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Rust doesn’t stop you from writing buggy code. However I do agree that it is easier to write memory safe code in Rust compared with C/C++.

masklinn 4y ago
TBF lots of formal tools don’t either, because you still have to translate the proven code by hand. It looks like Microsoft’s does the translation, so the only risk is bugs in the code generator, which seems nice. There’s also Google’s wuffs, but I don’t think that is formal, it just ensures memory safety.
- jnash 4y ago
  Yep if the translation is done by hand then that might be a potential problem. However a lot of modern proof tools have proven correct translators (based on the CompCert work).
staticassertion 4y ago
No amount of formal verification stops you from writing buggy code.
- jnash 4y ago
  I don't think you understand what "proven correct" means. If you formally prove a spec correct using modern proof tools then it is guaranteed to have zero bugs. Nobody has found even a single bug in the proven correct CompCert C compiler for example. However people find bugs in gcc all the time.
  - staticassertion 4y ago
    It means that a program has been verified against a model, and properties of that model are held true. What it does not mean is that the program is bug free. That would, of course, violate the halting problem, or its generalization Rice's Theorem - non-trivial programs can not have arbitrary properties proven about them. There's also Gödel's Incompleteness Theorem, obviously, as well as the fact that a proof is only as good as its model.
    - UncleMeat 4y ago
      It wouldn’t break Rice’s Thm to prove it for a particular program. All Rice’s Thm says is that any algorithm must fail on some programs.
      - staticassertion 4y ago
        It would if you're talking about arbitrary classes of bugs.
        UncleMeat 4y ago
        Not so. Consider the algorithm that reports "Yes" for all inputs for all program properties. For a large number of programs and properties, this algorithm will report the correct information.
    - mbrodersen 4y ago
      Not true for non-Turing complete languages. Which is why all proof tools require a proof of termination. Some do it automatically (essential ruling out Turing complete languages) others require a formal proof of termination by the user.
  - UncleMeat 4y ago
    Ages ago my advisor had a paper proving WEP to be correct. Many years later, a serious flaw was discovered. Even formal verification can be insufficient.
    - deterministic 4y ago
      Was it a proof by hand or using a formal proof tool? Never trust proofs that aren’t formally verified by a modern proof tool.
      - UncleMeat 4y ago
        Formal verification with model checking.
        deterministic 4y ago
        That frankly sounds hard to believe. Which proof tool did he use? Link to paper?

masklinn 4y ago

TBF lots of formal tools don’t either, because you still have to translate the proven code by hand. It looks like Microsoft’s does the translation, so the only risk is bugs in the code generator, which seems nice. There’s also Google’s wuffs, but I don’t think that is formal, it just ensures memory safety.

jnash 4y ago
Yep if the translation is done by hand then that might be a potential problem. However a lot of modern proof tools have proven correct translators (based on the CompCert work).

staticassertion 4y ago

No amount of formal verification stops you from writing buggy code.

jnash 4y ago
I don't think you understand what "proven correct" means. If you formally prove a spec correct using modern proof tools then it is guaranteed to have zero bugs. Nobody has found even a single bug in the proven correct CompCert C compiler for example. However people find bugs in gcc all the time.
- staticassertion 4y ago
  It means that a program has been verified against a model, and properties of that model are held true. What it does not mean is that the program is bug free. That would, of course, violate the halting problem, or its generalization Rice's Theorem - non-trivial programs can not have arbitrary properties proven about them. There's also Gödel's Incompleteness Theorem, obviously, as well as the fact that a proof is only as good as its model.
  - UncleMeat 4y ago
    It wouldn’t break Rice’s Thm to prove it for a particular program. All Rice’s Thm says is that any algorithm must fail on some programs.
    - staticassertion 4y ago
      It would if you're talking about arbitrary classes of bugs.
      - UncleMeat 4y ago
        Not so. Consider the algorithm that reports "Yes" for all inputs for all program properties. For a large number of programs and properties, this algorithm will report the correct information.
  - mbrodersen 4y ago
    Not true for non-Turing complete languages. Which is why all proof tools require a proof of termination. Some do it automatically (essential ruling out Turing complete languages) others require a formal proof of termination by the user.
- UncleMeat 4y ago
  Ages ago my advisor had a paper proving WEP to be correct. Many years later, a serious flaw was discovered. Even formal verification can be insufficient.
  - deterministic 4y ago
    Was it a proof by hand or using a formal proof tool? Never trust proofs that aren’t formally verified by a modern proof tool.
    - UncleMeat 4y ago
      Formal verification with model checking.
      - deterministic 4y ago
        That frankly sounds hard to believe. Which proof tool did he use? Link to paper?