this post was submitted on 14 May 2024
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Rust

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VOID WRONG

It seems like I misunderstood some stuff.


Over the years, up until Rust, there were 4 ways to have a 'safe' language:

  • A virtual machine (e.g. most languages today)

whether it's a high-level (scripting languages) one or low-level one (JVM, CLR)

  • What C++ and similar languages do, destructors, compile-time bound checks, make a global heap and shit in it, make a local stack and piss in it, etc (basically my AllocPPx.pl but baked into the language)
  • Bake the VM in with the machine code. That's what D and Go do, and I think Nim does that too. This will make the point of your language being compiled entirely moot imo.
  • Finally, the most 'controversial' decision for imperative-loving chuds: make a functional language. ML/Scheme/CLisp achieve memory safety through closures. Haskell achieves this through Monads. Functional languages have a property which allows them to be be both compiled into machine code and bytecode, and also, interpreted like an scripting language.

The problem with all these approaches is the trade-off between safety and speed. There's another factor, and that is low-level access to the system. Languages like OCaml came close to achieving a balance, and that's why Rust bassed itself on it.

Most imperative languages have 'operational semantics', but not 'denotational semantics'. You can't describe what C does with math. What C does depends on the platform, the CPU, etc.

Rust's safety is achieved by 'flattening out' the memory model of a functional language like Ocaml. OCaml is a language with denotational semantics, because it's a functional language. Rust is an imperative language but it has denotational semantics. At least when comes to memory management.

I am not going to even attempt to describe the denotational semantics of Rust because I am just an amatuer and i don't have a master's in LDT. But if someoen tries, they could.

I think people might have already done it. I am not sure.

If you tell me no, and Rust does not have denotational semantics, I stand by my great-great grandfather's barber's grave that yes, it does!

So why do I say Rust 'flattens out' the functional model for memory management? It does at least with lifetimes. So imagine this: lifetimes are just 'let' bindings, but with a different syntax.

OCaml:

let rec factorial = function
  | 0 -> 1
  | n -> n * factorial (n - 1);;

Scheme

; This uses `let` under the hood
(define (factorial n)
  (if (<= n 1)
      1
      (* n (factorial (- n 1)))))  

So these two in Rust would be:

fn factorial<'a>(n: u32) -> u32 {
    match n {
        0 => 1,
        _ => n * factorial(n - 1),
    }
}

I know 'a is almost useless here, but what I meant was, that 'a makes it similar to the 'let' bindings in the prior to examples!

Semantics here is clear. Right?

But C:

int factorial(int n)  {
   if (n == 0) return 1;
   else return n * factorial(n - 1);
}

We do have operational semantics here, but who's to say what are the denotational semantics? Right? What is a 'function' in C? Well most C compilers translate it to an Assembly subroutine, but what if our target does not support labels, or subroutines?

You see what I am arriving at?

Conclusion

Rust is a semi-functional, semi-imperative language, but the difference between Rust and other languages with functional aspects is: denotional semantics!

Note: A language having lambda closures does not make it 'functional', you can do that in GNU C too ffs! Denotational semantics make it functional.

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[–] [email protected] 0 points 6 months ago

First, tell me what you mean by 'closure'. I did not mean 'closure' as an operational function literal. Keep that in mind.