3 ms·
It was my impression that you can't do it even in unsafe because ownership rules still apply. But apparently I was wrong and there is a way to do it?
by rthomas6 5y ago
It was my impression that you can't do it even in unsafe because ownership rules still apply. But apparently I was wrong and there is a way to do it?
- School-Cotton 5y agoOwnership rules still apply _to references_, but not to raw pointers. This program prints "12". The only difficulty is that Rust's designers have intentionally made this pattern difficult by marking raw pointers as !Send, but you can get around that by wrapping them in a Send type. use std::{time::Duration, ops::{DerefMut, Deref}}; struct SendWrapper<T>(*mut T); impl<T> Deref for SendWrapper<T> { type Target = T; fn deref(&self) -> &Self::Target { unsafe { self.0.as_ref() }.unwrap() } } impl<T> DerefMut for SendWrapper<T> { fn deref_mut(&mut self) -> &mut Self::Target { unsafe { self.0.as_mut() }.unwrap() } } unsafe impl<T> Send for SendWrapper<T> {} fn main() { let mut x = 10; let p1: *mut usize = &mut x; let p2: *mut usize = &mut x; let mut s1 = SendWrapper(p1); let mut s2 = SendWrapper(p2); let j1 = std::thread::spawn(move || { *s1.deref_mut() += 1; }); let j2 = std::thread::spawn(move || { std::thread::sleep(Duration::from_secs(1)); *s2.deref_mut() += 1; }); j1.join().unwrap(); j2.join().unwrap(); println!("x: {}", x); }