Cell get returns a copy so why unsafe?

help
1

Hello,

Noob question: Cell::get returns a copy of T so why do we need an unsafe scope in the fonction? Copying T means that we could get a copy from UnsafeCell with no risk at all. At no point we are manipulating memory that is currenly being held or modified by somebody else.

2

It requires an unsafe block because all calls of UnsafeCell::get require an unsafe block. What you have written is simply an argument that the unsafe block is correct — and not that the unsafe block is unnecessary.

4 Likes
3

Actually, it is safe, the deref of the pointer is not. But the point is correct.

1 Like
4

In other words: you know that all you're doing with the pointer returned from UnsafeCell::get() is copying it, but the compiler doesn't know that (it could prove that in this case, but not in other lots of cases). You could also return a reference referencing it, which would be invalid.

1 Like
5

Right, it's the use of the pointer. The point still stands. It's unsafe because any use of a raw pointer is unsafe.

6

it's the dereference of the pointer that's unsafe, using a pointer can be perfectly safe:

pub fn f(ptr: *const u8) {
    let mut map = HashMap::new();
    map.insert(ptr, "My Pointer");
    dbg!(map);
}
7

This topic was automatically closed 90 days after the last reply. We invite you to open a new topic if you have further questions or comments.