Reputation: 442
fn main() {
let mut foo = 1;
let mut func = || foo += 1;
while foo < 5 {
func();
}
}
error[E0503]: cannot use `foo` because it was mutably borrowed
--> src/main.rs:5:11
|
4 | let mut func = || foo += 1;
| -- borrow of `foo` occurs here
5 | while foo < 5 {
| ^^^ use of borrowed `foo`
I understand why this isn't working, but I'm looking for a way to bypass the borrow checker somehow. Is there a way to use a closure here? Is there a good alternative besides using a function? I've got a situation where I have to change several variables.
Upvotes: 4
Views: 770
Reputation: 430358
No, you cannot do this. While the closure has the mutable borrow, nothing else may access that variable.
Instead...
You can use a Cell
or a RefCell
:
use std::cell::Cell;
fn main() {
let foo = Cell::new(1);
let func = || foo.set(foo.get() + 1);
while foo.get() < 5 {
func();
}
}
See also:
You can bake the comparison into the closure:
fn main() {
let mut foo = 1;
let mut func = || {
foo += 1;
foo < 5
};
while func() {}
}
struct Thing(i32);
impl Thing {
fn do_it(&mut self) {
self.0 += 1
}
fn test_it(&self) -> bool {
self.0 < 5
}
}
fn main() {
let mut foo = Thing(1);
while foo.test_it() {
foo.do_it();
}
}
Upvotes: 2
Reputation: 38576
One option you have is to pass a mutable reference to the closure rather than implicitly borrowing it by environment capture:
fn main() {
let mut foo = 1;
let func = |foo: &mut i32| *foo += 1;
while foo < 5 {
func(&mut foo);
}
}
Upvotes: 8