other languages. They
are effectively syntax sugar for traits. You'll want to make sure to have read
the [traits chapter][traits] before this one, as well as the chapter on [static
and dynamic dispatch][dispatch], which talks about trait objects.

[traits]: traits.html
[dispatch]: static-and-dynamic-dispatch.html

Got all that? Good.

The key to understanding how closures work under the hood is something a bit
strange: Using `()` to call a function, like `foo()`, is an overloadable
operator. From this, everything else clicks into place. In Rust, we use the
trait system to overload operators. Calling functions is no different. We have
three separate traits to overload with:

```rust
# mod foo {
pub trait Fn<Args> : FnMut<Args> {
    extern "rust-call" fn call(&self, args: Args) -> Self::Output;
}

pub trait FnMut<Args> : FnOnce<Args> {
    extern "rust-call" fn call_mut(&mut self, args: Args) -> Self::Output;
}

pub trait FnOnce<Args> {
    type Output;

    extern "rust-call" fn call_once(self, args: Args) -> Self::Output;
}
# }
```

You'll notice a few differences between these traits, but a big one is `self`:
`Fn` takes `&self`, `FnMut` takes `&mut self`, and `FnOnce` takes `self`. This
covers all three kinds of `self` via the usual method call syntax. But we've
split them up into three traits, rather than having a single one. This gives us
a large amount of control over what kind of closures we can take.

The `|| {}` syntax for closures is sugar for these three traits. Rust will
generate a struct for the environment, `impl` the appropriate trait, and then
use it.

# Taking closures as arguments

Now that we know that closures are traits, we already know how to accept and
return closures: just like any other trait!

This also means that we can choose static vs dynamic dispatch as well. First,
let's write a function which takes something callable, calls it, and returns
the result:

```rust
fn call_with_one<F>(some_closure: F) -> i32
    where F : Fn(i32) -> i32 {

    some_closure(1)
}

let answer = call_with_one(|x| x + 2);

assert_eq!(3, answer);
```

We pass our closure, `|x| x + 2`, to `call_with_one`. It just does what it
suggests: it calls the closure, giving it `1` as an argument.

Let's examine the signature of `call_with_one` in more depth:

```rust
fn call_with_one<F>(some_closure: F) -> i32
#    where F : Fn(i32) -> i32 {
#    some_closure(1) }
```

We take one parameter, and it has the type `F`. We also return a `i32`. This part
isn't interesting. The next part is:

```rust
# fn call_with_one<F>(some_closure: F) -> i32
    where F : Fn(i32) -> i32 {
#   some_closure(1) }
```

Because `Fn` is a trait, we can bound our generic with it. In this case, our closure
takes a `i32` as an argument and returns an `i32`, and so the generic bound we use
is `Fn(i32) -> i32`.

There's one other key point here: because we're bounding a generic with a
trait, this will get monomorphized, and therefore, we'll be doing static
dispatch into the closure. That's pretty neat. In many langauges, closures are
inherently heap allocated, and will always involve dynamic dispatch. In Rust,
we can stack allocate our closure environment, and statically dispatch the
call. This happens quite often with iterators and their adapters, which often
take closures as arguments.

Of course, if we want dynamic dispatch, we can get that too. A trait object
handles this case, as usual:

```rust
fn call_with_one(some_closure: &Fn(i32) -> i32) -> i32 {
    some_closure(1)
}

let answer = call_with_one(&|x| x + 2);

assert_eq!(3, answer);
```

Now we take a trait object, a `&Fn`. And we have to make a reference
to our closure when we pass it to `call_with_one`, so we use `&||`.

# Returning closures

It’s very common for functional-style code to return closures in various
situations. If you try to return a closure, you may run into an error. At
first, it may seem strange, but we'll figure it out. Here's how you'd probably
try to return a closure from a function:

```rust,ignore
fn factory() -> (Fn(i32) -> Vec<i32>) {
    let vec = vec![1, 2, 3];

    |n| vec.push(n)
}

let f = factory();

let answer = f(4);
assert_eq!(vec![1, 2, 3, 4], answer);
```

This gives us these long, related errors:

```text
error: the trait `core::marker::Sized` is not implemented for the type
`core::ops::Fn(i32) -> collections::vec::Vec<i32>` [E0277]
f = factory();
^
note: `core::ops::Fn(i32) -> collections::vec::Vec<i32>` does not have a
constant size known at compile-time
f = factory();
^
error: the trait `core::marker::Sized` is not implemented for the type
`core::ops::Fn(i32) -> collections::vec::Vec<i32>` [E0277]
factory() -> (Fn(i32) -> Vec<i32>) {
             ^~~~~~~~~~~~~~~~~~~~~
note: `core::ops::Fn(i32) -> collections::vec::Vec<i32>` does not have a constant size known at compile-time
fa ctory() -> (Fn(i32) -> Vec<i32>) {
              ^~~~~~~~~~~~~~~~~~~~~

```

In order to return something from a function, Rust needs to know what
size the return type is. But since `Fn` is a trait, it could be various
things of various sizes: many different types can implement `Fn`. An easy
way to give something a size is to take a reference to it, as references
have a known size. So we'd write this:

```rust,ignore
fn factory() -> &(Fn(i32) -> Vec<i32>) {
    let vec = vec![1, 2, 3];

    |n| vec.push(n)
}

let f = factory();

let answer = f(4);
assert_eq!(vec![1, 2, 3, 4], answer);
```

But we get another error:

```text
error: missing lifetime specifier [E0106]
fn factory() -> &(Fn(i32) -> i32) {
                ^~~~~~~~~~~~~~~~~
```

Right. Because we have a reference, we need to give it a lifetime. But
our `factory()` function takes no arguments, so elision doesn't kick in
here. What lifetime can we choose? `'static`:

```rust,ignore
fn factory() -> &'static (Fn(i32) -> i32) {
    let num = 5;

    |x| x + num
}

let f = factory();

let answer = f(1);
assert_eq!(6, answer);
```

But we get another error:

```text
error: mismatched types:
 expected `&'static core::ops::Fn(i32) -> i32`,
    found `[closure <anon>:7:9: 7:20]`
(expected &-ptr,
    found closure) [E0308]
         |x| x + num
         ^~~~~~~~~~~

```

This error is letting us know that we don't have a `&'static Fn(i32) -> i32`,
we have a `[closure <anon>:7:9: 7:20]`. Wait, what?

Because each closure generates its own environment `struct` and implementation
of `Fn` and friends, these types are anonymous. They exist just solely for
this closure. So Rust shows them as `closure <anon>`, rather than some
autogenerated name.

But why doesn't our closure implement `&'static Fn`? Well, as we discussed before,
closures borrow their environment. And in this case, our environment is based
on a stack-allocated `5`, the `num` variable binding. So the borrow has a lifetime
of the stack frame. So if we returned this closure, the function call would be
over, the stack frame would go away, and our closure is capturing an environment
of garbage memory!

So what to do? This _almost_ works:

```rust,ignore
fn factory() -> Box<Fn(i32) -> i32> {
    let num = 5;

    Box::new(|x| x + num)
}
# fn main() {
let f = factory();

let answer = f(1);
assert_eq!(6, answer);
# }
```

We use a trait object, by `Box`ing up the `Fn`. There's just one last problem:

```text
error: `num` does not live long enough
Box::new(|x| x + num)
         ^~~~~~~~~~~
```

We still have a reference to the parent stack frame. With one last fix, we can
make this work:

```rust
fn factory() -> Box<Fn(i32) -> i32> {
    let num = 5;

    Box::new(move |x| x + num)
}
# fn main() {
let f = factory();

let answer = f(1);
assert_eq!(6, answer);
# }
```

By making the inner closure a `move Fn`, we create a new stack frame for our
closure. By `Box`ing it up, we've given it a known size, and allowing it to
escape our stack frame.
