Some communities distinguish between lambdas and closures, but some don't. In the (mainstream) functional programming community it is unusual to make this distinction.

The recursion in open recursion refers to the ability of an object to refer to itself (the 'this' or 'self' value) and the open bit refers to late binding (the value to which 'this' or 'self' is bound is open to change).

Agreed on the duality between closures and objects. Closures are a poor man's objects and objects are a poor man's closures, as the joke goes.

> Which has nothing to do with recursion.

Had you read the article you might have realized this term is overloaded.

“When I use a word,” Humpty Dumpty said, in rather a scornful tone, “it means just what I choose it to mean—neither more nor less.” “The question is,” said Alice, “whether you can make words mean so many different things.” “The question is,” said Humpty Dumpty, “which is to be master—that's all.” - Lewis Carroll, a mathematician who wrote stories in his spare time.

Just because you didn't happen to not know this one doesn't mean that anyone just invented a new meaning.

https://en.m.wikipedia.org/wiki/This_(computer_programming)#...

A function calls a function with its own name-- there is a resemblance to recursion, where a function calls a function with its own name. This is the first I've ever heard of any controversy with the term "open recursion." Is it a common complaint, that the term "open recursion" abuses the notion of recursion?

I feel compelled to link the xkcd from the Friday before last:

https://xkcd.com/1860/

> Javascript often has closures which are not lambdas - that's the usual form of a callback function.

Can you clarify this further? To me a callback function is a lambda. It may or may not be a closure.

It seems like GP might have gotten it backwards; the usual form of a callback is a lambda that's not a closure. However, a callback function might not be either, e.g.:

   function foo(x) { return x + 1; }
   function bar(f, y) { return f(y); }

   // `foo` is a callback but neither a lambda nor a closure
   bar(foo, 2);

edit: adding the missing the "b" in lambda

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; }

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.

The only reason to distinguish between lambdas and closures is because some languages do not support closures (e.g. early versions of Emacs Lisp). Attempting to refer to other variables would always refer to a global variable with that name.

> 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".

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.

>A named lambda is still a lambda.

Most commonly a lambda is regarded as an anonymous function. [0] If you name it, it is no longer anonymous.

[0] - https://softwareengineering.stackexchange.com/questions/1307...

A named lambda isn't.

Yeah, your explanation is consistent with my understanding.