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Comment by saghm | Hacker News Reader

> foo and bar are both lambdas and closures. A named lambda is still a lambda. > foo = function (x) { return x + 1; } > is identical to > function foo (x) { return x + 1; }

I'll concede that for JavaScript (and a few other languages); some languages don't treat all functions as just a lambda with a binding, though.

> They are also closures. Since they don't refer to any outer variables, they are closures containing zero context variables. If foo were to refer to some other variable, the closure would contain one context variable.

I'm not very keen on the definition of "closure" including functions that close over zero variables, because then "closure" becomes indistinguishable from "function", and the word is no longer useful. I might be wrong about the technical definition, but I feel like having a word for closures isn't very useful if it doesn't mean "closing over one or more variables".

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Elisp disagrees:

  > (lambda () (print "hi"))
  (closure (t) nil (print "hi"))

  > (let ((a 1) (b 2)) (lambda () (print "hi")))
  (closure ((b . 2) (a . 1) t) nil (print "hi"))

  > (let ((a 1)) (defun foo (x) (+ x 1)))
  foo
  > (symbol-function 'foo)
  (closure ((a . 1) t) (x) (+ x 1))

  > (defun bar (x) (+ x 1))
  bar
  > (symbol-function 'bar)
  (closure (t) (x) (+ x 1))

So yes, every function is a closure.

A closure is a combination of a function and the environment containing the variables it is bound to (closed over) rather than left open (like old LISPs, like elisp).

Every function isn't a closure. In particular, elisp lambdas aren't closures because elisp is dynamically bound - the variables are open, they are not closed over. Or with another perspective, the variables are bound in the global context. Either way, elisp closures aren't.

From emacs scratch buffer:

    (setq f (let ((x 10))
      (lambda ()
        (progn
          (setq x (+ x 1))
          x))))

    (funcall f)
    11
    (funcall f)
    12
    (setq x 20)
    (funcall f)
    21

This is because elisp is dynamically bound - it only has one environment (in this example), the global environment (leave aside buffer-local etc. for the moment).

It doesn't matter if you try to create a new environment using a lambda, the same problem still occurs:

    (setq g
          (lambda (x)
            (lambda ()
              (progn
                (setq x (+ x 1))
                x))))
    (setq f (funcall g 1))
    (funcall f)
    22  ;; or whatever you last assigned to x, + 1

You're thinking of old elisp. Elisp hasn't been dynamically bound since around 2014. In modern elisp, every function is a closure.

If you pass your examples to `(eval ... t)`, you'll see what I mean. Or run `(setq-local lexical-binding t)` before evaluating them in the scratch buffer.

  > (setq g (lambda (x)
              (lambda ()
                   (progn
                  (setq x (+ x 1))
                  x))))
      (setq f (funcall g 1))
         (funcall f)
  2
  > f
  (closure ((x . 2) t) nil (progn (setq x (+ x 1)) x))
  > (setq x 20)
  20
  > (funcall f)
  3

It depends on your definition of the closure. Example: "A closure is the combination of a function and the lexical environment within which that function was declared." [0]

[0] https://developer.mozilla.org/en/docs/Web/JavaScript/Closure...

A lexical environment being syonymous with "a set of 0 or more variables accessible from within the function, but not passed as arguments." The key piece of that being 0 or more. The null set is still a set.

I do not think it is useful to interpret it as 0 or more. To me it only make sense to assume 1 or more. Otherwise you could say that the programming language that do not support closing variables, has nonetheless closures, as according to that notion every function is a closure that closes nothing.

The key difference here is whether the language allows closure over the lexical environment, not whether the lexical environment contains anything used.

You can see this in Ruby where methods don't close over the environment, but blocks do, clearly Ruby methods aren't closures, not even when reïfied into a Method, whereas blocks are closures.

I think this is a rectangle/square problem. When you say you accept/return a closure, you probably mean a "function + lexical scope of 0 or more variables". You can still refer to "non-closures" as functions, though.

It's like code expecting rectangles wouldn't mind to get squares, but a code expecting squares will definitely do mind getting rectangles.

>It's like code expecting rectangles wouldn't mind to get squares, but a code expecting squares will definitely do mind getting rectangles.

It is actually not that obvious since in OOP neither a square or a rectangle type is properly a subtype of the other. [0]

I understand how implementation-wise it could be convenient to treat a function as a closure. Although, I would prefer some other type, eg:

    functoid := closure | function

[0] - https://stackoverflow.com/questions/1030521/is-deriving-squa...

Well no, because a closure is a function and an environment. A function closing over nothing is different than a function being unable to close over something.

You might define the interface of a closure as `closure(f: Function, env: Mapping[str, Any])`, or in other words a closure is a function and a mapping of variable names to values. That mapping can be empty, but that's different than the different interface `not_a_closure(f: Function)`, which is incapable of closing over anything, including the empty set.

I would prefer other sources of information. Like papers from 60s to 80s.

http://repository.readscheme.org/ftp/papers/ai-lab-pubs/AIM-... (Lambda: The Ultimate Imperative, Steele & Sussman 1976)

Pay attention to page 17 and subsequent pages, talking about variable binding in a closed or open context - this is the closing that closure refers to. Page 31 talks about closure in the context of a lambda.

>So yes, every function is a closure [in Elisp].

-- only if lexical-binding is non-nil :)

No offense, but I'm not going to base my understanding of how programming language terminology works on Elisp.

> Rust enthusiast

What a surprise...

You're free to be as prejudiced as you want, but I gave an example of an actual implementation.

Correct me if I'm wrong, but isn't Elisp generally considered to be a rather poor dialect of Lisp? At the very least, there seems to be a desire in at least part of the Emacs community (i.e. the one place where Elisp is actually used) to replace it with Guile, which I can only assume is due to it being a superior language.

Regardless, though, I don't think what any one language calls something is going to go very far in convincing me. As for the Rust ad hominem, I've been specifically arguing against the terminology that Rust uses (i.e. all lambdas are called closures, and nothing is called lambdas) that is much more consistent to your viewpoint than mine, so I don't really see what point you're trying to make there, other than that you're as prejudiced against Rust as I am against Elisp. But that's fine though; people are free to like whichever languages they like, but I don't really think it's worth taking it personally when others disagree.

Elisp is not to be replaced with Guile. That's a myth.

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