Will the non-lexical lifetime borrow checker release locks prematurely?

问题

I've read What are non-lexical lifetimes?. With the non-lexical borrow checker, the following code compiles:

fn main() {
    let mut scores = vec![1, 2, 3];
    let score = &scores[0]; // borrows `scores`, but never used
                            // its lifetime can end here

    scores.push(4);         // borrows `scores` mutably, and succeeds
}

It seems reasonable in the case above, but when it comes to a mutex lock, we don't want it to be released prematurely.

In the following code, I would like to lock a shared structure first and then execute a closure, mainly to avoid deadlock. However, I'm not sure if the lock will be released prematurely.

use lazy_static::lazy_static; // 1.3.0
use std::sync::Mutex;

struct Something;

lazy_static! {
    static ref SHARED: Mutex<Something> = Mutex::new(Something);
}

pub fn lock_and_execute(f: Box<Fn()>) {
    let _locked = SHARED.lock(); // `_locked` is never used.
                                 // does its lifetime end here?
    f();
}

Does Rust treat locks specially, so that their lifetimes are guaranteed to extend to the end of their scope? Must we use that variable explicitly to avoid premature dropping of the lock, like in the following code?

pub fn lock_and_execute(f: Box<Fn()>) {
    let locked = SHARED.lock(); // - lifetime begins
    f();                        // |
    drop(locked);               // - lifetime ends
}

回答1:

There is a misunderstanding here: NLL (non-lexical lifetimes) affects the borrow-checks, not the actual lifetime of the objects.

Rust uses RAII¹ extensively, and thus the Drop implementation of a number of objects, such as locks, has side-effects which have to occur at a well-determined and predictable point in the flow of execution.

NLL did NOT change the lifetime of such objects, and therefore their destructor is executed at exactly the same point that it was before: at the end of their lexical scope, in reverse order of creation.

NLL did change the understanding of the compiler of the use of lifetimes for the purpose of borrow-checking. This does not, actually, cause any code change; this is purely analysis. This analysis was made more clever, to better recognize the actual scope in which a reference is used:

Prior to NLL, a reference was considered "in use" from the moment it was created to the moment it was dropped, generally its lexical scope (hence the name).
NLL, instead:
- Tries to defer the start of the "in use" span, if possible.
- Ends the "in use" span with the last use of the reference.

In the case of a Ref<'a> (from RefCell), the Ref<'a> will be dropped at the end of the lexical scope, at which point it will use the reference to RefCell to decrement the counter.

NLL does not peel away layers of abstractions, so must consider that any object containing a reference (such as Ref<'a>) may access said reference in its Drop implementation. As a result, any object that contains a reference, such as a lock, will force NLL to consider that the "in use" span of the reference extends until they are dropped.

¹Resource Acquisition Is Initialization, whose original meaning is that once a variable constructor has been executed it has acquired the resources it needed and is not in a half-baked state, and which is generally used to mean that the destruction of said variable will release any resources it owned.

回答2:

Does Rust treat locks specially, so that their lifetimes are guaranteed to extend to the end of their scope?

No. This is the default for every type, and has nothing to do with the borrow checker.

Must we use that variable explicitly to avoid premature dropping of the lock

No.

All you need to do is ensure that the lock guard is bound to a variable. Your example does this (let _lock = ...), so the lock will be dropped at the end of scope. If you had used the _ pattern instead, the lock would have been dropped immediately:

You can prove this for yourself by testing if the lock has indeed been dropped:

pub fn lock_and_execute() {
    let shared = Mutex::new(Something);

    println!("A");
    let _locked = shared.lock().unwrap();

    // If `_locked` was dropped, then we can re-lock it:
    println!("B");
    shared.lock().unwrap();

    println!("C");
}

fn main() {
    lock_and_execute();
}

This code will deadlock, as the same thread attempts to acquire the lock twice.

You could also attempt to use a method that requires &mut self to see that the immutable borrow is still held by the guard, which has not been dropped:

pub fn lock_and_execute() {
    let mut shared = Mutex::new(Something);

    println!("A");
    let _locked = shared.lock().unwrap();

    // If `_locked` was dropped, then we can re-lock it:
    println!("B");
    shared.get_mut().unwrap();

    println!("C");
}

error[E0502]: cannot borrow `shared` as mutable because it is also borrowed as immutable
  --> src/main.rs:13:5
   |
9  |     let _locked = shared.lock().unwrap();
   |                   ------ immutable borrow occurs here
...
13 |     shared.get_mut().unwrap();
   |     ^^^^^^^^^^^^^^^^ mutable borrow occurs here
...
16 | }
   | - immutable borrow might be used here, when `_locked` is dropped and runs the `Drop` code for type `std::sync::MutexGuard`