Comment by Asooka

To elaborate on that some more, safe Rust can guarantee that mutable aliasing never happens, without solving the halting program, because it forbids some programs that could've been considered legal. Here's an example of a function that's allowed:

    fn foo() {
        let mut x = 42;
        let mut mutable_references = Vec::new();
        let test: bool = rand::random();
        if test {
            mutable_references.push(&mut x);
        } else {
            mutable_references.push(&mut x);
        }
    }

Because only one if/else branch is ever allowed to execute, the compiler can see "lexically" that only one mutable reference to `x` is created, and `foo` compiles. But this other function that's "obviously" equivalent doesn't compile:

    fn bar() {
        let mut x = 42;
        let mut mutable_references = Vec::new();
        let test: bool = rand::random();
        if test {
            mutable_references.push(&mut x);
        }
        if !test {
            mutable_references.push(&mut x); // error: cannot borrow `x` as mutable more than once at a time
        }
    }

The Rust compiler doesn't do the analysis necessary to see that only one of those branches can execute, so it conservatively assumes that both of them can, and it refuses to compile `bar`. To do things like `bar`, you have to either refactor them to look more like `foo`, or else you have to use `unsafe` code.

It requires that the libraries you use do not have UB. If you have no unsafe, but your library does, you can get UB.

https://github.com/rust-lang/rust/pull/139553

This is why it may be a good idea to run MIRI on your Rust code, even when it has no unsafe, since a library like Rust stdlib might have UB.