alt Hacker News> True, but unsafe let's you conjure up any lifetime you want
The only thing unsafe does is let you have an unbounded lifetime. As I said, it doesn't check those:
fn get_str<'a>(s: *const String) -> &'a str {
unsafe { &*s }
}
> if you generously sprinkle pointer dereferences in unsafe code, you effectively disable the protection provided by the borrow checker
You don't disable anything. You wrote a "trust me compiler" block, and compiler trusted you.
Rust won't ever protect from all possible problems, just the ones the compiler handles.
Replies
What's the point of using Rust in the first place when you disable the compiler feature that protects you the most?
> The only thing unsafe does is let you have an unbounded lifetime.
No, you're wrong: You can create any lifetime. Proof:
fn oof<'desired>(x: &u32) -> &'desired u32 {
let ptr = x as *const u32;
unsafe { &*ptr }
}
> You don't disable anything.
I said "effectively disable". For example:
fn trust_me_bro<'a>(x: mut u32) -> &'a mut u32 { unsafe { &mut x } }
fn main() {
let mut x = 1_u32;
let reference_a = &mut x;
let reference_b = trust_me_bro(reference_a);
*reference_b = 2; -- Whoops
println!("reference_a: {reference_a} reference_b: {reference_b}");
}
> You don't disable anything. You wrote a "trust me compiler" block, and compiler trusted you.
That’s the same thing.