CODEWITHSUNDEEP
racketevaluation
user3855664
user3855664

Reputation: 69

How deterministic is Racket's evaluation order?

I would like to known how deterministic Racket's evaluation order is when set! is employed. More specifically,

  1. Does #%app always evaluates its arguments from left to right?
  2. If no, can the evaluation of different arguments be intertwined?

Take, for instance, this snippet:

#lang racket

(define a 0)
(define (++a) (set! a (add1 a)) a)

(list (++a) (++a)) ; => ?

Could the last expression evaluate to something different than '(1 2), such as '(1 1), '(2 2) or '(2 1)?

I failed to find a definite answer on http://docs.racket-lang.org/reference.

Upvotes: 0

Views: 185

Answers (1)

Sylwester
Sylwester

Reputation: 48745

Unlike Scheme, Racket is guaranteed left to right. So for the example call:

(proc-expr arg-expr ...)

You can read the following in the Guide: (emphasis mine)

A function call is evaluated by first evaluating the proc-expr and all arg-exprs in order (left to right).

That means that this program:

(define a 0)
(define (++a) (set! a (add1 a)) a)

(list (++a) (++a)) 
; ==> (1 2)

And it is consistent. For Scheme (2 1) is an alternative solution. You can force order by using bindings and can ensure the same result like this:

(let ((a1 (++ a)))
  (list a1 (++ a)))
; ==> (1 2)

Upvotes: 2

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