Talk:J operator

This article is rated Stub-class on Wikipedia's content assessment scale.
It is of interest to the following WikiProjects:

Computing

	This article is within the scope of WikiProject Computing, a collaborative effort to improve the coverage of computers, computing, and information technology on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.ComputingWikipedia:WikiProject ComputingTemplate:WikiProject ComputingComputing articles
???	This article has not yet received a rating on the project's importance scale.
	This article has been automatically rated by a bot or other tool as Stub-class because it uses a stub template. Please ensure the assessment is correct before removing the `\|auto=` parameter.

Electrical engineering High‑importance

	Engineering portal This article is within the scope of WikiProject Electrical engineering, a collaborative effort to improve the coverage of Electrical engineering on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.Electrical engineeringWikipedia:WikiProject Electrical engineeringTemplate:WikiProject Electrical engineeringelectrical engineering articles
High	This article has been rated as High-importance on the project's importance scale.
	This article has been marked as needing immediate attention.

What does it do and why?[edit]

Why did they invent the J operator? The article does not clearly describe what the purpose of the J operator is, or give any examples of how it would be used. What... is it? — Preceding unsigned comment added by 71.234.123.137 (talk) 04:52, 29 May 2012 (UTC)[reply]

I don't know why they invented J operator. I also think that it's misleading to call it that. The original name was "program-point". The paper that describes it, describes the operator as "the thing most closest to jump", which was basically a mechanism for cases where eager evaluation would not do. (This is my speculation). As far as I can understand, the below Common Lisp code tries to roughly represent the idea:

(defun j-operator (function continuation)
  (lambda (&rest args)
    (funcall continuation (apply function args))))

(defun j-operator-test (continuation)
  (j-operator (lambda (x) (+ x x)) continuation))

;; CL-USER> (j-operator-test (lambda (x) (format t "~&result: ~d" x)))

;; #<CLOSURE (LAMBDA (&REST ARGS) :IN J-OPERATOR) {100565428B}>
;; CL-USER> (funcall * 2)
;; result: 4

79.183.183.207 (talk) 11:05, 24 January 2014 (UTC)[reply]