5 ms·

Of course, but even the presence of CVEs does not make software worthless, it just makes it worth less.

• LtWorf 2mo ago
  You don't seem to understand what I'm talking about. bad axiom -> worthless theorem. bad unsafe section -> verified safe section is actually not safe.
  • whytevuhuni 2mo ago
    I’ve understood the distinction. I just don’t like you equating “actually not safe” with “worthless”.
    • LtWorf 2mo ago
      So you haven't understood the distinction or you're letting your feelings for software get the best of you. The proof that there are no (for rust's definition of "no") memory bugs becomes worthless, not the entire software itself.
      • whytevuhuni 2mo ago
        > So you haven't understood the distinction or you're letting your feelings for software get the best of you. Maybe? I'm not sure. > The proof that there are no (for rust's definition of "no") memory bugs becomes worthless I still don't think so, the proof that there are no memory bugs is definitely not worthless, since while the whole proof itself is binary (it is either fully correct or not), its effects act more like coverage, and the more of it that happens to be partially correct, the better. I think cve-rs is the perfect example of this. Its very existence proves that the borrow checker is already unsound, and with that, the very foundations of Rust's safety guarantees are currently flawed. TypeScript is another good example, its type system has been repeatedly proven to be unsound, so it's also flawed. But so what if it's flawed? Many smaller properties of these proofs still happen to hold, and they might prevent 90%+ of CVEs even as flawed as they are.