---------- Forwarded message --------- From: Stefan Israelsson Tampe <stefan.ita...@gmail.com> Date: Sat, Jun 19, 2021 at 1:23 PM Subject: Re: guile style To: Tim Van den Langenbergh <tmt_...@gmx.com>
I'm a big fan of named let's which are a very general functional construct and tons of commentaries about the functional style misses that named let's exists and thinks that functional programming is just map, fold, filter etc. It beats me why when other languages go functional, they consistently do not implement named let. On Sat, Jun 19, 2021 at 12:25 PM Tim Van den Langenbergh <tmt_...@gmx.com> wrote: > On Saturday, 19 June 2021 02:55:34 CEST jerry wrote: > > I am fairly new to guile and scheme. People tell me that I should use a > > functional style. > > > > I have 3 solutions for project euler problem #1. The first is > > functional, the second is imperative and the third is written in "Little > > Schemer" style. > > > > I was hoping other guile users would comment on preferences or the > > "correct way". Sorry in advance for any wrapping problems that may occur. > > > > #!/usr/local/bin/guile -s > > !# > > (use-modules (srfi srfi-1) (jpd stdio)) ;; for folds > > (define N 1000) > > > > (define ans > > (fold + 0 > > (filter > > (lambda (x) (or (= 0 (modulo x 3)) (= 0 (modulo x 5)))) > > (iota N)))) > > (print ans) > > > > (define ans 0) > > (for i N > > (if (or (= 0 (modulo i 3)) (= 0 (modulo i 5))) (set! ans (+ ans i)))) > > (print ans) > > > > (define ans > > (let loop ((i 1) (ans 0)) > > (cond > > ((>= i N) ans) > > ((or (= 0 (modulo i 3)) (= 0 (modulo i 5))) (loop (1+ i) (+ ans > i))) > > (else (loop (1+ i) ans)) ))) > > I'm not 100% sure about how Guile does it, but I know that some Scheme > implementations do some boxing for set! operations, which will make the > second > variation poorly optimised. Personally I would use combine the first and > third > answers by doing the divisible-by check during the fold, like this: > > (use-modules (srfi srfi-1)) > > (define (divisible-by? divident divisor) > ~~(zero? (modulo divident divisor))) > > (define N 1000) > > (define ans > ~~(fold (lambda (i res) > ~~~~~~~~~~(if (or (divisible-by? i 3) > ~~~~~~~~~~~~~~~~~~(divisible-by? i 5)) > ~~~~~~~~~~~~(+ i res) > ~~~~~~~~~~~~res)) > ~~~~~~~~0 > ~~~~~~~~(iota N))) > > Vale, > > -Tim > > > >