CODEWITHSUNDEEP
rust
Kai Sellgren
Kai Sellgren

Reputation: 30212

Is there a way to get two *mut pointers pointing to the same memory address?

I am trying to implement a Doubly Linked List for fun in Rust.

I faced an issue with the ownership system, the inability to alias two pointers. I realize the benefits of pointer aliasing, but I'd like to make an exception this time.

I'd like to know if there is any way to have two *mut A pointers pointing to the same memory address.

For example, right now I have this:

self.first = unsafe { mem::transmute(new_node) };
// How do I set self.last to point at the same new_node?

I am aware of Rc and RefCell types, but I'd like to avoid using them in data structures as they have run-time overhead.

Upvotes: 0

Views: 1368

Answers (2)

Shepmaster
Shepmaster

Reputation: 430634

While I agree with dbaupp that you should provide more context because this isn't what you should do, here's the answer to the exact question you asked:

fn main() {
    let mut a: u32 = 42;
    let b: *mut u32 = &mut a;
    let c: *mut u32 = &mut a;
    println!("{:?}, {:?}", b, c);
}

Now you have two *mut u32 pointers pointing at the same memory address. And without unsafe! Of course, if you want to use the pointers, you now have to do a lot more thinking to decide if you've completely broken all the assumptions and guarantees that the compiler expects! :-)

Upvotes: 2

Jorge Israel Peña
Jorge Israel Peña

Reputation: 38576

I'm not sure exactly what you mean. &mut is the one that has the aliasing enforcing. *mut can be copied like anything else, dereferencing a raw pointer of course is unsafe, but you should be able to do this:

fn main() {
    let mut x: i32 = 5i32;
    let xptr: *mut i32 = &mut x as *mut i32;

    let xptr_copy: *mut i32 = xptr;

    unsafe {
        println!("xptr: {}", *xptr);
        println!("xptr_copy: {}", *xptr_copy);
    }
}

playpen

Upvotes: 1

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