Sending type that is !Send

1

Hi guys,
How to send a type, that is not send? I've tried:
Arc<RwLock>//no go
Arc<Mutex>//no go

Which surprised me a lot, as I thought that it is the way to send types that do not implement Send trait.

2

The whole point of !Send is that you can't send it to another thread. Being able to send an !Send type to another thread is unsound. For example:

let a = Rc::new(1);
let b = a.clone();
thread::spawn(|| {
    b.clone(); // This is UB if it were to compile as there is a data-race with the other thread when incrementing the reference count.
});
a.clone(); // This would data-race with the other thread when incrementing the reference count.

This code doesn't actually compile as Rc is !Send to prevent concurrent (non-atomic) accesses to the reference count.

3

It's possible with something like fragile, i.e. with something that prevents this type from actually being accessed after sending. It's unsound otherwise, no matter the wrapper used.

Arc and RwLock/Mutex are used for another case - to make Send + !Sync type effectively Sync, i.e. to share between threads something that can be sent.

5 Likes
4

Thanks for the suggestion and explanation.

5

Would that also imply that let's say type that has !Send insite it, like raw ptr is !Send itself, and type that as a members has all members that are Send is also Send itself?

6

Yes, that's how the automatic implementations are set up.

Depending on how they're being used, however, it is sometimes OK to send something like a raw pointer to another thread if you're also providing some other access control mechanism. For these cases, you can unsafe impl Send for MyTypeWhichContainsAPointer {}— It's then up to you to ensure that whatever you're doing with that pointer can't result in a data race, use after free, etc. when used in a multi-threaded way.

7

Thanks for confirmation.

8

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