#### Note: this error code is no longer emitted by the compiler.
This error occurs when an attempt is made to mutate or mutably reference data
that a closure has captured immutably.
Erroneous code example:
```compile_fail
// 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/