Scheme::R5RS. Trying to make the absolute value procedure

Question

I'm trying to make the absolute value procedure and code got an error. I don't know why :/

(define (abs x)
   (cond ((> x 0) x)
       ((= x 0) 0)
       ((< x 0) (- x))))

Error message:

define-values: assignment disallowed;
 cannot change constant
  constant: abs

The `abs` function is already there in the library. Rename your `abs` to something else. Racket won't allow you to mutate it with `set!` too. — Vlad, Feb 14 '20 at 08:45
Is this [Racket R5RS](https://docs.racket-lang.org/r5rs/index.html)? — Flux, Feb 15 '20 at 13:01

score 0 · Accepted Answer · answered Feb 14 '20 at 09:06

0

The abs procedure is already part of the language you're using and you can't create another procedure with the same name. Simply rename it (and the implementation can be simplified a bit, by the way):

(define (myabs x)
  (cond ((>= x 0) x)
        (else (- x))))

answered Feb 14 '20 at 09:06

Óscar López

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386

Thanks! +Why do we have to put brackets around -x (line3)? I've tried without, and it's not showing correct result. – Honolulu Feb 14 '20 at 09:32
@Jayjay We need the brackets because we're calling the `-` procedure with a single argument: `x`. If you wrote `-x` it wouldn't work, Scheme thinks `-x` is a variable name which doesn't exist, because a dash can be part of a variable name – Óscar López Feb 14 '20 at 09:36
@Jayjay And to add to that `-` and `abs` are variables too and they evaluate to procedure values. – Sylwester Feb 14 '20 at 22:04

score 0 · Answer 2 · answered May 10 '20 at 16:21

0

Here is a simple solution to your question.

(define (abs x)
(if (< x 0) (* x -1) (* x 1)))

Here is the results in DrRacket:

> (abs -4)
4

> (abs 3)
3

answered May 10 '20 at 16:21

Bosphoran

21
7

Scheme::R5RS. Trying to make the absolute value procedure

2 Answers2