4 ms·

Rust unsafe has pretty much the same functionality as plain C, with more vervose syntax. So I would expect this model to refuse to give any examples of C code w

• zoky 3y ago
  To be fair, C is basically one giant footgun…
  • meindnoch 3y ago
    Yawn. This is such a tired trope.
    • zoky 3y ago
      Is it not true, though? Is there some other language that has a list of banned (commonly used) functions[0] in a major project? [0] https://github.com/git/git/blob/master/banned.h https://github.com/git/git/blob/master/banned.h
• TheCoreh 3y ago
  C is a much simpler language, so it's far easier to reason about the semantics of "unsafe" code. For example: Rust has destructors (Drop trait) that run automatically when a value goes out of scope or is overwritten. If memory for a struct with a Drop trait is manually allocated (and initialized with garbage) and it is assigned to, the `drop()` method will run for the previous value of it which will cause undefined behavior. That's just one feature: Rust also has references, tagged unions, virtual method calls (&dyn Trait), move semantics, `Pin<T>`, closures, async/await, and many more, all of which make it harder to reason about safety without the guardrails provided by the language for regular, "safe" code—for which barring a compiler bug it is actually _impossible_ to shoot yourself in the foot like this. This is actually why it's so incredibly hard to write C++ code that is provably correct: It has even more features that could cause problems than Rust, and is _always_ in "unsafe" mode, with no guardrails.
  • blibble 3y ago
    gcc C has destructors (cleanup attr), nested functions/closures you can do tagged unions with cpp tricks, and so-on (sadly I have seen most of these used...) regardless, what makes C hard is undefined behaviour
• junon 3y ago
  Not entirely true. You can't bypass the borrow checker for example, and you have to maintain Rust invariants when you use it. Hence the name.