00:00:00 --- log: started forth/04.04.02 00:25:23 --- quit: ASau () 00:53:48 --- quit: yeoh ("User pushed the X - because it's Xtra, baby") 01:38:51 --- join: ASau (~asau@158.250.48.196) joined #forth 01:46:52 Privet 01:58:51 God dag! 02:45:35 --- join: qFox (C00K13S@cp12172-a.roose1.nb.home.nl) joined #forth 02:58:13 --- quit: jewel (Read error: 54 (Connection reset by peer)) 03:10:13 --- join: crc (crc@1Cust178.tnt1.levittown.pa.da.uu.net) joined #forth 03:14:16 --- quit: crc (Nick collision from services.) 03:14:48 --- join: crc (crc@1Cust191.tnt1.levittown.pa.da.uu.net) joined #forth 03:21:54 --- join: crc_ (crc@1Cust198.tnt1.levittown.pa.da.uu.net) joined #forth 03:22:05 --- quit: crc (Nick collision from services.) 03:22:23 --- nick: crc_ -> crc 03:38:25 --- quit: crc ("Off to work...") 04:56:25 --- join: yeoh (~Yeoh@219.95.25.34) joined #forth 05:01:05 Herkamire 06:00:09 bye :) 06:00:12 --- part: yeoh left #forth 06:06:21 --- join: tathi (~josh@pcp02123722pcs.milfrd01.pa.comcast.net) joined #forth 06:12:59 --- quit: Robert ("brb") 06:25:49 --- quit: tathi ("leaving") 06:39:20 --- join: Robert (~snofs@c-f55a71d5.17-1-64736c10.cust.bredbandsbolaget.se) joined #forth 07:15:25 --- join: networm (~networm@L0626P08.dipool.highway.telekom.at) joined #forth 07:28:13 --- quit: ASau (kornbluth.freenode.net irc.freenode.net) 07:28:33 --- join: ASau (~asau@158.250.48.196) joined #forth 07:49:32 --- join: kc5tja (~kc5tja@66-91-231-74.san.rr.com) joined #forth 07:49:42 --- mode: ChanServ set +o kc5tja 08:00:13 --- join: Teratogen (~leontopod@intertwingled.net) joined #forth 08:00:20 FORTH UBER ALLES! 08:02:43 Unfortunately, I do not speak German. 08:02:49 Teratogen: you loon 08:04:35 ianp, you crazy person! 08:05:25 thats me 08:08:38 --- join: tathi (~josh@pcp02123722pcs.milfrd01.pa.comcast.net) joined #forth 08:09:31 --- quit: ASau (Read error: 60 (Operation timed out)) 08:40:57 --- nick: Teratogen -> {}o_0{} 08:40:57 --- quit: networm (Read error: 104 (Connection reset by peer)) 08:41:42 --- nick: {}o_0{} -> Teratogen 08:44:31 qFox 08:46:06 --- join: madwork (~madgarden@derby.metrics.com) joined #forth 08:46:29 yea i'm trying to figure it out myself 08:46:35 having trouble with mul and div 08:48:13 --- join: networm (~networm@L0632P24.dipool.highway.telekom.at) joined #forth 08:50:19 Greetings. 09:00:35 hi madwork! 09:06:08 hi! 09:12:01 Yay for enormously bloated and complexifuckated code bases! 09:16:53 heh 09:17:00 that's what you are working on? 09:17:45 I'm still not 100% sure. :P 09:17:56 It's a huge print engine. 09:18:23 Part of it is an interpreter that converts from some AS400 printer code format to PCL. 09:18:52 hm.. PCL.. should I know it? 09:19:27 Printer Control Language. 09:19:34 ah 09:41:15 hm 09:45:13 qFox: I understand the MUL instruction and DIV instruction in the p24 manual. Although I don't entirely understand Why (on a bit level) they produce a multiply and devision when repeated 09:54:21 oh oh oh oh I get it. I now grok the MUL instruction. 09:55:18 when I read this: "The T-A register pair is now shifted to the right by one bit." I was picturing T on the right. but on the left it works out :) 09:55:46 --- quit: networm (Remote closed the connection) 10:01:38 Otherwise they'd have called it the A-T register pair ;-) 10:02:47 yeah :) seems rather obvious now :) 10:03:18 hm well i think i have it visually, but i have to write it out in order to actually prove to myself that it is correct 10:04:05 * Herkamire tries long devision in binary with a pencil 10:04:19 :p 10:05:41 heh :) gosh, that's easy 10:06:26 but when you shift right 10:06:43 does bit 0 of T shift into bit 24 of A, or bit 23 of A? 10:06:55 no, the div rotates left 10:07:02 sorry i meant for mul 10:07:14 although same thing for div i guess 10:07:19 is the carry ignored? 10:08:24 I don't think the carry should ever get set 10:08:36 the low bit in T is shifted to the high bit of A 10:09:29 yes but they are 25bit registers 10:09:40 and bit 24 (the 25th bit) is the carry 10:10:50 thing is 10:10:55 i have all the instructions scripted now 10:11:00 except for mul and div 10:11:09 so i just copied that example to the memmory 10:11:19 the one on that page 10:11:30 and it worked etc 10:11:36 but the result is the same as what i started with :s 10:14:27 qFox: You're still getting hung up on that 25th bit. 10:14:35 yes 10:14:35 :) 10:14:38 The "Carry" bit is just the 25th bit of an addition or subtraction. 10:14:40 it'll haunt me forever 10:14:40 That's it.b 10:14:46 It's just a 25th bit. 10:14:58 so ignored with the shl/shr in mul/div 10:15:01 It has no more or no less significance than that. 10:15:04 No! 10:15:05 oki 10:15:07 thanks :) 10:15:09 It's a proper 25th bit. 10:15:22 i hate it 10:15:26 and it hates me 10:15:34 kc5tja: is the stack 25 bit also? 10:15:39 Herkamire: Yes. 10:15:41 memory is 24-bit right? 10:15:44 P24 is an offspring of the P21. 10:15:48 Yes, memory is 24 bit. 10:16:17 Anyway, I need to haul off to work. 10:16:39 --- quit: kc5tja ("THX QSO ES 73 DE KC5TJA/6 CL ES QRT AR SK") 10:17:08 rrrrr... kc5tja, stand by for questioning! 10:17:55 hahaha 10:18:22 I really don't think the carry bit of A is used in multiplication 10:18:30 thats just what he said. 10:18:36 it says a 48 bit product. not a 50 bit product 10:18:46 it says to call MUL 24 times, not 25 times. 10:19:35 if the carry bit of A was used, and you call MUL 24 times like they say, I don't see how the low bit the product (in A) would ever get set. 10:19:53 well i have a picture in my head about how it should go 10:20:53 101 x 101 should be 11001 10:21:07 unfortunatly i cant look forward so i cant do it in my head :) 10:21:17 (any product of two odd numbers comes out odd. eg with the low bit set) 10:24:54 i think i'll write it out with 5 bit registers 10:25:15 i dont feel like actually writing down 24bits a gazlion times :p 10:25:52 heh :) 10:29:26 --- join: I440r (~mark4@64.47.44.254) joined #forth 10:48:27 Hey 10:49:43 --- quit: tathi ("leaving") 11:29:32 herk 11:29:34 Herkamire 11:29:37 you there? 11:31:25 Hey Herk! Hey Herk! Here comes Daedalus! 11:31:41 wrong herk :p 11:50:30 --- quit: madwork (Read error: 104 (Connection reset by peer)) 12:29:21 --- quit: Teratogen ("SKYKING, SKYKING, DO NOT ANSWER") 12:46:10 --- join: madwork (~madgarden@derby.metrics.com) joined #forth 12:49:43 --- join: wossname (wossname@HSE-MTL-ppp61506.qc.sympatico.ca) joined #forth 12:51:52 --- quit: madwork (Read error: 104 (Connection reset by peer)) 13:05:20 --- quit: wossname ("irc is dead? :(") 13:13:14 bah i dont get the mul or div. either the explanation is incorrect or i'm looking at it from the wrong view or whatever. 13:14:08 qFox: ok, I'm back 13:14:17 my hero 13:14:22 hehe 13:14:28 no i really cant figure it out 13:15:07 especially the div 13:15:22 only add S to T if it produces a carry? 13:15:23 work on mul then 13:15:32 cant figure out either 13:15:39 but div is just plain wrong :) 13:15:55 but with mul 13:15:59 div is confusing. I have sortof a vague theoretical understanding of div. 13:16:07 mul makes sense to me 13:16:09 well lets take mul 13:16:18 add S to T, if bit 0 of S is 1 13:16:23 ok, thats fine, np 13:16:41 but then shift T-A right 13:16:41 no 13:16:51 i'm sorry 13:16:55 if bit 0 of A is 1 13:16:55 if bit 0 of A is 1 13:17:03 tahts what i mean 13:17:06 sorry 13:17:08 but ok 13:17:12 then you shift T-A 13:17:41 now am i mistaken or is T-A a bitwise concat like this: ttttttttttttttttttttttttaaaaaaaaaaaaaaaaaaaaaaaa 13:17:51 correct 13:18:04 and the product of the shr will be 0ttttttttttttttttttttttttaaaaaaaaaaaaaaaaaaaaaaa 13:18:09 ? 13:18:18 right 13:18:23 (23 a's, 24 t's, and the 0) 13:18:38 right 13:18:54 um 13:19:09 if you could look at the expression on my face right now... 13:19:18 :) 13:19:26 one side of my brain says its right, and the other side doesnt feel right about it :) 13:19:52 qFox: take a pencil, and do multiplication on paper in binary of these numbers: 11001 x 1011 13:20:08 i'll try again 13:20:22 i have a feeling i'm mixing up the shl/shr with mul/div 13:20:44 shr just moves all the bits to the right one place 13:20:58 --- join: madwork (~madgarden@derby.metrics.com) joined #forth 13:24:17 11001 13:24:18 x 1011 13:24:18 ------- 13:24:18 11001 13:24:18 11001 13:24:20 + 11001 13:24:22 ---------- 13:24:25 100010011 13:25:44 the top number is in S, the next in A 13:26:21 the answer is accumulated in T which is shifted down into A each line 13:29:34 for each digit in A (starting from the right) we add in S (but shifted left once for each digit in A we have already done) 13:29:53 yes, sorry, i "see" the concept, it just has to come thru in my head 13:30:04 the only way i'll remember it 13:30:05 --- quit: I440r ("Leaving") 13:30:58 on that machine, instead of shifting S up to add it in in the right place, it shifts the answer down 13:31:29 after you call MULL 24 times, it has been shifted all the way down into A 13:32:14 yea 13:33:22 kinda cool how the 1011 fully dissapears at the same time that the answer is fully shifted into A 13:34:00 ye :) 13:35:24 --- join: networm (~networm@L0632P24.dipool.highway.telekom.at) joined #forth 13:39:55 --- join: Downix (~downix@adsl-2-24-31.bct.bellsouth.net) joined #forth 13:47:38 --- quit: fridge (Remote closed the connection) 13:54:17 qFox: MUL make sense to you now? 13:54:40 yes, even though on paper it didnt work out with 5 registers, but i'm gonna code it anyways 13:54:49 5 bits 13:54:51 god i'm tired 13:55:59 ok, now lits DIV in 4 bits 13:56:05 we'll do 10/3 13:56:18 k 13:56:25 ie 1010/0011 13:57:28 I'm following the /MOD example 13:57:47 ok 13:57:53 we negate 3 and place it in S. So S is 1101 (-3) 13:58:12 your lsb is left? 13:58:26 x000 is your least significant bit? 13:59:38 no. lsb is on right 13:59:49 always 13:59:57 except in japenese maybe 14:00:26 numbers are always written with the least significant digits on the right 14:00:57 anyway, the 1010 goes in A, and we start T off with a zero 14:01:16 we add S into T. no carry bit. T= 1101 14:02:05 ok, the carry is dropped? 14:02:06 shift T-A left (t's cary (0) goes to A[0]) 14:02:40 so T: 1011, A: 0100 14:02:52 oh 14:02:55 that's one DIV. now we do again 14:04:19 add S into T. T: 11000 (carry set) 14:05:27 rotate T-A left (T's carry into A[0]): T: 0000, A: 1001 14:05:36 --- part: madwork left #forth 14:06:06 again add S into T. T: 1101 (no carry) 14:06:24 oh so it is actually rotated, not shifted 14:06:43 --- quit: warp0b00 ("Tried to warn you about Chino and Daddy Gee, but I can't seem to get to you through the U.S. Mail.") 14:06:54 rotate T-A left (T's lack of carry into A[0]): T: 1011, A: 0010 14:07:19 qFox: don't get lost on the terminology, it just seems like a rotate because the T's cary bit (which seems like the highest bit) is put into A 14:07:35 well if thats not rotating what is... 14:10:08 --- join: warpzero (~warpzero@dsl.142.mt.onewest.net) joined #forth 14:11:32 ok, is the carry condition valid or not? like its explained, i keep tripping over it :( 14:11:45 "If the addition produces a carry place the sum in T, otherwise leave T unchanged." 14:12:08 Oh crap, I forgot about that. 14:12:13 because if this is true, and you mul with a clear T, this condition is NEVER true... 14:12:16 no wonder my answer was coming out wrong :) 14:12:30 as you cant possibly create a carry when adding 0 to a number 14:12:47 it will be true when a high enough value from A is shifted up into T 14:13:09 oh, yea, so basicly all first div's never add 14:13:23 when you do long devision in binary, it iether goes in or it doesn't. if it goes it you put a 1 in your answer and subtract the divisor. 14:13:45 if it doesn't go in you move over a digit and repeat 14:14:23 --- join: I440r (~mark4@64.47.44.254) joined #forth 14:15:56 --- join: I440r_ (~mark4@64.47.44.254) joined #forth 14:17:31 Hi 14:17:41 init S: 1101 T: 0000 A: 1010 14:17:41 add T: 0000 A: 1010 14:17:41 rot T: 0001 A: 0100 14:17:41 add T: 0001 A: 0100 14:17:41 rot T: 0010 A: 1000 14:17:42 add T: 0010 A: 1000 14:17:44 rot T: 0101 A: 0000 14:17:46 add T: 10010 A: 0000 14:17:48 rot T: 0100 A: 0010 14:17:53 add T: 10001 A: 0010 14:17:54 rot T: 0010 A: 1010 14:17:55 is that correct? 14:19:08 --- quit: I440r (Read error: 60 (Operation timed out)) 14:20:19 the first part is 14:20:31 add T: 10010 A: 0000 \ good to here 14:20:40 what a surprise 14:20:48 then what do i do with the carry? 14:20:51 rot T: 0100 A: 0001 14:20:59 what if its this 14:21:04 add T: 11010 A: 0000 14:21:09 how would you rot that? 14:21:22 the carry bit of T goes into the low bit of A 14:21:41 yes but what happens with the high bit of T? 14:21:53 add T: 11010 A: 0000 14:22:08 where does that shift to? 14:22:10 carry? 14:22:41 dunno 14:22:54 :\ 14:23:20 let's play it through that it just dissapears and see if we get the right answer 14:23:47 alright.. 14:24:48 not a good example for that 14:25:01 add T: 10010 A: 0000 14:25:01 rot T: 0100 A: 0001 14:25:01 add T: 10001 A: 0001 14:25:01 rot T: 0010 A: 0010 14:26:21 looks correct to me 14:26:48 shr T: 0001 14:26:58 what's with this though: "1 ldi xor shr" 14:27:12 1 xor would change the low bit. then shr would drop it 14:27:26 what's the point of changing the low bit if you're just going to shr? 14:27:48 (Note: I think that this last shr undoes the most recent shl that is 14:27:48 part of div, aligning the remainder properly in T. Also I think 14:27:48 this division actually only works properly for 47 bit unsigned 14:27:48 numbers in T-A. -- WRC) 14:27:59 oh thats actually for the above, nevermind 14:28:14 but you're asking me? :) 14:28:44 ok, so the answer to your question about the high bit in T shifting out, is that it should always be a zero 14:29:00 I'm not sure about that 14:29:16 not at all... 14:29:26 i have a suggestion 14:29:36 we wait for the human cpu dude to return :p 14:29:45 kct5ja :) 14:29:54 I don't think that that bit will ever have a 1 in it. 14:30:01 ever? 14:30:01 (when the rotate happens) 14:30:07 i doubt that :\ 14:30:21 it only would if you devide by a negative number (one with the top bit set 14:30:24 must be some occurrence when that happens? 14:30:53 I don't know 14:31:36 well, I think I've given you as much as I understand 14:32:07 I gotta do some work now 14:32:30 k, later 14:56:47 on the bright side, my mul works properly now, leaving the high and low part of the result in T and S 14:56:55 (high on tos) 15:01:10 nice :) 15:01:33 that's great. I'm very tempted to write a p24 emulator myself 15:01:46 but unfortunately I have to work now 15:03:08 :) 15:03:36 i suddenly do realize that i have no idea how negative numbers are treated... whatever it is i dont think properly 15:03:40 but i'm not quite sure 15:11:52 with addition it doesn't matter (except that the carry bit often get's set) 15:13:38 I think it does matter with dividing and probably multiplication. we'll have to wait for kc5tja or someone on that one. 15:23:22 where do i go on from here... i cant seem to get specific details on what's next... coding the metacompiler, sure, but where to start? 15:23:52 the cpu emulator works fine, i think. it reads from the "memmory", and executes its instructions accordingly 15:32:48 great :) 15:38:00 next... probably you need to write a cross compiler to compile come code for it to execute. 15:38:12 s/come/some 15:38:42 and you'll probably want to make some way to interract with the p24 15:38:51 yea but i'm not sure how i start. all i have now is a nice cpu that can process memmory, but the memmory is empty :) 15:39:06 right. now you need to put some instructions in memory 15:39:37 use a forth, and define a word for each of the p24 instructions (make a word for each, that compiles that instruction) 15:39:59 so you can paste in the definitions from the p24 manual to compile definitions such as the /MOD one we were trying 15:40:14 yes can understand that, but the problem lies in "use a forth" 15:40:21 --- quit: networm ("Client exiting") 15:40:31 where do i start building a meta compiler... i mean, thats my next step, aint it? 15:41:19 I think cross-compiler is the term. I think meta-compiler is when you write the compiler in the language that the compiler implements 15:41:32 eg, herkforth contains code to compile herkforth. 15:41:52 yes, but thats what i'll want. 15:42:11 define RET which puts 0000010 or whatever the opcode is, and packs it into the next intsruction word 15:42:17 i only want to define the barest of words in code, for the rest i want a threaded forth 15:42:32 oop, dinner. gotta go 15:42:35 k 15:43:25 the other problem is this: how will you interact with the p24? 15:43:40 it has no concept of input and output 15:43:43 i'm using a mirc binvar to emulate the memmory 15:44:18 i read that the forth most basic word is QUIT, which simply loops like : QUIT check-input recurse ; 15:44:24 you could emulate a frame buffer or something 15:44:31 (was recurse the correct word for a word that calls itself?) 15:44:43 looks good 15:44:47 ok, really leaving now 15:44:50 hehe 16:03:38 : QUIT BEGIN QUERY EVAL AGAIN ; 16:07:27 --- quit: ianp (Remote closed the connection) 16:23:15 --- join: Sonarman (1000@adsl-67-113-235-181.dsl.snfc21.pacbell.net) joined #forth 16:30:01 is the word lookup table a linked list structure type? 16:30:11 single or double chain? 16:38:09 < nutbar> [root@linux!/usr/src/bind] grep "{" named.conf.newer | wc -l 16:38:09 < nutbar> 19314 16:38:09 < nutbar> [root@linux!/usr/src/bind] grep "}" named.conf.newer | wc -l 16:38:09 < nutbar> 19313 16:38:09 < nutbar> [root@linux!/usr/src/bind] 16:38:11 < nutbar> great 16:38:14 * nutbar fumes 16:38:17 < nutbar> one fucking missing } 17:00:41 ok here's what i do now. in bytes, the binvar starts at address 1, not 0 so i'll adopt that, hence the first address of the memmory is 1. 17:00:56 1-3 : jump 81 17:01:28 4-6 : user pointer (value where the current dictionary is at, basicly HERE) 17:02:05 7-9 : address of first dictionary entry (this will be a double chained linked list) 17:02:26 10 : number of bytes in input buffer (if 0, there is no input) 17:02:40 11-80 : input buffer 17:02:55 81 : QUIT 17:04:17 hm thats flawed isnt it... should be 91 17:04:44 ok its not flawed, but i mean the standard is 80 char input, so to that it is flawed :) 17:04:49 but ok, QUIT is at 91 17:18:49 oh, i guess you cant have a double chained dictionary lookup table... because then you either had to remember where you started, or you would loop endless if the word didnt exist... oopsie :p 17:20:05 hm or, you can, just a broken chain. 17:20:24 sorry if i bore anyone :) 17:39:26 --- quit: I440r_ ("Leaving") 17:55:27 well that works, yay :) 17:56:03 scripted in a BYE so it wont continously check the buffer (mirc processes scripts serially and "hangs" as long as its processing a script) 18:00:02 next is creating a word that reads in one space delimited word, that looks it up in the table, and returns the adr if a match is found, or 0 if not. 18:00:52 ohwell 4am, bed time for me 18:00:58 you are creating a forth in mIRC? 18:03:36 yep 18:03:48 :) 18:06:39 what about the p24? :) 18:06:56 what do you think is the basis for it? 18:07:35 i've completely emulated the p24 now, with the exception of DIV, which i just cant figure out. will need some help from someone who does know eventually 18:07:57 and just started to create some "boot" code for it 18:08:38 wow :) 18:08:46 so the forth kernel is for your emulated p24? 18:08:50 yep 18:09:21 the goal is to actually only use the p24, and nothing else (no quick mirc hacks to accomplish something) 18:09:39 the next goal is to define a minimal set of words in CODE, and the rest in forth. a threaded forth 18:10:00 so once i coded that minimal set, i'm completely in forth and wont bother about the cpu anymore 18:10:30 all the instructions listed, except for mul now, act exactly like they're supposed to 18:10:47 although i'm seriously thinking of adding a instruction that can retrieve a byte from the memmory :\ 18:10:57 but i kinda dont want to 18:11:24 ohwell 18:11:26 sleep time 18:11:26 nite 18:11:51 nite 18:12:11 --- quit: qFox ("if at first you dont succeed, quit again") 18:17:58 --- join: TheBlueWizard (TheBlueWiz@207.111.96.94) joined #forth 18:17:58 --- mode: ChanServ set +o TheBlueWizard 18:18:03 hiya all 18:29:49 hi 18:31:03 --- join: kc5tja (~kc5tja@66-91-231-74.san.rr.com) joined #forth 18:31:09 --- mode: ChanServ set +o kc5tja 18:31:57 hiya kc5tja 18:32:35 howdy 18:32:50 * kc5tja just purchased a primer book on Verilog. 18:33:24 --- join: thin (thin@bespin.org) joined #forth 18:33:36 A CAD? 18:35:06 * thin installed firefox, opera, kmeleon.. though firefox/kmeleon load fastest at 10 to 12 seconds.. bah 18:36:57 kmeleon loads the pages faster than firefox but firefox has a more userfriendly set up 18:37:07 so i guess i'm going with firefox 18:45:16 Verilog is a hardware description language. 18:45:21 It's used to program FPGA and CPLD chips. 18:45:33 I see 18:46:49 I am hoping to write some behavioral models of the Kestrel's custom logic, and hopefully simulate it in Icarus, an open source Verilog simulator. 18:50:43 I love Icarus 18:51:00 re Downix 18:51:03 hey Sam 18:51:05 Hey, didn't know you were here. 18:51:15 hehe 18:51:17 it's ok 18:51:18 I'm going to be testing the waters of custom hardware design for my Kestrel. 18:51:23 *nods* 18:51:31 I put Andy into an FPGA finally 18:51:50 the video-chip 18:51:56 Although my up-front costs are higher, the end result will end up being cheaper, and hopefully will let me make (a) higher profits, and (b) sell more units. 18:52:00 *nods* 18:52:15 hope things work out for you 18:52:20 The Kestrel's newest hardware plans is to be a micro-Amiga. 18:52:33 Nice 18:52:48 Eddas' current design looks more like a Jaguar raped an Amiga 18:52:48 The bus controller chip will have 8 DMA channels: 4 for video, 2 for audio, and 1 for the PIB, and 1 left unused. 18:52:50 8) 18:53:17 Unlike the Amiga, though, I don't steal cycles from the CPU. 18:53:24 *nods* 18:53:37 I actually use alternate clock phases (one more reason to use the 65816 over a Z-80). 18:53:43 *nods* 18:53:57 I'm making it more a system-in-a-system 18:54:02 communicating over PCI 18:54:04 Video resolutions planned are: 320xV at 16 colors, and 640xV at 4 colors, max. 18:54:09 not bad 18:54:26 And yes, I use bit PLANES thank you very much. 18:54:30 There *IS* a blitter. 18:54:32 no problem 18:54:35 *nods* 18:54:36 But it's CPU-fed, not DMA-fed. 18:54:43 Still much faster than a pure software-only solution. 18:55:19 2 blitter sources, so only 15 logical operations. But carefully designed software can get the same effect as a 3-operand blitter. 18:55:25 *nods* 18:55:31 not bad 18:55:39 mine is GPU-controlled 18:55:48 but DMA-fed 18:56:21 Downix: are you building something too? 18:56:34 *nods* 18:56:49 kc5tja here helped me with some of my hardware work awhile back 18:58:40 I see 18:58:48 but he's designing a much more plausible machine. I'm just having fun 18:58:49 8) 18:59:36 Downix: Well, it's supposed to be a kit. 18:59:54 It'll end up with me pre-soldering more than half the components needed for it though. 18:59:59 *nods* 19:00:00 (for most people, I would imagine) 19:00:07 right 19:00:19 I highly doubt I'll ever make a dime from my work 19:00:21 but it is fun 19:00:26 kc5tja: amiga steals cycles from the cpu? for what? monitor refreshing? 19:00:37