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multithreading - How to ensure a piece of code is always used by one thread at any given time?

I'm writing a function that needs to use mutable static variables to work (a weird implementation of a message loop). In order to allow only one writer to access those variables at any given time, I need to make the access to this function exclusive (to the first thread that accesses it).

Using AtomicBool

My first guess was to use an AtomicBool:

use std::sync::atomic::{Ordering, AtomicBool};

static FLAG: AtomicBool = AtomicBool::new(false);

fn my_exclusive_function() {
    if FLAG.load(Ordering::SeqCst) {
        panic!("Haha, too late!")
    }
    
    FLAG.store(true, Ordering::SeqCst);

    /* Do stuff */
}

This code has an obvious flaw: if two threads happen to read FLAG at the same time, they would both think that this is ok for them to continue.

Using Mutex<()>

Then I thought about using a Mutex<()> for its lock.

use std::sync::Mutex;

static FLAG: Mutex<()> = Mutex::new(());

fn my_exclusive_function() {
    let _lock = FLAG.try_lock() {
        Ok(lock) => lock,
        Err(_) => panic!("Haha, too late!"),
    };

    /* Do stuff */

    // `_lock` gets dropped here, another thread can now access the function
}

There are two problems here:

  1. Mutex::new is not a const function which makes me unable to initialize the Mutex. I could use a library like lazy_static here but if I can avoid an additional dependency, it would be great.
  2. Even if I use lazy_static, a user-defined function (the handler inside of the message loop actually) could panic and poison the mutex. Here again, I could use a thread-party library like parking_lot but that would be yet anther additional dependency.

I'm willing to use those if there is no alternatives but if I can avoid it, that is better.

What would the right way to do this look like? Is it just a matter of Ordering?

Related question that did not help me


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