[nemetroid]

> The effect of referring to a copy of the object when locking, unlocking, or destroying it is undefined.

https://pubs.opengroup.org/onlinepubs/9699919799/functions/V...

I.e., if I pthread_mutex_init(&some_addr, ...), I cannot then copy the bits from some_addr to some_other_addr and then pthread_mutex_lock(&some_other_addr). Hence not movable.

> Moving a mutex is otherwise non-sensical once the mutex is visible

What does "visible" mean here? In Rust, in any circumstance where a move is possible, there are no other references to that object, hence it is safe to move.

[tialaramex]

Well, technically if you only have a mutable borrow (it's not your object) then you can't move from it unless you replace it somehow. If you have two such borrows you can swap them, if the type implements Default you can take from one borrow and this replaces it with its default and if you've some other way to make one you can replace the one you've got a reference to with that one, but if you can't make a new one and don't have one to replace it with, then too bad, no moving the one you've got a reference to.

[nemetroid]

You're right and I edited my comment.

[wahern]

> What does "visible" mean here? In Rust, in any circumstance where a move is possible, there are no other references to that object, hence it is safe to move.

And other than during construction or initialization (of the mutex object, containing object, or related state), how common is it in Rust to pass a mutex by value? If you can pass by value then the mutex isn't (can't) protect anything. I'm struggling to think of a scenario where you'd want to do this, or at least why the inability to do so is a meaningful impediment (outside construction/initialization, that is). I understand Rust is big on pass-by-value, but when the need for a mutex enters the fray, it's because you're sharing or about to share, and thus passing by reference.

[MindSpunk]

Depends on the program, and it can be a very useful tool.

Rust has Mutex::get_mut(&mut self) which allows getting the inner &mut T without locking. Having a &mut Mutex<T> implies you can get &mut T without locks. Being able to treat Mutex<T> like any other value means you can use the whole suite of Rust's ownership tools to pass the value through your program.

Perhaps you temporarily move the Mutex into a shared data structure so it can be used on multiple threads, then take it back out later in a serial part of your program to get mutable access without locks. It's a lot easier to move Mutex<T> around than &mut Mutex<T> if you're going to then share it and un-share it.

Also It's impossible to construct a Mutex without moving at least once, as Rust doesn't guarantee return value optimization. All moves in Rust are treated as memcpy that 'destroy' the old value. There's no way to even assign 'let v = Mutex::new()' without a move so it's also a hard functional requirement.

[mastax]

You can pass the mutex by value and it does continue to protect its value.

https://play.rust-lang.org/?version=stable&mode=debug&editio...