[REBOL] Re: bitset help
From: greggirwin:mindspring at: 26-Sep-2002 12:26
Hi Romano,
<< I do not understand how can be used all this new stuff from the new
betas. Can
someone make an useful example to me? I knows only how to use bitset in
parse
rule for char, but this seems something different. >>
Bitsets can be very handy if you have a really large number of boolean flags
to keep track of and want to use minimal space and processing power (in
general). The idea, of course, is that each flag only takes one bit, which
means you can store boatloads of them in little memory when compared to,
say, a normal boolean/logic! value that probably uses 32 bits (or more in
the future :).
Here is something I just hacked together very quickly (really, truly, very
little testing, probably lots of problems, you have been warned, watch for
wrap, requires latest betas...I think, blah blah blah).
--Gregg
REBOL [
Title: "Bit Tools"
Author: "Gregg Irwin"
EMail: [greggirwin--acm--org]
File: %bit-tools.r
]
bit: make object! [
; Quick hack! No real thought about naming issues. May be jumping
; through a lot of unnecessary hoops, but some things that I
; thought should work didn't. :\
; Could/should CLEAR change bitset in place?
; Should bit indexes be treated as 0 based or 1 based?
; Note XOR naming issue in comment.
; Add TOGGLE function to flip bits?
; Some functions could easily take blocks to check the status of
; multiple bits (e.g. SET?).
; This is a special version just for this purpose, and always rounds up.
; So, why pass in the interval parameter? Good question. :)
round-to-interval: func [value interval /local whole-val diff] [
whole-val: to integer! value / interval
diff: either 0 = remainder value interval [0][1]
whole-val + diff * interval
]
bitset-from-bits: func [
bits [integer! block!] "Bit indices."
/local result
][
bits: compose [(bits)]
result: make bitset! round-to-interval max 1 first maximum-of bits 8
foreach bit bits [insert result bit]
]
and: func [set1 [bitset!] set2 [bitset!]] [
intersect set1 set2
]
or: func [set1 [bitset!] set2 [bitset!]] [
union set1 set2
]
; Just using to binary! on bitsets doesn't work. You get junk
; at the end if the bitset is short than 256 bits.
; This is called BXOR because calling it XOR and then referencing
; the standard XOR function as SYSTEM/WORDS/XOR fails. Must be
; missing something obvious, but I'm not seeing it. Need more coffee.
; There has *got* to be a better way than this!
bxor: func [set1 [bitset!] set2 [bitset!] /local b1 b2] [
(head system/words/clear at to binary! set1 add 1 divide length?
set1 8)
xor
(head system/words/clear at to binary! set2 add 1 divide length?
set2 8)
]
not: func [bitset [bitset!]] [
complement bitset
]
get: func [
bitset [bitset!]
index [integer!] "Which bit (0 to (length? value) - 1)"
][
either empty? and bitset bitset-from-bits index [0][1]
]
set: func [
bitset [bitset!]
index [integer!] "Which bit (0 to (length? value) - 1)"
][
insert bitset index
]
set?: func [
bitset [bitset!]
index [integer!] "Which bit (0 to (length? value) - 1)"
][
either empty? and bitset bitset-from-bits index [false][true]
]
clear: func [
bitset [bitset!]
index [integer!] "Which bit (0 to (length? value) - 1)"
][
and bitset complement bitset-from-bits index
]
]
; Paste this into the console to see the results in the conext of calls.
print items: make bitset! 40
print bit/set items 0
print bit/set items 39
print bit/set? items 0
print bit/set? items 1
print bit/set? items 38
print bit/set? items 39
print bit/get items 0
print bit/get items 1
print bit/get items 38
print bit/get items 39
print items2: complement items
print bit/and items items2
print bit/or items items2
print bit/bxor items items2
print items: bit/clear items 0
print items: bit/clear items 39
halt
