[REBOL] Re: Rebol Memory Allocation Strategy?
From: al:bri:xtra at: 22-Sep-2002 11:01
Tim wrote:
> ==>>If that is so then would not the following be more
> efficient?
>
> my-obj: make object![ ; untested code
> str: make string! 8192
> _my-f: func[][
> clear str
> ; appends much data to string...
> ] ; end function
> ] ;end object
>
> Would not 'str be allocated just once and that would be during evaluation
of my-obj?
Usually, one wants the string value for use in other things. So the first
function:
my-f: func[][
str: make string! 8192 ; untested code
; appends much data to str....
return str
]
is better if one wants to use the value returned by evaluationg 'my-f
multiple times in the same 'compose. Otherwise you'll end up with the same
value repeated several times. For example:
>> F: does [S: "" clear S append S random 100 S]
>> f ; Just to show it works seemingly well enough...
== "95"
>> f
== "52"
>> f
== "80"
>> f
== "96"
>> f
== "67"
>> x: compose [(F) (F) (F)] ; One chance in 1000000...?
== ["12" "12" "12"]
>>>> x: compose [(F) (F) (F)] ; and again... :(
== ["71" "71" "71"]
And now with new string! value each time:
>> F: does [S: make string! 2 append S random 100 S]
>> x: compose [(F) (F) (F)]
== ["32" "67" "30"]
>> x: compose [(F) (F) (F)] ; See! All different. :)
== ["57" "86" "35"]
I hope that helps!
Andrew Martin
ICQ: 26227169 http://valley.150m.com/
