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Re: processing time & stack overflow
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Re: processing time & stack overflow


  • Subject: Re: processing time & stack overflow
  • From: monk <email@hidden>
  • Date: Wed, 29 Aug 2001 05:15:17 -0400

-- 8/28/01 12:06 PM: email@hidden said:

> On 8/28/01 8:49 AM, "Paul Skinner" <email@hidden> wrote:
>
>> on 8/28/01 11:05 AM, Arthur J Knapp wrote:
>>
>>> on NewOrder(lst)
>>> set {s1, s2, len} to ,
>>> {random number from 1 to 100000, 31415, lst's length}
>>> repeat with i from 1 to len
>>> set {j, s1} to {(s1 mod len) + 1, (s1 * s2) mod len}
>>> set {lst's item i, lst's item j} to ,
>>> {lst's item j, lst's item i}
>>> end repeat
>>> end NewOrder
>>>
>>> set lst to {"a", "b", "c", "d"}
>>>
>>> NewOrder(lst)
>>>
>>
>> Nice. I'm still digesting it. You do want to return the value of 'lst'
>> at the end though.
>> I don't see any OSAX calls. Was that in the previous version you mention?
>
>
> 'random number' is an osax call from Standard Additions. It's not a
> 3rd-party osax, but it is an osax nevertheless. Arthur mentions it because
> osax calls, to Standard Additions as to any other osax, are somewhat
> time-and-memory-demanding if you are going to repeat it thousands of times,
> and if you're counting (Arthur is always counting).

well that helps clarify why it was log-jamming along, is there a way around
this, like assigning every item to a distribution formula, that might not be
random but give me close enough results then?

thanks

h 'monk' elmer
--
            
http://www.assemblage.org
            


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