[REBOL] word Re:(2)
From: tim:johnsons-web at: 14-Aug-2000 8:53
I have a question about the code that Russell
wrote to illustrate:
Let's contrast the two "files" below:
REBOL[]
memb: none ; "declared" at beginning of file, global
block: ["a" "b" "c"]
code: [print memb]
foreach member block
[
memb: member
do code
]
; EOF
; Now without the initial declaration
REBOL[]
; no global
block: ["a" "b" "c"]
code: [print memb]
foreach member block
[
memb: member
do code
]
; Is there a difference in machine overhead here? Is 'memb being "redeclared"
; in each loop or is it sort of cached. This may seem to be a subtle
distinction
'; in these days of "just do it", but in the case of several million loops,
it could
; make a difference.
Just wondering :)
Tim
memb is a "global" word. In each loop, it's value is set equal to the value
