with_lock

Deadlock freedom

Docs

Example

Say you have this code:

use std::sync::Mutex;

fn main() {
    let a = Mutex::new(2);
    let b = Mutex::new(3);
    let a_lock = a.lock().unwrap();
    let b_lock = b.lock().unwrap();
    println!("{:?}", *a_lock + *b_lock);
    let a_lock_2 = a.lock().unwrap();
    let b_lock_2 = b.lock().unwrap();
    println!("{:?}", *a_lock_2 + *b_lock_2);
}

That code will log 5 once, when it should log twice. As you can see here, it is deadlocking.

However, we can prevent this by replacing our manual calls of .lock with .with_lock. Code that wouldn't error would look something like:

use std::sync::Mutex;
use with_lock::WithLock;

fn main() {
    let a = WithLock::<i64>::new(Mutex::new(2));
    let b = WithLock::<i64>::new(Mutex::new(3));
    let a_lock = a.with_lock(|s| *s);
    let b_lock = b.with_lock(|s| *s);
    println!("{:?}", a_lock + b_lock);
    let a_lock_2 = a.with_lock(|s| *s);
    let b_lock_2 = b.with_lock(|s| *s);
    println!("{:?}", a_lock_2 + b_lock_2);
}

That code would log 5 twice. This is an example of how it can prevent deadlocks.

Minimal code changes

For the people that want little to no code changes, with_lock exposes a custom MutexCell type.

Code that would produce deadlocks would look like this:

use std::sync::Mutex;

fn main() {
    let a = Mutex::new(2);
    let b = Mutex::new(3);
    let a_lock = a.lock().unwrap();
    let b_lock = b.lock().unwrap();
    println!("{:?}", *a_lock + *b_lock);
    let a_lock_2 = a.lock().unwrap();
    let b_lock_2 = b.lock().unwrap();
    println!("{:?}", *a_lock_2 + *b_lock_2);
}

And using the custom MutexCell type that wouldn't deadlock would look like:

use with_lock::MutexCell;

fn main() {
    let a = MutexCell::new(2);
    let b = MutexCell::new(3);
    let a_locked = a.get();
    let b_locked = b.get();
    println!("{:?}", a_locked + b_locked);
    let a_lock_2 = a.get();
    let b_lock_2 = b.get();
    println!("{:?}", a_lock_2 + b_lock_2);
}