Bonus: recent talk from Ralf Jung on his group's efforts to precisely specify Rust's operational semantics in executable form in a dialect of Rust: https://youtube.com/watch?v=yoeuW_dSe0o
How true is this really?
Torvalds has argued for a long time that strict aliasing rules in C are more trouble than they're worth, I find his arguments compelling. Here's one of many examples: https://lore.kernel.org/all/CAHk-=wgq1DvgNVoodk7JKc6BuU1m9Un... (the entire thread worth reading if you find this sort of thing interesting)
Is Rust somehow fundamentally different? Based on limited experience, it seems not (at least, when unsafe is involved...).
And it seams to not be the case on the stable compiler version?
fn write(x: &mut i32) {*x = 10}
fn main() {
let x = &mut 0;
let y = x as *mut i32;
//write(x); // this should use the mention implicit twophase borrow
*x = 10; // this should not and therefore be rejected by the compiler
unsafe {*y = 15 };
}
[1] https://github.com/Voultapher/sort-research-rs/blob/main/wri... Miri column
[2] https://github.com/rust-lang/rust/blob/6b3ae3f6e45a33c2d95fa...
> The problem with unsafe code is that it can do things like this:
fn main() {
let mut x = 42;
let ptr = &mut x as *mut i32;
let val = unsafe { write_both(&mut *ptr, &mut *ptr) };
println!("{val}");
}
No it can't? Using pointers to coexist multiple mutable references to the same variable is undefined behavior. Unless I'm just misunderstanding the point they're trying to make here."I want my fuckin' money back."
"Hoom, hmm, let us not be hasty!"
"You got 48 hours to deliver or the sapling gets it, Treebeard."