racket/lazy

placeholder appears in result

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#3 opened on Feb 19, 2016

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Description

This program, due to @jasonhemann, produces the output '((constant) . #<placeholder>).

#lang lazy
(require lazy/force)

(define (pull t$) (if (eq? (car t$) 'tag1) (pull (cdr t$)) t$))
(define (relρ a) (cons 'tag1 (cons 'tag2 (cons a (relρ a)))))
;; change either a in the body to 'const 

(!! 
 (let (($ (cdr (pull (relρ '(constant)))))) ;; s/'(constant)/'constant/
   (cons (car $) (cadr (pull (cdr $)))))) ;; change (car $) to 'other-constant
      ;; or replace the above with (cadr (pull (cdr $)))
      ;; or in-line the let in both places
      ;; also interesting, swap the order of args to cons

Contributor guide

Research direction
Reproduce the issue by running the provided code in Racket with the lazy language. Trace the evaluation of the `pull` function and the use of `cons` and `cdr` to understand why a placeholder appears. Focus on how lazy evaluation handles the infinite stream and the manipulation within `pull`.
Tech stack
None
Domain
compilers
Issue type
Bug
DifficultyEstimated implementation difficulty for a new contributor, from 1 for very small changes to 5 for expert-level work.
3
Estimated timeA rough time range for an experienced contributor to investigate, implement, test, and prepare a pull request.
Half day
Activity statusHow available the issue appears right now: fresh, active, stale, blocked, or waiting on maintainer input.
Active
ClarityHow clearly the issue explains the expected change, acceptance criteria, and next step.
Needs investigation
Prerequisites
Racketlazy evaluation
Newbie friendlinessA 1-100 score estimating how approachable this issue is for first-time contributors.
30