Weird Expressions in Rust

Posted by lukastyrychtr 5 days ago

Weird Expressions in Rust(www.wakunguma.com)

192 points | 150 commentspage 3

JoeOfTexas 5 days ago|

Bruh, I started learning Rust yesterday. Why do you do this to me. Now I don't know anything I studied.

steveklabnik 5 days ago|

You don't need to know any of this. It's just a parsing stress test, with meaningless programs. It's fun trivia.

b0a04gl 5 days ago||

they exist because whole language built to treat expressions as firstclass citizens : blocks, ifs, matches, even macros as expressions that return values. so once you internalize that, all these weirdo one liners are artifacts. just artifact of a system where expressions compose infinitely. the syntax tree runs deeper than most people's habbits allow. you hit that depth and brain says this is wrong but compiler's allowing.

derriz 5 days ago||

That sounds superficially reasonable to me and I'm all for regularity in programming language semantics but on thinking about it further, I actually think it's a design flaw.

It makes no more sense to me for "return <expr>" to have a type than it does to make "if <expr>" or "break" or "{" or any other keyword to have a type. These are syntactic elements.

Rust's type system is clearly inspired by Hindley-Milner and most languages using such a type system either don't even have a return keyword.

Even if you disagree with this argument, this design decision has resulted in all these weird/confusing but absolutely useless code examples and there is no upside that I can see to this decision in terms of language ergonomics. What practical value is it to users to allow "return <expr>" to itself be an expression? That you can use such an "expression" as arguments to function calls with hilarious wtf consequences? It's a piece of syntactic sugar.

steveklabnik 5 days ago|||

Respectfully, "it makes no sense to me" isn't an argument. if and break both have types in Rust as well.

> don't even have a return keyword.

This is because they are not procedural languages, it has nothing to do with the type system.

> there is no upside that I can see to this decision in terms of language ergonomics.

There's tremendous upside! That's why lots of languages choose this. For example, there is no need for the ternary in Rust: if can just do that.

> What practical value is it to users to allow "return <expr>" to itself be an expression?

Code like this just works:

        let guess: u32 = match guess.trim().parse() {
            Ok(num) => num,
            Err(_) => return,
        };

That is, if return wasn't an expression, we'd have a type error: the two arms would have incompatible types.

nine_k 4 days ago|||

This makes sense because the `match` returns a union of u32 and `never`.

Assigning values of expressions that are purely `never`, or having values that are purely `never` or `()` as the condition in a conditional operator, should be marked as an error, like unreachable code.

derriz 5 days ago|||

See my comment above, your example only "just works" if the enclosing function has the appropriate return type (in this case none).

So the syntactic element "return" is not just an expression - unlike other sub-expressions, it involves action at a distance - i.e. it must not just agree with it's context as part of an expression but it must agree with the enclosing fn signature.

steveklabnik 5 days ago||

I replied over there, let's keep it to that sub-tree so we both don't have to duplicate comments :)

bobbylarrybobby 5 days ago||||

The issue with `return expr` not having a type is that you lose the ability to write something like

let y = match option { Some(x) => x, None => return Err("whoops!"), };

Without a type, the None branch loses the ability to unify with the Some branch. Now you could say that Rust should just only require branches’ types to unify when all of them have a type, but the ! never type accomplishes that goal just fine.

derriz 5 days ago||

I'm responding here because so many replies are making the same point.

In your particular example, let's put your example into a context. Is

  fn foo(option: Option<i32>) -> i32 {
     let y = match option { Some(x) => x, None => return Err("whoops!"), };
     return 1;
  }

well typed? It should be if we are to believe that "return <expr>" is an expression of type () - but, naturally, it causes a compilation error because the compiler specifically treats "return <expr>" unlike other expressions. So there is no improvement in regularity, while it admits all sorts of incomprehensible "puzzlers".

I don't see why you'd lose this ability if you removed the claim that "return <expr>" is itself an expression. Most/many languages have mechanisms to allow expressions to affect flow control - e.g. with exceptions, yield, etc. - which do not these constructs (for example "throw x") to have a type.

Rust could just as easily supported the syntax you use above without making "return <expr>" a tapeable expression.

steveklabnik 5 days ago|||

> Is ... well typed?

It's not, but not due to the return, it's because you're trying to return a Result from a function that returns an i32. This works:

  fn foo(option: Option<i32>) -> Result<i32, &'static str> {
     let y = match option { Some(x) => x, None => return Err("whoops!"), };
     return Ok(1);
  }

> It should be if we are to believe that "return <expr>" is an expression of type ()

It is not, it is an expression of type !. This type unifies with every other type, so the overall type of y is i32. return is not treated in a special way.

> if you removed the claim that "return <expr>" is itself an expression

This code would no longer work, because blocks that end in an expression evaluate to (), and so you would get the divergent, not well typed error, because one arm is i32 and the other is ().

derriz 5 days ago||

Sorry for the confusion - I meant to use ! and not ().

"It's not, but not due to the return, it's because you're trying to return a Result from a function that returns an i32."

That's exactly my point. "return <expr>" is not just an expression which can be typed. If you tell me the types of all the identifiers used, I can look at any expression in Rust which does not include a return, and tell you if it's well typed or not. If the expression includes a return, then I cannot tell you whether the expression is well-formed.

steveklabnik 5 days ago|||

> "return <expr>" is not just an expression which can be typed.

Yes, it is, and it can. It has the type !, no matter the type of <expr>.

derriz 5 days ago||

It only has type !, if the return type of the lexically enclosing function declaration has the same type as that of <expr>, otherwise it's illformed.

For any expression NOT involving "return", I can write, for example:

const Z = <expr>

but I cannot if <expr> contains a return embedded somewhere. The existence of a "return" somewhere in an expression changes the character of the entire expression.

I.e. there are two classes of "expressions". Those NOT containing returns (which are equivalent to the notion of "expression" in the languages that Rust was inspired by) and those containing a return somewhere in them which are subject to further rules about wellformedness.

My point is that none of this is necessary at all - you don't need to provide type rules for every lexical feature of your language to have a language with a powerful expressive type system (like Rust's).

kelnos 5 days ago|||

> For any expression NOT involving "return", I can write, for example:

> const Z = <expr>

> but I cannot if <expr> contains a return embedded somewhere.*

Sure, but that's not special about this case at all. I also can't write 'break' or 'continue' when I'm not inside a loop. When declaring a 'const', I am lexically not inside a function body, so I can't use 'return', which makes sense (the compiler will even tell you, "return statement outside of function body").

Particular statements being allowed in some contexts but not in others is entirely normal.

> My point is that none of this is necessary at all

Maybe it's not necessary, but I like the consistency this provides ("everything has a type"), and I imagine the implementation of the type checker/inferer is more straightforward this way.

Sure, you could define the language such that "a 'return' in a position that expects a typed expression will not affect other type that need to match with it" (or something else, in better, formal language). Or you can just define those statements to have the 'never' type, and not worry about it.

But ok, let's agree that it's not necessary. Then we're just talking about personal preferences, so there's no right or wrong here, and there's no point in arguing.

int_19h 5 days ago||||

You can write it just fine if `const Z` is itself nested inside a function definition.

And this isn't really any different from variable references, if you think about it. If you have an expression (x + 1), you can only use it somewhere where there's an `x` in scope. Similarly, you can only use `return` somewhere where there's a function to return from in scope. Indeed, you could even make this explicit when designing the language! A function definition already introduces implicit let-definitions for all arguments in the body. Imagine if we redefined it such that it also introduces "return" as a local, i.e. given:

   fn foo(x: i32, y: i32) -> i32 {
     ...
   }

the body of the function is written as if it had these lines prepended:

   let x = ...;
   let y = ...;
   let return = ...;
   ...

where "return" is a function that does the same thing as the statement. And similarly for break/continue and loops.

The thing that actually makes these different from real variables is that they cannot be passed around as first-class values (e.g. having the function pass its "return" to another function that it calls). Although this could in fact be done, and with Rust lifetime annotations it would even be statically verifiable.

kelnos 5 days ago||

> You can write it just fine if `const Z` is itself nested inside a function definition.

You can't, actually: 'const' is special in that it's not considered by the compiler to be inside a function definition, even if it is (and the compiler will tell you, "return statement outside of function body").

But that doesn't invalidate your point; in a way it supports it: 'return' can only be used in function contexts, just like 'continue' or 'break' can only be used in loop contexts.

tialaramex 4 days ago||

You can do this:

   const ONE: i32 = { const fn foolish() -> i32 { return 1 } foolish() };

But yes, your larger point is exactly correct, the constant, even if it happens to be defined inside a function body, is not itself inside a function body and so we obviously can't return from it. It is also not inside an expression we can break out of (Rust allows you to break out of any expression, not just loops). It's a constant, like 5 is a constant, or 'Z' is a constant - this is not C or C++ where "const" means "actually a variable".

steveklabnik 5 days ago|||

Okay, I think we are indeed talking past each other and I see what you are saying here. I am not sure that I agree, exactly, but I appreciate your point. I'm going to have to think about it a bit more.

Timwi 5 days ago|||

The same is true if `return` is a statement, so this doesn't seem to have anything to do with `return` being an expression.

bobbylarrybobby 5 days ago||||

The type of return is !, not (). Meaning there are zero instances of this type (whereas there is one instance of ()). ! can coerce to any type.

Also, the type of return is a separate matter from the type of the thing being returned. You obviously can't return Result from a function returning i32. The point of type coercion is that you can yield `return Err(...)` in one branch of a match and have it type check with the other branch(es).

rtpg 5 days ago||||

When discussing well typed-ness, and with more complex languages where you have weird undecideable components, you can end up with a notion of "Well typed" as follows:

e: T is well typed _if_ the end result of e would be of type T

(end result being hand-wave-y)

It's not a guarantee that e is a value of a certain type, but a guarantee that if e is a value in the first place, then it will be a certain type. You sidestep having to prove the halting nature of e.

This leaves a nice spot for computation that doesn't complete!

    let y = return 1
    f(y)

y could be any type, and it's well typed, because you're never in a secnario where f(y) will be provided a value of the wrong type.

Well-typed-ness, by my understanding in more complex type system, is not a guarantee of control flow, but a guarantee that _if_ we evaluate some expression, then it will be fine.

And so... you can put `!` as a type in your system, treat return as an expression, and have a simpler semantic model, without really losing anything. Less moving parts, etc.... that's my read of it anyways.

efnx 5 days ago|||

Well now we’re just talking about personal preferences, then.

NobodyNada 5 days ago||||

In practice, it's quite a useful feature, because you can write things like this:

    let day_number = match name {
        "Sunday" => 0,
        "Monday" => 1,
        "Tuesday" => 2,
        "Wednesday" => 3,
        "Thursday" => 4,
        "Friday" => 5,
        "Saturday" => 6,
        _ => return Err("invalid day")
    };

pornel 5 days ago||||

This makes the language more uniform. Instead of having a special ternary ?: operator for if-else in expression position, you have one syntax everywhere.

It makes generic code work without need to add exceptions for "syntactic elements". You can have methods like `map(callback)` that take a generic `fn() -> T` and pass through `T`. This can work uniformly for functions that do return values as well as for functions that just have `return;`. Having nothingness as a real type makes it just work using one set of rules for types, rather than having rules for real types plus exceptions for "syntactic elements".

deathanatos 5 days ago||||

We can use match to do pattern matching:

  let name = match color_code {
    0 => "red",
    1 => "blue",
    2 => "green",
    _ => "unknown",
  };

The RHS of the `=>` has to be an expression, since we're assigning it to a variable. Here, you should already see one "useful" side-effect of what you're calling "syntactic elements" (I'd perhaps call them "block statements", which I think is closer to the spirit of what you're saying.) The whole `match … {}` in the example above here is an expression (we assign the evaluation of it to a variable).

> What practical value is it to users to allow "return <expr>" to itself be an expression?

Now, what if I need to return an error?

  let name = match color_code {
    0 => "red",
    1 => "blue",
    2 => "green",
    _ => return Err("unknown color"),
  };

The expression arms need to be the same type (or what is the type of `name`?). So now the type of the last branch is !. (Which as you hopefully learned from TFA, coerces to any type, here, to &str.)

There's more ways this "block statements are actually expressions" is useful. The need not be a ternary operator / keyword (like C, C++, Python, JS, etc.):

  let x = if cond { a } else { b };

In fact, if you're familiar with JavaScript, there I want this pattern, but it is not to be had:

  const x;  // but x's value will depend on a computation:
  // This is illegal.
  if(foo) {
    x = 3;
  } else {
    x = 4;
  }
  // It's doable, but ugly:
  const x = (function() { if(foo) { return 3; } else { return 4; }})();
  // (Yes, you can do this example with a ternary.
  // Imagine the if branches are a bit more complicated than a ternary,
  // e.g., like 2 statements.)

Similarly, loops can return a value, and that's a useful pattern sometimes:

  let x = loop {
    // e.g., find a value in a datastructure. Compute something. Etc.
    if all_done {
      break result;
    }
  };

And blocks:

  let x = {
    // compute x; intermediate variables are properly scoped
    // & cleaned up at block close.
    //
    // There's also a slight visual benefit of "here we compute x" is
    // pretty clearly denoted.
  };

> Even if you disagree with this argument, this design decision has resulted in all these weird/confusing but absolutely useless code examples

I think one can cook up weird code examples in any language.

wredcoll 5 days ago||

I appreciate, so much, that rust is slowly evolving into perl.

trealira 4 days ago||

What Rust's syntax really reminds me of is Algol 68, or BLISS, both of them being these old procedural languages where everything is an expression. The "loop { ... break expr; ... }" thing reminds me of BLISS's "exitloop expr" construct.

wredcoll 3 days ago||

There's so many programming languages (low barrier to create) that there's a ton of overlap and evolutionary changes/similarities between them. I was thinking of perl's "do { x } while foo" style constructs in this particular case.

I am incredibly amused that I got downvoted to -1 for mentioning perl though. People here are Weird.

trealira 3 days ago||

Oh, sorry, I didn't downvote you. Just gave you an upvote.

dathinab 5 days ago|||

it doesn't need to make sense on a higher abstraction level of logic/semantics

I mean you don't see any of the nonsense in the blog post in any realistic PR (so they don't matter),

but you would run into subtle edge case issues if some expressions where more special then other expressions (so that does matter),

especially in context of macros/proc macros or partial "in-progress" code changes (which is also why `use` allows some "strange" {-brace usage or why a lot of things allow optional trailing `,` all of that makes auto code gen simpler).

AIPedant 5 days ago|||

Hmm my read is this is a slight overstatement - Rust was always built with the idea of expressions as first class citizens, but practicality and performance requires expression-breaking keywords like “return” which don’t fit neatly in an ML-ish language and have a few plain old hacks associated with implementing them (not “hack” as in lacking robustness; I mean theoretically/formally inelegant). Likewise there’s some stuff (u8) which is a simple syntax quirk. But a lot of these return/etc oddities are because Rust is ultimately an imperative language with strong influence from functional programming.

steveklabnik 5 days ago|||

return is an expression in Rust, and it fits in well.

There are very few statements: https://doc.rust-lang.org/stable/reference/statements.html

and a lot of expressions: https://doc.rust-lang.org/stable/reference/expressions.html

AIPedant 5 days ago|||

We're speaking past each other since there's "expression" as defined in the Rust specification vs "expression" as in ordinary computer science, and Rust's use of return is certainly not an expression in the latter sense. It is shoehorned into being called an expression but it has no semantically meaningful type, it is an effect. A type is (carefully but somewhat arbitrarily) assigned to it, which is why some of those examples involving "return" are particularly goofy. It is not material for most programs since it only comes up with intentional misuse of the keyword. But "return" does not make sense in functional languages with true first-class expressions - functions don't return values, they get evaluated and the frame destruction / etc are all abstracted away. It makes sense in Rust because expressions in the CS sense of the term are ultimately not first class.

steveklabnik 5 days ago|||

I do think we're speaking past each other. I don't fully agree with your "CS sense of the term," as Rust does have a semantically meaningful type: !. This is all pretty bog-standard stuff. Rust isn't doing anything weird or novel here.

rtpg 5 days ago||

I do wonder how many languages have the "never returns" type explicitly available. Typescript and Rust.... Haskell has bottom but I wonder semantically how much space there is between bottom and "never return". Obviously laziness makes things weird.

This is what I find interesting in this generation of languages though. Any C programmer understands the notion of an infinite loop, and the value of conditional expressions like ternary ops. But now languages are realizing that when you start treating more and more things as expressions, you really want to start giving names to things that you wouldn't name in the past.

valenterry 5 days ago|||

Scala and Haskell are there and I think they inspired this in Kotlin and Rust. In Haskell it's "bottom" and in Scala it's "Nothing".

In Scala no one uses "return" (mostly because we don't care about performance in the same way), but if you do, the way it is internally implemented is by throwing exceptions, so in a sense it suffers from the same problems as Rust.

It's actually very important to have that type in a language that uses immutable collections. Imagine this pseudocode:

    // List() creates an immutable list
    let emptyList = List() 
    let listWithAnInteger = emptyList.add(42)
    let listWithAString = emptyList.add("foo")

This works in Scala. But how can the compiler know that `emptyList.add(42)` is allowed? After all, you can only add things to a list where the added element matches the type of the other elements right?

The reason this works is because the type of emptyList will be List<Nothing> and since Nothing a subtype of every other type, the type of listWithAnInteger will become List<Integer>. You can annotate these types explicitly if you want.

Every language without such a bottom type has a failed type-system in my opinion. (looking at you Golang and many others)

wk_end 5 days ago||

? This works fine in a type system without an explicit bottom type. In Haskell or ML or whatever `emptyList` would be given a polymorphic type, `List a` or `'a list`.

There's issues around doing this with mutable collections (i.e. the value restriction) but that's not what you're referring to...

valenterry 5 days ago||

> In Haskell or ML or whatever `emptyList` would be given a polymorphic type, `List a` or `'a list`.

That alone would not work. Think about it: `List a` means "A list that contains values of type `a` and `a` can be any type whatsoever". Now imagine you combine that list with a list of integers. That obviously cannot work, since `(++) :: [a] -> [a] -> [a]` as you see, the types must align.

The way Haskell fixes that is (apparently) by doing something called `Let-generalisation` (https://ghc.gitlab.haskell.org/ghc/doc/users_guide/exts/let_...)

To me that feels like hacky way to exactly resolve the problem that I described, and if you turn it off then that code would stop working and fail to compile as expected.

octachron 4 days ago||

You are misreading the quantification, a value l of type List a means that for all type a, the element of the list has type a. In other words, this is an universal quantification whereas your interpretation is an existential quantification.

This is obviously only possible if the list itself has no elements, and indeed a simple proof is that the statement above is valid for the empty type: all elements of a list of type List a have type empty (among other types). Thus there are no elements in this list.

And both Haskell or OCaml can prove it:

     type empty = | (* this is defining a never type *)
     type polymorphic_list = { l: 'a. 'a list } 
     (* OCaml require to explicit construct polymorphic type *)

     let polymorphic_lists_are_empty ({l} : polymorphic_list ) =
     match (l:empty list) with
     | [] -> () (* this is the empty list *)
     | _ -> . 
     (* this clause requires to the OCaml typechecker to prove that the remaining cases are unreachable *)

valenterry 3 days ago||

> this is an universal quantification whereas your interpretation is an existential quantification

Indeed, I stand corrected. Interesting! I'll look into that a bit more, thank you.

kuschku 5 days ago|||

Kotlin also has this type, it's called "Nothing".

https://kotlinlang.org/api/core/kotlin-stdlib/kotlin/-nothin...

As you can never get a value of type nothing, it can coerce into anything, just like rust's ! or () or typescripts never.

nialv7 5 days ago||

I recently realized (by playing with Lean) that coercing of "!" is because of the principle of explosion [1]. Basically propositions are types in Lean, and proofs are instances of those types. A proposition that is false doesn't have any instances, so they are like "!". Principle of explosion says ∀ P, False -> P, which is exactly the type signature for "!" coercing.

[1]: https://leanprover-community.github.io/mathlib4_docs/Init/Pr...

int_19h 5 days ago||||

Pure functional languages have the equivalent of "never" - it's the bottom type. Indeed, the return type of `error` in Haskell is that, but also cases like predictable infinite recursion. But this semantics works great for cases like "return" and other forms of control transfer - the expression in which they appear also "never finishes" (but some other expression which contains that one as a subexpression does).

Now, yes, ideally you'd have effects in the type system so that you can express this kind of stuff with more precision. But if you restrict this to stuff like return/break/continue where the destination is statically known and can be validated, you can treat those effect types as been there, just inferred for all expressions and forbidden to cross the function boundary.

For exceptions specifically this trick no longer works because the whole point is for them to cross that boundary. But the amount of complexity this stuff adds to typing even trivial generic code is arguably too much for practical use (see also: checked exceptions in Java). In any case, in Rust you use Result types instead so those exceptions produce regular values. And although panics can be handled, they are certainly not meant to be used as a generic mechanism for transfer of control, so adding effect types for them alone is just not worth it.

octachron 5 days ago||||

Return (or other effects) does make sense as an expression in a functional language. Typically, OCaml has `raise Exception` which is also an expression, with the same type as `return` or any never returning function. And exceptions can also be used to implement a user-defined `return` function.

nextaccountic 4 days ago|||

Return has the ! type. It's a type with no values, similar to an enum without variants, like this:

    enum Never {
    }

Languages like OCaml, Haskell as well as Rust have types with no values (called uninhabited types)

missinglugnut 5 days ago|||

Steve, I know you're an authority on the language but you've dismissed the point being made here without engaging with it.

Return is a statement in the minds of most programmers, but an expression in the language. That was a very pragmatic decision that required an unintuitive implementation. As a result, we've got this post full of code that is valid to the compiler but doesn't make a lick of sense to most programmers reading it.

steveklabnik 5 days ago|||

> Return is a statement in the minds of most programmers

I would take issue with this, sure, for a lot of people, they may be bringing assumptions over from languages where assignment is a statement. That doesn't make them correct.

> required an unintuitive implementation

To some people, sure. To others, it is not unintuitive. It's very regular, and people who get used to "everything is an expression" languages tend to prefer it, I've found.

saurik 3 days ago|||

> ...people who get used to "everything is an expression" languages tend to prefer it, I've found.

This feels awkward as my mental model is that in "everything is an expression" languages you simply DO NOT offer "return" (and, if you do, it must be mapped to bottom and do something insane like throw an exception... but, like, if you are really used to using such a language, you'd never let yourself type a "return", as the entire concept feels icky and wrong in such a language).

steveklabnik 1 day ago||

Ruby is an "everything is an expression" language and it has return, and idiomatically it's used identically to Rust: for early returns.

It is of course not statically typed.

hansvm 5 days ago|||

> people who get used to "everything is an expression" languages tend to prefer it, I've found

I.e., if we bias our sample to the data points proving our point then our point is proven. It's like that quip about how every car insurance company can simultaneously claim "people who switched saved hundreds of dollars in average."

I also like "everything is an expression" languages, but I don't think that's a fantastic argument.

saghm 5 days ago||

The original claim that the was responding to in this thread was that `return` as an expression didn't fit in well with Rust, and he said that it did. He also cited how far more things in Rust are expressions than statements, so it stands to reason that people who program in Rust are familiar with those styles of language. It sounds like you're arguing that it makes more sense to judge whether return makes sense as an expression in Rust based on the expectations of people who aren't as familiar with expression-based languages (and therefore aren't super familiar in Rust), which doesn't make a ton of sense to me.

int_19h 5 days ago|||

We've been going down this road for a long time now. E.g. "throw" is a (void-typed) expression in C++ already for similar reasons, although it doesn't go far enough without a proper bottom type. C# took it further and added the type so that you can write things like e.g. `x = y ?? throw new Error(...)`. There's no obvious reason why "return" should be conceptually different.

A better question at this point, arguably, is why there should even be an expression/statement distinction in the first place. All imperative statements can be reasonably and sensibly represented as expressions that produce either () or "never". Semicolon then is just a sequencing operator, like comma in C++.

Timwi 5 days ago||

I considered whether to mention C# in this thread but initially decided against it because it doesn't actually have a bottom type. You can't assign a throw expression to an implicitly-typed variable, or anywhere else where it is needed to infer a type. You can only use it in places where a type is already known so the throw expression can be coerced to it.

In fact, I recently ran into the finding that you can't use it with logical operators either: `return myBool && throw...` doesn't work. I assume that's because && can be used with many types even if the first operand is a bool, but the compiler error message doesn't explain that, it just says throw is an invalid token here, and if you parenthesize it, it says a throw expression can't be used in this context. I was very surprised by this seemingly arbitrary limitation.

int_19h 4 days ago||

Yes, this is a good example of how not having the bottom type actually makes things messier overall. Without it you have to make those case-by-case hacks. With it, all the stuff that people actually want to write and that makes sense "just works", and sure, there's more stuff that you could write that doesn't make sense as well, but it's not something that people might end up writing accidentally by mistake and get wrong behavior.

efnx 5 days ago||||

I’ve found rust to be an ML in C’s clothing.

The main difference from other MLs is the lack of higher kinded types, so it’s difficult to express things like Functor, Monad, Arrow, etc

saghm 5 days ago||||

I'm confused about your parenthetical about "u8". What does the name of the unsigned 8-bit integer type have to do with whether the language is imperative?

GrantMoyer 5 days ago|||

Haskell has `bottom`[1] (see also [2]), which acts like Rust's `return` from a type checking perspective.

I wouldn't call using a uninhabited type for the type of a return expression theoretically inelegant. On the contrary, I find it quite pleasing.

[1]: https://wiki.haskell.org/Bottom

[2]: https://en.wikipedia.org/wiki/Bottom_type

dmkolobov 5 days ago||

On the more mainstream side of things, Typescript also has a bottom type called `never` which is used to type unreachable/exceptional code.

saghm 5 days ago||

Rust has this too: https://doc.rust-lang.org/std/primitive.never.html

nine_k 4 days ago|||

This is logically sound, but pragmatically not so. I wish the compiler could issue a warning or even an error if an expression of type `never` is used in a logical condition, like that of an `if`. While such cases might have rare legitimate uses (e.g. some edge cases of macro expansion, etc), I'd like it to be marked explicitly, similar to `unsafe`, e.g. with some `allow_never_as_condition` marker.

Likely the same should apply to expressions of type `()`.

throwawaymaths 5 days ago|||

its not just that some things you would usually think are control flow are expressions, its also that there are unusual rules around coercing the `noreturn` type.

tialaramex 5 days ago||

The only "unusual" rule here is that Rust offers the zero type addition, but does not provide the (much more complicated) other type additions

So Rust does have: String + ! = String

But Rust doesn't have: String + i32 = Either<String,i32>

Note that the never type ! isn't special here, Rust will also cheerfully: String + Infallible = String or if you were to define your own empty type like so:

    enum MyEmptyType {} // MyEmptyType has no possible values

Now under type arithmetic String + MyEmptyType = String and indeed that works in Rust.

Edited: Syntax fix

gmueckl 5 days ago||

If a language that claims to be security focused is easily able to express constructs that human minds find barely comprehensible, or worse, then this is itself arguably a security issue: it's impossible to check the correctness of logic that is incomprehensible.

pornel 5 days ago|||

What's the threat model? If you're reviewing untrusted or security-critical code and it's incomprehensible, for any reason, then it's a reject.

Syntax alone can't stop sufficiently determined fools. Lisp has famously simple syntax, but can easily be written in an incomprehensible way. Assembly languages have very restrictive syntax, but that doesn't make them easy to comprehend.

Rust already has a pretty strong type system and tons of lints that stop more bad programs than many other languages.

gmueckl 5 days ago|||

Many modern language designers focus on shaping expressibility rather than providing the maximum possible flexibility because their designers learned from C, Lisp and other languages that made mistakes. Examples lamguages are Java, C#, D, Go... some arguably with more success than others. But language design that gave ultimate expressive power to the the programmer is a relic of the past.

pornel 5 days ago||

???

"Expressibility" and "expressive power" are vague and subjective, so it's not clear what you mean.

I suppose you object to orthogonality in the syntax? Golang and Java definitely lack it.

But you also mention C in the context of "maximum possible flexibility"? There's barely any in there. I can only agree it has mistakes for others to learn from.

There's hardly any commonality between the languages you list. C# keeps adding clever syntax sugar, while Go officially gave up on removing its noisiest boilerplate.

D has fun stuff like UFCS, template metaprogramming, string mixins, lambdas — enough to create "incomprehensible" code if you wanted to.

You're talking about modern languages vs relics of the past, but all the languages you mention are older than Rust.

gmueckl 5 days ago||

Have you ever seen submissions to IOCCC or Underhanded C Code Contest? That is what too much syntactic flexibility looks like (if taken to the extreme).

If you want your code to be secure, you need it to be correct. And in order for it to be correct, it needs to be comprehensible first. And that requires syntax and semantics devoid of weird surprises.

kelnos 5 days ago|||

Eh, I'm not sure I agree here. This feels sort of along the lines of, "well C is safe because someone can review your code and reject it if you try to dereference a possibly-NULL pointer".

The point of a language that is "safe" along some axes is that it makes those unsafe things impossible to represent, either by omitting an unsafe feature entirely, or making it a compile-time error to do unsafe/unsound things.

I will admit that this is something of a grey area, since we're talking about logic errors here and not (for example) memory-safety bugs. It's a bit muddier.

In general, though, I do agree that people should write code that is reasonable to read, and if a reviewer thinks some code in a PR is incomprehensible, they should reject it.

pornel 4 days ago||

I think these situations are very different, because Weird Rust affects only weird code, while unsafety of C affects regular C code.

The difficulty in reviewing pointer dereferences is in reasoning about potential program's states and necessary preconditions, which C won't do for you. You can have neatly written C using very simple syntax, and still have no idea if it's safe or not. Solving that lack of clarity requires much than syntax-level changes.

OTOH the Weird Rust examples are not a problem you get in your own code. It's a local syntax problem, and it doesn't require complex whole-program reasoning. The stakes are also lower, because you still have the same safety checks, type checks, automatic memory management, immutability. The compiler aggressively warns about unreachable code and unused/unread variables, so it's not easy to write undetected Weird code.

Rust tried having Underhanded Code Contest, but it has been very Underwhelming.

steveklabnik 5 days ago||||

> If a language that claims to be security focused

Rust does not claim to be particularly security-focused, only memory safe.

Also, this means that you'd consider any expression-based language to be inherently a security problem.

gmueckl 5 days ago||

"Memory safety" is an aspect of computer security. And security is the first listed value in rust's mission statement.

Rust is not written as a pure expression based language. And as we all know very well from the experience with C and JS, any unexpected and weird looking code has the potential to hide great harm. Allowing programmers to stray too much from expected idioms is dangerous.

kelnos 5 days ago|||

I think you're looking at it a little backward. Or rather, with superset/subset confusion. Rust can say "we care about memory safety" without being security-focused. But Rust cannot say "we are security-focused" without also caring about memory safety.

Being security-focused requires you to care about a laundry list of things, including memory safety. But on its own, caring about memory safety just means... you care about memory safety.

steveklabnik 5 days ago||||

It is an aspect, but it Rust does not promise your core is secure.

It’s not purely expression based but it is very close to it, there’s only a few kinds of statements, the vast majority of things are expressions.

keybored 5 days ago|||

We would need an example of puzzling code that can easily hide (security) bugs.

The submission shows weird program snippets. I don’t think it shows weird snippets that can also easily hide bugs?

int_19h 5 days ago||||

"never" is an easily comprehensible concept once you start asking the right questions.

But also, all examples in TFA are very artificial convoluted code. Meaning that you can write things like these just like you can write something like &&&...x - but why would you? Actual real-world uses of this feature are all quite readable.

PaulHoule 5 days ago|||

It is a critique of macros. 500 lines of Common Lisp replaces 50,000 lines of C++ but those 500 lines make no sense at all the first time you see them.

creative1122 4 days ago||

[dead]

xyst 5 days ago||

These “weird expressions” probably get used code golf.

steveklabnik 5 days ago|

Basically none of them are actually useful, or even do anything, it's mostly a parser stress test.

nikolayasdf123 5 days ago|

this is why I like Go

nemo1618 5 days ago||

I wonder, what's the "weirdest" expression in Go? Here's one:

   type Foo struct{}
   func (Foo) Bar() { println("weird...") }
   func main() {
    ([...]func(){^^len(`
   
   
   `): (&Foo{}).Bar})[cap(append([]any(nil),1,2,3))]()
   }

assbuttbuttass 5 days ago|||

Makes sense to me, [...]func() is an array of functions, and [...]T{index: value} is uncommon but still perfectly comprehensible

nemo1618 5 days ago|||

Many people aren't aware that you can use key: val declarations in arrays

jenadine 5 days ago|||

That's why I like Rust /s

yencabulator 4 days ago||||

I have a personal fondness for silly variations of

  type __ *[]*__

nikolayasdf123 5 days ago|||

this the worst? not too bad. fairly comprehensible.

techbrovanguard 5 days ago|||

this is not the gotcha you think it is, you just dropped your gluestick

timeon 5 days ago||

It does not have stress tests for parser?