3 ms·

Rust doesn't really have a way to control for effects in functions. The type of the function doesn't define what the function can do. This I think is the bigges

• MrBuddyCasino 8y ago
  How does that work in Haskell, is it the pure/impure dichotomy regarding I/O that is encoded in the type signature? Because one could argue that Rusts ownership is a primitive version of that, but yielding most of the gains ("if it compiles, it works").
  • lomnakkus 8y ago
    Kind of, but not really. The combination of type classes (traits in Rust) and HKTs in Haskell mean that you can quite easily restrict a function to only be able to do a subset of what's possible in "the world". A trivial example would be foo :: Logger m => Int -> m () where Logger is a type class (trait in Rust-speak) which implements the Monad type class (thus allowing 'imperative'-style programming), but also has a "log" method which allows the program to emit a string to the log. The thing is that given the restrictions of parametric polymorphism that function 'foo' has to work for any Logger instance we give it, so it absolutely cannot do anything other than pure computation or invoking the "log" method. It cannot do any other side effects at all. So, it's not "pure", but it's also much more restricted than "impure". That's a very powerful tool for modeling highly effectful systems.