New upstream version 1.20.0+dfsg1

[rustc.git] / src / librustc_borrowck / diagnostics.rs

// Copyright 2014 The Rust Project Developers. See the COPYRIGHT
// file at the top-level directory of this distribution and at
// http://rust-lang.org/COPYRIGHT.
//
// Licensed under the Apache License, Version 2.0 <LICENSE-APACHE or
// http://www.apache.org/licenses/LICENSE-2.0> or the MIT license
// <LICENSE-MIT or http://opensource.org/licenses/MIT>, at your
// option. This file may not be copied, modified, or distributed
// except according to those terms.

#![allow(non_snake_case)]

register_long_diagnostics! {

E0373: r##"
This error occurs when an attempt is made to use data captured by a closure,
when that data may no longer exist. It's most commonly seen when attempting to
return a closure:

```compile_fail,E0373
fn foo() -> Box<Fn(u32) -> u32> {
    let x = 0u32;
    Box::new(|y| x + y)
}
```

Notice that `x` is stack-allocated by `foo()`. By default, Rust captures
closed-over data by reference. This means that once `foo()` returns, `x` no
longer exists. An attempt to access `x` within the closure would thus be
unsafe.

Another situation where this might be encountered is when spawning threads:

```compile_fail,E0373
fn foo() {
    let x = 0u32;
    let y = 1u32;

    let thr = std::thread::spawn(|| {
        x + y
    });
}
```

Since our new thread runs in parallel, the stack frame containing `x` and `y`
may well have disappeared by the time we try to use them. Even if we call
`thr.join()` within foo (which blocks until `thr` has completed, ensuring the
stack frame won't disappear), we will not succeed: the compiler cannot prove
that this behaviour is safe, and so won't let us do it.

The solution to this problem is usually to switch to using a `move` closure.
This approach moves (or copies, where possible) data into the closure, rather
than taking references to it. For example:

```
fn foo() -> Box<Fn(u32) -> u32> {
    let x = 0u32;
    Box::new(move |y| x + y)
}
```

Now that the closure has its own copy of the data, there's no need to worry
about safety.
"##,

E0381: r##"
It is not allowed to use or capture an uninitialized variable. For example:

```compile_fail,E0381
fn main() {
    let x: i32;
    let y = x; // error, use of possibly uninitialized variable
}
```

To fix this, ensure that any declared variables are initialized before being
used. Example:

```
fn main() {
    let x: i32 = 0;
    let y = x; // ok!
}
```
"##,

E0382: r##"
This error occurs when an attempt is made to use a variable after its contents
have been moved elsewhere. For example:

```compile_fail,E0382
struct MyStruct { s: u32 }

fn main() {
    let mut x = MyStruct{ s: 5u32 };
    let y = x;
    x.s = 6;
    println!("{}", x.s);
}
```

Since `MyStruct` is a type that is not marked `Copy`, the data gets moved out
of `x` when we set `y`. This is fundamental to Rust's ownership system: outside
of workarounds like `Rc`, a value cannot be owned by more than one variable.

If we own the type, the easiest way to address this problem is to implement
`Copy` and `Clone` on it, as shown below. This allows `y` to copy the
information in `x`, while leaving the original version owned by `x`. Subsequent
changes to `x` will not be reflected when accessing `y`.

```
#[derive(Copy, Clone)]
struct MyStruct { s: u32 }

fn main() {
    let mut x = MyStruct{ s: 5u32 };
    let y = x;
    x.s = 6;
    println!("{}", x.s);
}
```

Alternatively, if we don't control the struct's definition, or mutable shared
ownership is truly required, we can use `Rc` and `RefCell`:

```
use std::cell::RefCell;
use std::rc::Rc;

struct MyStruct { s: u32 }

fn main() {
    let mut x = Rc::new(RefCell::new(MyStruct{ s: 5u32 }));
    let y = x.clone();
    x.borrow_mut().s = 6;
    println!("{}", x.borrow().s);
}
```

With this approach, x and y share ownership of the data via the `Rc` (reference
count type). `RefCell` essentially performs runtime borrow checking: ensuring
that at most one writer or multiple readers can access the data at any one time.

If you wish to learn more about ownership in Rust, start with the chapter in the
Book:

https://doc.rust-lang.org/book/first-edition/ownership.html
"##,

E0383: r##"
This error occurs when an attempt is made to partially reinitialize a
structure that is currently uninitialized.

For example, this can happen when a drop has taken place:

```compile_fail,E0383
struct Foo {
    a: u32,
}
impl Drop for Foo {
    fn drop(&mut self) { /* ... */ }
}

let mut x = Foo { a: 1 };
drop(x); // `x` is now uninitialized
x.a = 2; // error, partial reinitialization of uninitialized structure `t`
```

This error can be fixed by fully reinitializing the structure in question:

```
struct Foo {
    a: u32,
}
impl Drop for Foo {
    fn drop(&mut self) { /* ... */ }
}

let mut x = Foo { a: 1 };
drop(x);
x = Foo { a: 2 };
```
"##,

E0384: r##"
This error occurs when an attempt is made to reassign an immutable variable.
For example:

```compile_fail,E0384
fn main() {
    let x = 3;
    x = 5; // error, reassignment of immutable variable
}
```

By default, variables in Rust are immutable. To fix this error, add the keyword
`mut` after the keyword `let` when declaring the variable. For example:

```
fn main() {
    let mut x = 3;
    x = 5;
}
```
"##,

/*E0386: r##"
This error occurs when an attempt is made to mutate the target of a mutable
reference stored inside an immutable container.

For example, this can happen when storing a `&mut` inside an immutable `Box`:

```compile_fail,E0386
let mut x: i64 = 1;
let y: Box<_> = Box::new(&mut x);
**y = 2; // error, cannot assign to data in an immutable container
```

This error can be fixed by making the container mutable:

```
let mut x: i64 = 1;
let mut y: Box<_> = Box::new(&mut x);
**y = 2;
```

It can also be fixed by using a type with interior mutability, such as `Cell`
or `RefCell`:

```
use std::cell::Cell;

let x: i64 = 1;
let y: Box<Cell<_>> = Box::new(Cell::new(x));
y.set(2);
```
"##,*/

E0387: r##"
This error occurs when an attempt is made to mutate or mutably reference data
that a closure has captured immutably. Examples of this error are shown below:

```compile_fail,E0387
// Accepts a function or a closure that captures its environment immutably.
// Closures passed to foo will not be able to mutate their closed-over state.
fn foo<F: Fn()>(f: F) { }

// Attempts to mutate closed-over data. Error message reads:
// `cannot assign to data in a captured outer variable...`
fn mutable() {
    let mut x = 0u32;
    foo(|| x = 2);
}

// Attempts to take a mutable reference to closed-over data.  Error message
// reads: `cannot borrow data mutably in a captured outer variable...`
fn mut_addr() {
    let mut x = 0u32;
    foo(|| { let y = &mut x; });
}
```

The problem here is that foo is defined as accepting a parameter of type `Fn`.
Closures passed into foo will thus be inferred to be of type `Fn`, meaning that
they capture their context immutably.

If the definition of `foo` is under your control, the simplest solution is to
capture the data mutably. This can be done by defining `foo` to take FnMut
rather than Fn:

```
fn foo<F: FnMut()>(f: F) { }
```

Alternatively, we can consider using the `Cell` and `RefCell` types to achieve
interior mutability through a shared reference. Our example's `mutable`
function could be redefined as below:

```
use std::cell::Cell;

fn foo<F: Fn()>(f: F) { }

fn mutable() {
    let x = Cell::new(0u32);
    foo(|| x.set(2));
}
```

You can read more about cell types in the API documentation:

https://doc.rust-lang.org/std/cell/
"##,

E0388: r##"
E0388 was removed and is no longer issued.
"##,

E0389: r##"
An attempt was made to mutate data using a non-mutable reference. This
commonly occurs when attempting to assign to a non-mutable reference of a
mutable reference (`&(&mut T)`).

Example of erroneous code:

```compile_fail,E0389
struct FancyNum {
    num: u8,
}

fn main() {
    let mut fancy = FancyNum{ num: 5 };
    let fancy_ref = &(&mut fancy);
    fancy_ref.num = 6; // error: cannot assign to data in a `&` reference
    println!("{}", fancy_ref.num);
}
```

Here, `&mut fancy` is mutable, but `&(&mut fancy)` is not. Creating an
immutable reference to a value borrows it immutably. There can be multiple
references of type `&(&mut T)` that point to the same value, so they must be
immutable to prevent multiple mutable references to the same value.

To fix this, either remove the outer reference:

```
struct FancyNum {
    num: u8,
}

fn main() {
    let mut fancy = FancyNum{ num: 5 };

    let fancy_ref = &mut fancy;
    // `fancy_ref` is now &mut FancyNum, rather than &(&mut FancyNum)

    fancy_ref.num = 6; // No error!

    println!("{}", fancy_ref.num);
}
```

Or make the outer reference mutable:

```
struct FancyNum {
    num: u8
}

fn main() {
    let mut fancy = FancyNum{ num: 5 };

    let fancy_ref = &mut (&mut fancy);
    // `fancy_ref` is now &mut(&mut FancyNum), rather than &(&mut FancyNum)

    fancy_ref.num = 6; // No error!

    println!("{}", fancy_ref.num);
}
```
"##,

E0499: r##"
A variable was borrowed as mutable more than once. Erroneous code example:

```compile_fail,E0499
let mut i = 0;
let mut x = &mut i;
let mut a = &mut i;
// error: cannot borrow `i` as mutable more than once at a time
```

Please note that in rust, you can either have many immutable references, or one
mutable reference. Take a look at
https://doc.rust-lang.org/book/first-edition/references-and-borrowing.html
for more information. Example:


```
let mut i = 0;
let mut x = &mut i; // ok!

// or:
let mut i = 0;
let a = &i; // ok!
let b = &i; // still ok!
let c = &i; // still ok!
```
"##,

E0500: r##"
A borrowed variable was used in another closure. Example of erroneous code:

```compile_fail
fn you_know_nothing(jon_snow: &mut i32) {
    let nights_watch = || {
        *jon_snow = 2;
    };
    let starks = || {
        *jon_snow = 3; // error: closure requires unique access to `jon_snow`
                       //        but it is already borrowed
    };
}
```

In here, `jon_snow` is already borrowed by the `nights_watch` closure, so it
cannot be borrowed by the `starks` closure at the same time. To fix this issue,
you can put the closure in its own scope:

```
fn you_know_nothing(jon_snow: &mut i32) {
    {
        let nights_watch = || {
            *jon_snow = 2;
        };
    } // At this point, `jon_snow` is free.
    let starks = || {
        *jon_snow = 3;
    };
}
```

Or, if the type implements the `Clone` trait, you can clone it between
closures:

```
fn you_know_nothing(jon_snow: &mut i32) {
    let mut jon_copy = jon_snow.clone();
    let nights_watch = || {
        jon_copy = 2;
    };
    let starks = || {
        *jon_snow = 3;
    };
}
```
"##,

E0501: r##"
This error indicates that a mutable variable is being used while it is still
captured by a closure. Because the closure has borrowed the variable, it is not
available for use until the closure goes out of scope.

Note that a capture will either move or borrow a variable, but in this
situation, the closure is borrowing the variable. Take a look at
http://rustbyexample.com/fn/closures/capture.html for more information about
capturing.

Example of erroneous code:

```compile_fail,E0501
fn inside_closure(x: &mut i32) {
    // Actions which require unique access
}

fn outside_closure(x: &mut i32) {
    // Actions which require unique access
}

fn foo(a: &mut i32) {
    let bar = || {
        inside_closure(a)
    };
    outside_closure(a); // error: cannot borrow `*a` as mutable because previous
                        //        closure requires unique access.
}
```

To fix this error, you can place the closure in its own scope:

```
fn inside_closure(x: &mut i32) {}
fn outside_closure(x: &mut i32) {}

fn foo(a: &mut i32) {
    {
        let bar = || {
            inside_closure(a)
        };
    } // borrow on `a` ends.
    outside_closure(a); // ok!
}
```

Or you can pass the variable as a parameter to the closure:

```
fn inside_closure(x: &mut i32) {}
fn outside_closure(x: &mut i32) {}

fn foo(a: &mut i32) {
    let bar = |s: &mut i32| {
        inside_closure(s)
    };
    outside_closure(a);
    bar(a);
}
```

It may be possible to define the closure later:

```
fn inside_closure(x: &mut i32) {}
fn outside_closure(x: &mut i32) {}

fn foo(a: &mut i32) {
    outside_closure(a);
    let bar = || {
        inside_closure(a)
    };
}
```
"##,

E0502: r##"
This error indicates that you are trying to borrow a variable as mutable when it
has already been borrowed as immutable.

Example of erroneous code:

```compile_fail,E0502
fn bar(x: &mut i32) {}
fn foo(a: &mut i32) {
    let ref y = a; // a is borrowed as immutable.
    bar(a); // error: cannot borrow `*a` as mutable because `a` is also borrowed
            //        as immutable
}
```

To fix this error, ensure that you don't have any other references to the
variable before trying to access it mutably:

```
fn bar(x: &mut i32) {}
fn foo(a: &mut i32) {
    bar(a);
    let ref y = a; // ok!
}
```

For more information on the rust ownership system, take a look at
https://doc.rust-lang.org/book/first-edition/references-and-borrowing.html.
"##,

E0503: r##"
A value was used after it was mutably borrowed.

Example of erroneous code:

```compile_fail,E0503
fn main() {
    let mut value = 3;
    // Create a mutable borrow of `value`. This borrow
    // lives until the end of this function.
    let _borrow = &mut value;
    let _sum = value + 1; // error: cannot use `value` because
                          //        it was mutably borrowed
}
```

In this example, `value` is mutably borrowed by `borrow` and cannot be
used to calculate `sum`. This is not possible because this would violate
Rust's mutability rules.

You can fix this error by limiting the scope of the borrow:

```
fn main() {
    let mut value = 3;
    // By creating a new block, you can limit the scope
    // of the reference.
    {
        let _borrow = &mut value; // Use `_borrow` inside this block.
    }
    // The block has ended and with it the borrow.
    // You can now use `value` again.
    let _sum = value + 1;
}
```

Or by cloning `value` before borrowing it:

```
fn main() {
    let mut value = 3;
    // We clone `value`, creating a copy.
    let value_cloned = value.clone();
    // The mutable borrow is a reference to `value` and
    // not to `value_cloned`...
    let _borrow = &mut value;
    // ... which means we can still use `value_cloned`,
    let _sum = value_cloned + 1;
    // even though the borrow only ends here.
}
```

You can find more information about borrowing in the rust-book:
http://doc.rust-lang.org/book/first-edition/references-and-borrowing.html
"##,

E0504: r##"
This error occurs when an attempt is made to move a borrowed variable into a
closure.

Example of erroneous code:

```compile_fail,E0504
struct FancyNum {
    num: u8,
}

fn main() {
    let fancy_num = FancyNum { num: 5 };
    let fancy_ref = &fancy_num;

    let x = move || {
        println!("child function: {}", fancy_num.num);
        // error: cannot move `fancy_num` into closure because it is borrowed
    };

    x();
    println!("main function: {}", fancy_ref.num);
}
```

Here, `fancy_num` is borrowed by `fancy_ref` and so cannot be moved into
the closure `x`. There is no way to move a value into a closure while it is
borrowed, as that would invalidate the borrow.

If the closure can't outlive the value being moved, try using a reference
rather than moving:

```
struct FancyNum {
    num: u8,
}

fn main() {
    let fancy_num = FancyNum { num: 5 };
    let fancy_ref = &fancy_num;

    let x = move || {
        // fancy_ref is usable here because it doesn't move `fancy_num`
        println!("child function: {}", fancy_ref.num);
    };

    x();

    println!("main function: {}", fancy_num.num);
}
```

If the value has to be borrowed and then moved, try limiting the lifetime of
the borrow using a scoped block:

```
struct FancyNum {
    num: u8,
}

fn main() {
    let fancy_num = FancyNum { num: 5 };

    {
        let fancy_ref = &fancy_num;
        println!("main function: {}", fancy_ref.num);
        // `fancy_ref` goes out of scope here
    }

    let x = move || {
        // `fancy_num` can be moved now (no more references exist)
        println!("child function: {}", fancy_num.num);
    };

    x();
}
```

If the lifetime of a reference isn't enough, such as in the case of threading,
consider using an `Arc` to create a reference-counted value:

```
use std::sync::Arc;
use std::thread;

struct FancyNum {
    num: u8,
}

fn main() {
    let fancy_ref1 = Arc::new(FancyNum { num: 5 });
    let fancy_ref2 = fancy_ref1.clone();

    let x = thread::spawn(move || {
        // `fancy_ref1` can be moved and has a `'static` lifetime
        println!("child thread: {}", fancy_ref1.num);
    });

    x.join().expect("child thread should finish");
    println!("main thread: {}", fancy_ref2.num);
}
```
"##,

E0505: r##"
A value was moved out while it was still borrowed.

Erroneous code example:

```compile_fail,E0505
struct Value {}

fn eat(val: Value) {}

fn main() {
    let x = Value{};
    {
        let _ref_to_val: &Value = &x;
        eat(x);
    }
}
```

Here, the function `eat` takes the ownership of `x`. However,
`x` cannot be moved because it was borrowed to `_ref_to_val`.
To fix that you can do few different things:

* Try to avoid moving the variable.
* Release borrow before move.
* Implement the `Copy` trait on the type.

Examples:

```
struct Value {}

fn eat(val: &Value) {}

fn main() {
    let x = Value{};
    {
        let _ref_to_val: &Value = &x;
        eat(&x); // pass by reference, if it's possible
    }
}
```

Or:

```
struct Value {}

fn eat(val: Value) {}

fn main() {
    let x = Value{};
    {
        let _ref_to_val: &Value = &x;
    }
    eat(x); // release borrow and then move it.
}
```

Or:

```
#[derive(Clone, Copy)] // implement Copy trait
struct Value {}

fn eat(val: Value) {}

fn main() {
    let x = Value{};
    {
        let _ref_to_val: &Value = &x;
        eat(x); // it will be copied here.
    }
}
```

You can find more information about borrowing in the rust-book:
http://doc.rust-lang.org/book/first-edition/references-and-borrowing.html
"##,

E0506: r##"
This error occurs when an attempt is made to assign to a borrowed value.

Example of erroneous code:

```compile_fail,E0506
struct FancyNum {
    num: u8,
}

fn main() {
    let mut fancy_num = FancyNum { num: 5 };
    let fancy_ref = &fancy_num;
    fancy_num = FancyNum { num: 6 };
    // error: cannot assign to `fancy_num` because it is borrowed

    println!("Num: {}, Ref: {}", fancy_num.num, fancy_ref.num);
}
```

Because `fancy_ref` still holds a reference to `fancy_num`, `fancy_num` can't
be assigned to a new value as it would invalidate the reference.

Alternatively, we can move out of `fancy_num` into a second `fancy_num`:

```
struct FancyNum {
    num: u8,
}

fn main() {
    let mut fancy_num = FancyNum { num: 5 };
    let moved_num = fancy_num;
    fancy_num = FancyNum { num: 6 };

    println!("Num: {}, Moved num: {}", fancy_num.num, moved_num.num);
}
```

If the value has to be borrowed, try limiting the lifetime of the borrow using
a scoped block:

```
struct FancyNum {
    num: u8,
}

fn main() {
    let mut fancy_num = FancyNum { num: 5 };

    {
        let fancy_ref = &fancy_num;
        println!("Ref: {}", fancy_ref.num);
    }

    // Works because `fancy_ref` is no longer in scope
    fancy_num = FancyNum { num: 6 };
    println!("Num: {}", fancy_num.num);
}
```

Or by moving the reference into a function:

```
struct FancyNum {
    num: u8,
}

fn main() {
    let mut fancy_num = FancyNum { num: 5 };

    print_fancy_ref(&fancy_num);

    // Works because function borrow has ended
    fancy_num = FancyNum { num: 6 };
    println!("Num: {}", fancy_num.num);
}

fn print_fancy_ref(fancy_ref: &FancyNum){
    println!("Ref: {}", fancy_ref.num);
}
```
"##,

E0507: r##"
You tried to move out of a value which was borrowed. Erroneous code example:

```compile_fail,E0507
use std::cell::RefCell;

struct TheDarkKnight;

impl TheDarkKnight {
    fn nothing_is_true(self) {}
}

fn main() {
    let x = RefCell::new(TheDarkKnight);

    x.borrow().nothing_is_true(); // error: cannot move out of borrowed content
}
```

Here, the `nothing_is_true` method takes the ownership of `self`. However,
`self` cannot be moved because `.borrow()` only provides an `&TheDarkKnight`,
which is a borrow of the content owned by the `RefCell`. To fix this error,
you have three choices:

* Try to avoid moving the variable.
* Somehow reclaim the ownership.
* Implement the `Copy` trait on the type.

Examples:

```
use std::cell::RefCell;

struct TheDarkKnight;

impl TheDarkKnight {
    fn nothing_is_true(&self) {} // First case, we don't take ownership
}

fn main() {
    let x = RefCell::new(TheDarkKnight);

    x.borrow().nothing_is_true(); // ok!
}
```

Or:

```
use std::cell::RefCell;

struct TheDarkKnight;

impl TheDarkKnight {
    fn nothing_is_true(self) {}
}

fn main() {
    let x = RefCell::new(TheDarkKnight);
    let x = x.into_inner(); // we get back ownership

    x.nothing_is_true(); // ok!
}
```

Or:

```
use std::cell::RefCell;

#[derive(Clone, Copy)] // we implement the Copy trait
struct TheDarkKnight;

impl TheDarkKnight {
    fn nothing_is_true(self) {}
}

fn main() {
    let x = RefCell::new(TheDarkKnight);

    x.borrow().nothing_is_true(); // ok!
}
```

Moving a member out of a mutably borrowed struct will also cause E0507 error:

```compile_fail,E0507
struct TheDarkKnight;

impl TheDarkKnight {
    fn nothing_is_true(self) {}
}

struct Batcave {
    knight: TheDarkKnight
}

fn main() {
    let mut cave = Batcave {
        knight: TheDarkKnight
    };
    let borrowed = &mut cave;

    borrowed.knight.nothing_is_true(); // E0507
}
```

It is fine only if you put something back. `mem::replace` can be used for that:

```
# struct TheDarkKnight;
# impl TheDarkKnight { fn nothing_is_true(self) {} }
# struct Batcave { knight: TheDarkKnight }
use std::mem;

let mut cave = Batcave {
    knight: TheDarkKnight
};
let borrowed = &mut cave;

mem::replace(&mut borrowed.knight, TheDarkKnight).nothing_is_true(); // ok!
```

You can find more information about borrowing in the rust-book:
http://doc.rust-lang.org/book/first-edition/references-and-borrowing.html
"##,

E0508: r##"
A value was moved out of a non-copy fixed-size array.

Example of erroneous code:

```compile_fail,E0508
struct NonCopy;

fn main() {
    let array = [NonCopy; 1];
    let _value = array[0]; // error: cannot move out of type `[NonCopy; 1]`,
                           //        a non-copy fixed-size array
}
```

The first element was moved out of the array, but this is not
possible because `NonCopy` does not implement the `Copy` trait.

Consider borrowing the element instead of moving it:

```
struct NonCopy;

fn main() {
    let array = [NonCopy; 1];
    let _value = &array[0]; // Borrowing is allowed, unlike moving.
}
```

Alternatively, if your type implements `Clone` and you need to own the value,
consider borrowing and then cloning:

```
#[derive(Clone)]
struct NonCopy;

fn main() {
    let array = [NonCopy; 1];
    // Now you can clone the array element.
    let _value = array[0].clone();
}
```
"##,

E0509: r##"
This error occurs when an attempt is made to move out of a value whose type
implements the `Drop` trait.

Example of erroneous code:

```compile_fail,E0509
struct FancyNum {
    num: usize
}

struct DropStruct {
    fancy: FancyNum
}

impl Drop for DropStruct {
    fn drop(&mut self) {
        // Destruct DropStruct, possibly using FancyNum
    }
}

fn main() {
    let drop_struct = DropStruct{fancy: FancyNum{num: 5}};
    let fancy_field = drop_struct.fancy; // Error E0509
    println!("Fancy: {}", fancy_field.num);
    // implicit call to `drop_struct.drop()` as drop_struct goes out of scope
}
```

Here, we tried to move a field out of a struct of type `DropStruct` which
implements the `Drop` trait. However, a struct cannot be dropped if one or
more of its fields have been moved.

Structs implementing the `Drop` trait have an implicit destructor that gets
called when they go out of scope. This destructor may use the fields of the
struct, so moving out of the struct could make it impossible to run the
destructor. Therefore, we must think of all values whose type implements the
`Drop` trait as single units whose fields cannot be moved.

This error can be fixed by creating a reference to the fields of a struct,
enum, or tuple using the `ref` keyword:

```
struct FancyNum {
    num: usize
}

struct DropStruct {
    fancy: FancyNum
}

impl Drop for DropStruct {
    fn drop(&mut self) {
        // Destruct DropStruct, possibly using FancyNum
    }
}

fn main() {
    let drop_struct = DropStruct{fancy: FancyNum{num: 5}};
    let ref fancy_field = drop_struct.fancy; // No more errors!
    println!("Fancy: {}", fancy_field.num);
    // implicit call to `drop_struct.drop()` as drop_struct goes out of scope
}
```

Note that this technique can also be used in the arms of a match expression:

```
struct FancyNum {
    num: usize
}

enum DropEnum {
    Fancy(FancyNum)
}

impl Drop for DropEnum {
    fn drop(&mut self) {
        // Destruct DropEnum, possibly using FancyNum
    }
}

fn main() {
    // Creates and enum of type `DropEnum`, which implements `Drop`
    let drop_enum = DropEnum::Fancy(FancyNum{num: 10});
    match drop_enum {
        // Creates a reference to the inside of `DropEnum::Fancy`
        DropEnum::Fancy(ref fancy_field) => // No error!
            println!("It was fancy-- {}!", fancy_field.num),
    }
    // implicit call to `drop_enum.drop()` as drop_enum goes out of scope
}
```
"##,

E0596: r##"
This error occurs because you tried to mutably borrow a non-mutable variable.

Example of erroneous code:

```compile_fail,E0596
let x = 1;
let y = &mut x; // error: cannot borrow mutably
```

In here, `x` isn't mutable, so when we try to mutably borrow it in `y`, it
fails. To fix this error, you need to make `x` mutable:

```
let mut x = 1;
let y = &mut x; // ok!
```
"##,

E0597: r##"
This error occurs because a borrow was made inside a variable which has a
greater lifetime than the borrowed one.

Example of erroneous code:

```compile_fail,E0597
struct Foo<'a> {
    x: Option<&'a u32>,
}

let mut x = Foo { x: None };
let y = 0;
x.x = Some(&y); // error: `y` does not live long enough
```

In here, `x` is created before `y` and therefore has a greater lifetime. Always
keep in mind that values in a scope are dropped in the opposite order they are
created. So to fix the previous example, just make the `y` lifetime greater than
the `x`'s one:

```
struct Foo<'a> {
    x: Option<&'a u32>,
}

let y = 0;
let mut x = Foo { x: None };
x.x = Some(&y);
```
"##,

}

register_diagnostics! {
//    E0385, // {} in an aliasable location
    E0524, // two closures require unique access to `..` at the same time
    E0594, // cannot assign to {}
    E0595, // closure cannot assign to {}
    E0598, // lifetime of {} is too short to guarantee its contents can be...
}