On Sat, Feb 11, 2012 at 15:07, Rodolfo Carvalho <rhcarva...@gmail.com>wrote:
> Hello,
>
> On Sat, Feb 11, 2012 at 14:55, Laurent <laurent.ors...@gmail.com> wrote:
>
>> ...
>>
>> (define (bar [arg2 (get-default-value foo arg2)])
>> (foo 5 arg2))
>>
>>
>>
> Maybe someone will have a better idea (or a brighter implementation), but
> if not, here's my contribution:
>
> ....
>
If we use #f for arg2 in foo, we can write:
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
#lang racket
(require rackunit)
(define (foo arg1 [arg2 #f])
(+ arg1 (or arg2 10)))
(define (bar [arg2 #f])
(let ([arg1 5])
(foo arg1 arg2)))
(check-= (bar 2) 7 0)
(check-= (bar) 15 0)
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
This way it is not necessary to accommodate for calling foo with different
number of args. All we need is to propagate the #f default value.
[]'s
Rodolfo
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