> There's an implicit assumption here that all the values returned from the iterator at

> for i in &mut v

> are disjoint. But iterators, in general, do not guarantee disjoint outputs. An iterator which returned each reference value twice, for example, is a valid iterator. But used in the context above, two threads would receive mutable access to the same element of a vector. That seems to violate a core Rust safety assumption.

Your assumptions here are wrong. "An iterator which returned each reference value twice" could only be implemented using unsafe Rust, and code emitting multiple co-occuring mutable references to the same value are instant-UB. In other words, a "safe abstraction" being able to do this is actually not a safe abstraction at all. This would be a bug in the implementation of the custom iterator.

In fact, if you tried to prove such an implementation using the formal tools (Iris) from RustBelt you wouldn't be able to do it. I know, because I've done it as part of my course work under Lars Birkedal.

I'm not sure if you misunderstanding how `&mut v` turns into an iterator, which values it spits out etc, or what is happening. Let's look at how that code is translated into a program and typechecked. First, let's look at the for-loop itself. Anything on the form of

for x in y

must have an implementation of the `IntoIterator` trait for the type of the expression `y`. Whatever that type might be. In this case `y` is actually `&mut v`, where `v`'s type is `Vec<i32>`. Luckily enough, an implementation of `IntoIterator` exists for any `&mut Vec<T>`. [0]

Looking at the expanded signature we can see that the associated type `Item` is `&mut T`. This tells us that the types the resulting iterator will produce are `&mut i32`. Keep in mind that this reference is into the backing memory store of the `Vec<i32>` of `v`. This is where Rust's semantic help us.

You say the following:

> But iterators, in general, do not guarantee disjoint outputs.

The issue with that statement is that iterators don't have to. Rust guarantees that condition. Safe Rust can't return multiple mutable references to the same value. Further, any unsafe implementation, even when exposed through a safe interface, giving this behavior is a bug. It would be great if the compiler could statically verify unsafe code too, but if it could there wouldn't be a need to call it unsafe.

> The question is whether a piece of encapsulated code is always memory-safe for all uses. There's no guarantee of that from the language.

That is ultimately impossible to prove completely and statically. Turing and Gödel made sure of that. However, some things that the compiler can't prove we can still prove as outside observers. Often these things will boil down to controlling and checking the states of several variables before performing an unsafe operation, but knowing that in a given context the operation is safe.

The first part of the above quote is partially true. However, if we assume a given unsafe implementation is indeed safe then the rules of Rust's semantics make the composition of any number of safe APIs a completely safe composition. This is a strong, useful statement because it means the very few places where unsafe is required can be easily checked, and every other piece of code doesn't have to spend mental energy for the programmer to consider the safety of their implementation.

[0] https://doc.rust-lang.org/std/iter/trait.IntoIterator.html#impl-IntoIterator-19