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program does and how it functions when they reuse the code a year after they wrote it.This also helps the person reading the code and gives them insight into why the author wrote the routine the way they did. You are allowed to use the Bellterm code on a few conditions. The main ones being the signing of a license for the code if used commercially. It is covered in the "README" file within the SFX files. >It seems to me that ML is very difficult to edit. That is the value of the source code! I uploaded some ML tutor programs a year or so back. Check them out. ------------ Category 5, Topic 33 Message 20 Thu Aug 19, 1993 CBM-ED [e.g.bell] at 08:49 EDT Well put John! ------------ Category 5, Topic 33 Message 21 Tue Aug 24, 1993 M.SEABRUM at 20:35 EDT Ok. Ok. I have done some reading on a trip to Kansas and on the way back and everything is kinda falling into place. I have Jim Butterfields book and it's helping me out alot. Tell me this. I guess I will have to buy a SYMBOLIC ASSEMBLER from TENEX or somewhere. If I had the source code for say this ML that I am working with, I can change the address from within the source and it will load at it's new location changing all the JMPS and JSR's along the way. Right? I guess this might depend on the programmers style. I would like to thank everyone for introducing me to ML. Now I have ran into something that I'm total confused about. It's the Logical operators. AND, OR, and EOR. I understand that AND turns bits off, OR does the opposite, and EOR flips the bit (0 becomes 1, 1 becomes 0). What's the purpose of these operators. I have read and read and still don't understand them. I know that they are dealing with bits and are very confusing. I'm writing ML programs [Hand on Experience] thanks to Mr. Butterfield. Little things, not biggies yet. I'm curious. If I change the error vectors and point them to a area of memory where I will store a ML program to reboot my BBS if a error occrs, how will I say Load "Filename",8 in ML? I know I must disable the interrupts before my routine and enable them after it is complete. Just before I say load the program I guess. Any help? ------------ Category 5, Topic 33 Message 22 Tue Aug 24, 1993 C128.JBEE at 20:56 EDT >M.SEABRUM >If I had the source code for say this ML that I am working with, I can >change the address from within the source and it will load at it's new >location changing all the JMPS and JSR's along the way. Right? Yes, it is usually as simple as that. ------------ Category 5, Topic 33 Message 23 Wed Aug 25, 1993 R.KNOP1 [Rob Knop] at 01:17 EDT Re: AND and OR and EOR. The easiest way to understand these is with a truth table. Consider a single bit (with apologies to those in 40 columns): input2 input2 AND ------ OR ------ 1 0 1 0 input1: 1 1 0 input1: 1 1 1 0 0 0 0 1 0 EOR input2 ------ 1 0 input1: 1 0 1 0 1 0 When you do this to the whole byte, you do it bit-by-bit, so, for example: $35 AND $6D = %0011 1001 AND %0110 1101 ---------- %0010 1001 = $29 $35 OR $6D = %0011 1001 OR %0110 1101 ---------- %0111 1101 = $7D As to why would you use this. Well, once you learn how they work and get a feel for them, you will start seeing circumstances in which they will be useful. For intance, suppose you wanted to test to see if the joystick firebutton was held down. However, the register for the joystick actually has 5 bits controlled by the 'stick: for directions (up, down, left, right) and the button. Given that the button is controlled by the highest bit, bit 7, you could extract the state of the button using: lda $DC00 ;Control register for Joystick port 2 and #$80 ;$80 = %1000000 At this point, the accumulator holds $80 if the button is up, $00 if the button is pressed, regardless of the direction in which the joystick is being pushed. Hope this is of some help. Re: buying a symbolic assembler, before you do so be sure to ask folks here for advise. There are some that are better than others. Once you get to the good ones, though, you will never get all of us here to agree as to which is the best! Some will swear by Buddy, others won't budge from Merlin, while others are addicted to LADS and derivatives. Are you on a 64 or a 128? -Rob ------------ Category 5, Topic 33 Message 24 Wed Aug 25, 1993 CBM-ED [e.g.bell] at 08:50 EDT Michael: You have already gotten some good answers from John and Rob, but I'll add a few logs too... MR> I know that they are dealing with bits and are very confusing. MR> I'm writing ML programs [Hand on Experience] thanks to Mr. MR> Butterfield. Little things, not biggies yet. Rob's answer was very good. I was going to use sprites as an example. Sprites are bit controlled... for example, the sprite control register turns the eight sprites on and off using a corresponding bit for each sprite. To turn on sprite 1 without affecting the rest of the sprites you would, for example, use the instructions: lda 53269 ora %00000001 sta 53269 To turn it off, you would use the instruction: lda 53269 and %11111110 sta 53269 The AND and OR are used so that you can affect just the desired bit or bits and leave the ones you don't want to change alone. EOR has a more practical use, or one that I use all the time. I use it to toggle bits. For example, if I want something to happen when a flag is set (greater than zero) and something else when it is not set (equal to zero) I use EOR to toggle a bit in the flag. This way, I only need 1 routine with 1 entry point to toggle the flag. For example: lda flag eor %00000001 sta flag Every time you used this bit of code, the variable 'flag' would change from 0 to 1 (assuming no other bits are being used as flags). MR> I'm curious. If I change the error vectors and point them to a MR> area of memory where I will store a ML program to reboot my MR> BBS if a error occurs, how will I say Load "Filename",8 in ML? Depends on whether you are doing it in BASIC or ML. If in BASIC, you will just execute your BASIC program at the appropriate line number, something beyond the scope of this answer (tho not all that difficult). If in ML, an example might be: lda length-of-name:ldx #<name.low:ldy #>name.high:jsr setnam lda #0:tax:jsr setbnk; (c128 mode only) lda channel:ldx drive number:ldy #$01:jsr setlfs:jsr load This will do a 'relocating' load to the address in the file header, which should work ok for most of your program files. However, this will not do all that you want if the file being loaded is a BASIC file, because you will still have to RUN it. But what you are asking about the error vectors will work if you do it right. MR> I know I must disable the interrupts before my routine and MR> enable them after it is complete. Just before I say load the MR> program I guess. Any help? Not sure why you have to do this. The vectors are not really involved with the interrupts, at least not the error vector. & Y N6 j ^ I don't think it would hurt tho. ------------ Category 5, Topic 33 Message 25 Wed Aug 25, 1993 R.KNOP1 [Rob Knop] at 22:26 EDT Another clever little use of ora is to check multi-byte values vs. zero. Suppose you want to test if a 2-byte variable is zero; the most boneheaded way to do this is (pardon my geoProgrammer accent): lda var cmp #$00 bne 10$ lda var+1 cmp #$00 bne 10$ ; -- code for var=0 rts 10$ ; -- code for var<>0 rts Now, of course, the cmp's are wholly gratuitous here, since lda will set the Z flag, so you can shave off four bytes by cutting the code to: lda var bne 10$ lda var+1 bne 10$ ; -- code for var=0 rts 10$ ; -- code for var<>0 rts However, with the use of the ORA opcode, you can cut of another two bytes: lda var ora var+1 bne 10$ ; -- code for var=0 rts 10$ ; -- code for var<>0 If you then go to variables longer than two bytes, the savings becomes more significant. I have used this method routinely myself in code that needed to repeatedly check various four-byte variables vs. 0. Instead of four lda's and four bne's, I have one lda, three ora's, and one bne -- a savings of three bne's, or six bytes. -Rob ------------ Category 5, Topic 33 Message 26 Fri Aug 27, 1993 M.SEABRUM at 00:29 EDT Thank you guys for all of your input! I really appreciate it. I have saved all the comments/suggestion concerning this matter and will save it to disk to try to make some sense out of it. I have a new question now. Don't I always :)! Here we go. As mentioned in a previous statement, I'm basically thru with GM- BBS V4.0 and it has been released. Since the release, I have noticed a few minor bug. These are dealing with LINEFEEDS of all things. I was not responsible enough to check the ML to see if it sends out linefeeds after each carriage return. The linefeeds are added automatically if the characters going out is sent from within the program. However, no linefeeds are going out if I read something from a disk. Like a SEQ file or something. I know if you open the modem channel above 127 linefeeds will follow carriage return. Is that correct? It really does not matter because it will be too much trouble to go back and change it open statement. Wait a minute, it should not be that much trouble after all. So I can get that open as a option. Anyway, the BBS is located at 40960 (A000) and ends at 42040 (a438). As you can see, the ML is located at the bottom of basic! How in the world can he get a prgram to enter the bottom of basic [Basic Rom]. I thought ROM meant 'Read Only Memory' and that it could not be written to. Somehow there's a trick that lets him do this. Another question, why when I go to disassemble this ML it does not work. My procedure for editing this ML is as follows: 1) Turn off computer 2) Load Supermon + (w/,8) 3) Type Run [Once in Supermon] l "gm-bbsml-2",08 [Calls the BBSML into memory]. Remember, it's at the bottom of BASIC. After the ML has loaded, I do the following: d +40960 It's suppose to disassemble the ML, instead it gives me a whole diffrent reading... For example: Address 40960 (A000) in the BBS ML read ... brk Ok.. But when I disassemble that address I get something else not the BRK. In other words, I have printed out the ML and when I disassemble it things are not matching as they suppose to. All the address when disassembled are something diffrent from what's on the printer paper... Now the second half of the problem ... LINEFEEDS!! When the ML reads a seq or something of the disk LINEFEEDS are not coming thru or being sent to the modem. I will include the section of the ML that reads from disk and sends it to the modem. Here we go... 41298 jsr clrchn 41301 ldy #0 41303 sty 41297 41306 sty 41296 41309 ldx #8 41311 lda #0 41313 jsr chkin 41316 cmp #8 41318 beq 41321 41320 rts 41321 jsr getin 41324 sta 41256,y 41327 iny 41328 cmp #13 41330 beq 41344 41332 cpy #40 41334 beq 41344 41336 ldx +144 [?] 41338 stx +155 [?] 41340 cpx #64 41342 bne 41321 41344 ldx +144 [?] 41346 stx +255 [?] 41348 sty 41297 41351 jsr clrchn 41354 ldy #0 41356 lda 41256,y 41359 sta 41019 41362 jsr chrout 41365 ldx 831 send2 modem 0=n 255=y 41368 cpx #0 41370 bne 41395 41372 ldx #5 [Modem] 41374 jsr chkout 41377 lda 41019 41380 ldx 832 [0=graphics 255=ASC] 41383 cpx #0 41385 cpx #0 [Twice, why?] 41387 beq 41392 41389 jsr 40961 41392 jsr chrout 41395 jsr clrchn 41398 ldx 41296 41401 cpx #0 41403 bne 41408 41405 jsr 41432 41408 iny 41409 cpy 41297 41412 bne 41356 41414 ldx +255 41416 cpx #64 41418 beq 41426 41420 jmp 41471 [* not included] 41423 jmp 41301 41426 lda #8 41428 jsr close 41431 close 41432 sty 40960 41435 jsr getin 41438 cmp #0 41440 bne 41461 41442 ldx 831 41445 cpx #0 41447 bne 41467 41449 ldx #5 41451 jsr chkin 41454 jsr getin 41457 cmp #0 41459 beq 41467 41461 sta 41296 41464 jsr clrchn 41467 ldy 40960 41470 rts 40960 brk 40961 cmp #65 40963 bcc 40973 40965 cmp #91 40967 bcs 40973 40969 ora #32 [?] 40971 bne 40983 40973 cmp #93 40975 bcc 40983 40977 cmp #219 40979 bcs 40983 40981 and #127 40983 RTS Ok, there you have it. I was thinking if I relocated the area at location 40960 and placed it somewhere after the ML at 42040 say 42045 and added a routine to check the accumlator for a return key ($0D) then print it. A linefeed ($0A) will loaded into the accumulator and printed right after the return ($0D). I guess I will have to stash the contents of the accumulator away before I load it with the carriage return and linefeed. This area we are working in is a CALL FROM JSR. Therefore I was thinking the accumulator should be exactly what it was when the routine was called. So I'll stash the accumulator away when the routine is first called and will load it contents back in from memory when this routine is finished. This way, the accumulator is not altered. How does this all sound? You think it will work? Please help. DON'T FORGET to instruct me how to disassmeble this thing from the BOTTOM OF BASIC. Notes on ML program listed above: [?]= What's the function of the statement. All of this just to add a LINEFEED to data going out from disk. Is there a easier way? P.S. I know I will have to ADD JMP at address 40961. The JMP will point to the new address location (42045) Any help. I know it's quite a shopping list. ------------ Category 5, Topic 33 Message 27 Mon Aug 30, 1993 CBM-MARK at 23:46 EDT Whenever you want to send a formatted list, like the source listing in the previous message, use '*sn' to send the message. The formatting won't be all messed up. Mark ;) ------------ Category 5, Topic 33 Message 28 Tue Aug 31, 1993 M.SEABRUM at 03:35 EDT I see. You can delete the message because the problem has been solved. Thanks.. ------------ 5