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See #12882 and rust-lang/rust#125897. When r: &mut T, from_ref(r) is equivalent to from_ref(r as &T). Thus, it will no longer be safe to later cast the resultant *const T into a *mut T. Instead, when r: &mut T, the user should use from_mut(r).const_cast() to get a *const T.
Granted, it usually doesn't matter, as we usualy don't cast *const T to *mut T, but when we do, it matters a lot.
Advantage
This is safe:
let p = ptr::from_mut(r).const_cast();
...
let mut_p = p as *mut T;
Whereas this may not be safe:
let p = ptr::from_ref(r);
let mut_p = p as *mut T;
Drawbacks
None that I'm aware of.
Example
Original code:
use core::ptr;
fn main() {
let mut x = 123u8;
let r = &mut x;
let p = ptr::from_ref(r);
let p_mut = p as *mut T; // Potential UB from this point.
}
Improved code:
- let p = ptr::from_ref(r);+ let p = ptr::from_mut(r).const_cast();
The text was updated successfully, but these errors were encountered:
nitpick: the word for well-behaved unsafe code is "sound" (specifically unsafe code that cannot be made to invoke undefined behavior by any safe code is considered "sound")
What it does
See #12882 and rust-lang/rust#125897. When
r: &mut T
,from_ref(r)
is equivalent tofrom_ref(r as &T)
. Thus, it will no longer be safe to later cast the resultant*const T
into a*mut T
. Instead, whenr: &mut T
, the user should usefrom_mut(r).const_cast()
to get a*const T
.Granted, it usually doesn't matter, as we usualy don't cast
*const T
to*mut T
, but when we do, it matters a lot.Advantage
This is safe:
Whereas this may not be safe:
Drawbacks
None that I'm aware of.
Example
Original code:
Improved code:
The text was updated successfully, but these errors were encountered: