OSAllocatedUnfairLock has two different methods for executing a block of code with the lock held: withLock() and withLockUnchecked(). The only difference between the two is that the former has the closure and return type marked as Sendable. withLockUnchecked() has a comment (that is for reasons I do not understand not visible in the documentation) saying: /// This method does not enforce sendability requirement /// on closure body and its return type. /// The caller of this method is responsible for ensuring references /// to non-sendables from closure uphold the Sendability contract. What I do not understand here is why Sendable conformance would be needed. These function should call this block synchronously in the same context, and return the return value through a normal function return. None of this should require the types to be Sendable. This seems to be supported by this paragraph of the documentation for OSAllocatedUnfairLock: /// This lock must be unlocked from the same thread that locked it. As such, it /// is unsafe to use `lock()` / `unlock()` across an `await` suspension point. /// Instead, use `withLock` to enforce that the lock is only held within /// a synchronous scope. So, why does this Sendable requirement exist, and in practice, if I did want to use withLockUnchecked, how would I "uphold the Sendability contract"? To summarise the question in a more concise way: Is there an example where using withLockUnchecked() would actually cause a problem, and using withLock() instead would catch that problem?