将Option <＆mut T>转换为* mut T.

Question

I'm writing a Rust wrapper around a C library and while doing so I'm trying to take advantage of the "nullable pointer optimization" mentioned in The Book , but I can't find a good way to convert Option<&T> to *const T and Option<&mut T> to *mut T like what they're describing.

What I really want is to be able to call Some(&foo) as *const _ . Unfortunately that doesn't work, so the next best thing I can think of is a trait on Option<T> that enables me to call Some(&foo).as_ptr() . The following code is a working definition and implementation for that trait:

use std::ptr;

trait AsPtr<T> {
    fn as_ptr(&self) -> *const T;
}

impl<'a, T> AsPtr<T> for Option<&'a T> {
    fn as_ptr(&self) -> *const T {
        match *self {
            Some(val) => val as *const _,
            None => ptr::null(),
        }
    }
}

Now that I can call Some(&foo).as_ptr() to get a *const _ , I want to be able to call Some(&mut foo).as_ptr() to get a *mut _ . The following is the new trait I made to do this:

trait AsMutPtr<T> {
    fn as_mut_ptr(&self) -> *mut T;
}

impl<'a, T> AsMutPtr<T> for Option<&'a mut T> {
    fn as_mut_ptr(&self) -> *mut T {
        match *self {
            Some(val) => val as *mut _,
            None => ptr::null_mut(),
        }
    }
}

The problem is, the AsMutPtr trait won't compile. When I try, I get the following error:

error[E0507]: cannot move out of borrowed content
  --> src/lib.rs:22:15
   |
22 |         match *self {
   |               ^^^^^
   |               |
   |               cannot move out of borrowed content
   |               help: consider removing the `*`: `self`
23 |             Some(val) => val as *mut _,
   |                  --- data moved here
   |
note: move occurs because `val` has type `&mut T`, which does not implement the `Copy` trait
  --> src/lib.rs:23:18
   |
23 |             Some(val) => val as *mut _,
   |                  ^^^

I don't see what changed between the two traits that causes it to fail — I didn't think adding mut would make that big a difference. I tried adding a ref , but that just causes a different error, and I wouldn't expect to need that anyway.

Why doesn't the AsMutPtr trait work?

Answer 1

Unfortunately, writing the trait impl for &mut T instead of &T does make a big difference. &mut T , as opposed to &T , is not Copy , therefore you cannot extract it out of a shared reference directly:

& &T      --->  &T
& &mut T  -/->  &mut T

This is fairly natural - otherwise aliasing of mutable references would be possible, which violates Rust borrowing rules.

You may ask where that outer & comes from. It actually comes from &self in as_mut_ptr() method. If you have an immutable reference to something, even if that something contains mutable references inside it, you won't be able to use them to mutate the data behind them. This also would be a violation of borrowing semantics.

Unfortunately, I see no way to do this without unsafe. You need to have &mut T "by value" in order to cast it to *mut T , but you can't get it "by value" through a shared reference. Therefore, I suggest you to use ptr::read() :

use std::ptr;

impl<'a, T> AsMutPtr<T> for Option<&'a mut T> {
    fn as_mut_ptr(&self) -> *mut T {
        match *self {
            Some(ref val) => unsafe { ptr::read(val) as *mut _ },
            None => ptr::null_mut(),
        }
    }
}

val here is & &mut T because of ref qualifier in the pattern, therefore ptr::read(val) returns &mut T , aliasing the mutable reference. I think it is okay if it gets converted to a raw pointer immediately and does not leak out, but even though the result would be a raw pointer, it still means that you have two aliased mutable pointers. You should be very careful with what you do with them.

Alternatively, you may modify AsMutPtr::as_mut_ptr() to consume its target by value:

trait AsMutPtr<T> {
    fn as_mut_ptr(self) -> *mut T;
}

impl<'a, T> AsMutPtr<T> for Option<&'a mut T> {
    fn as_mut_ptr(self) -> *mut T {
        match self {
            Some(value) => value as *mut T,
            None => ptr::null_mut()
        }
    }
}

However, in this case Option<&mut T> will be consumed by as_mut_ptr() . This may not be feasible if, for example, this Option<&mut T> is stored in a structure. I'm not really sure whether it is possible to somehow perform reborrowing manually with Option<&mut T> as opposed to just &mut T (it won't be triggered automatically); if it is possible, then by-value as_mut_ptr() is probably the best overall solution.

Answer 2

The problem is that you are reading an &mut out of an & , but &mut s are not Copy so must be moved - and you can't move out of a const reference. This actually explains Vladimir Matveev insight about &&mut → & s in terms of more fundamental properties.

This is actually relatively simply solved. If you can read a *const _ , you can read a *mut _ . The two are the same type, bar a flag that says "be careful, this is being shared". Since dereferences are unsafe either way, there's actually no reason to stop you casting between the two.

So you can actually do

match *self {
    Some(ref val) => val as *const _ as *mut _,
    None => ptr::null_mut(),
}

Read the immutable reference, make it an immutable pointer and then make it a mutable pointer. Plus it's all done through safe Rust so we know we're not breaking any aliasing rules.

That said, it's probably a really bad idea to actually use that *mut pointer until the &mut reference is gone. I would be very hesitant with this, and try to rethink your wrapper to something safer.

Answer 3

Will this do what you expect?

trait AsMutPtr<T> {
    fn as_mut_ptr(self) -> *mut T;
}

impl<T> AsMutPtr<T> for Option<*mut T> {
    fn as_mut_ptr(self) -> *mut T {
        match self {
            Some(val) => val as *mut _,
            None => ptr::null_mut(),
        }
    }
}

Answer 4

To avoid unsafe code, change the trait to accept &mut self instead of either self or &self :

trait AsMutPtr<T> {
    fn as_mut_ptr(&mut self) -> *mut T;
}

impl<'a, T> AsMutPtr<T> for Option<&'a mut T> {
    fn as_mut_ptr(&mut self) -> *mut T {
        match self {
            Some(v) => *v,
            None => ptr::null_mut(),
        }
    }
}

You could also reduce the implementation to one line, if you felt like it:

fn as_mut_ptr(&mut self) -> *mut T {
    self.as_mut().map_or_else(ptr::null_mut, |v| *v)
}

This can be used to give you multiple mutable raw pointers from the same source. This can easily lead you to causing mutable aliasing , so be careful:

fn example(mut v: Option<&mut u8>) {
    let b = v.as_mut_ptr();
    let a = v.as_mut_ptr();
}

I would recommend not converting the immutable reference to a mutable pointer as this is very likely to cause undefined behavior.