* numArgs of a lambda procedure
@ 2012-08-12 7:10 Chuah Teong Leong
2012-08-12 7:22 ` Per Bothner
0 siblings, 1 reply; 2+ messages in thread
From: Chuah Teong Leong @ 2012-08-12 7:10 UTC (permalink / raw)
To: kawa
Is there a simple way to get the number of required arguments from a
procedure object?
I tried the following
(define lamb0 (lambda () #f)
(define lamb1 (lambda (a) #f)
(define lamb2 (lambda (a b) #f)
(define lamb3 (lambda (a b c) #f)
(define lamb4 (lambda (a b c d) #f)
(display lamb0:numArgs)(newline)
(display lamb1:numArgs)(newline)
(display lamb2:numArgs)(newline)
(display lamb3:numArgs)(newline)
(display lamb4:numArgs)(newline)
0
4097
8094
12291
16388
Through this i can just match the value for 0 to 4 but I'm thinking
there must be a more orthodox means to do this.
Is there something I can do like
(procedure-property lamb0 'num-args)
to get the number of arguments?
Thank you in advance
^ permalink raw reply [flat|nested] 2+ messages in thread
* Re: numArgs of a lambda procedure
2012-08-12 7:10 numArgs of a lambda procedure Chuah Teong Leong
@ 2012-08-12 7:22 ` Per Bothner
0 siblings, 0 replies; 2+ messages in thread
From: Per Bothner @ 2012-08-12 7:22 UTC (permalink / raw)
To: Chuah Teong Leong; +Cc: kawa
On 08/12/2012 12:10 AM, Chuah Teong Leong wrote:
> Is there a simple way to get the number of required arguments from a
> procedure object?
Use minArgs() and MaxArgs().
numArgs is just an (unfriendly) "encoding".
--
--Per Bothner
per@bothner.com http://per.bothner.com/
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