Session Start (hexnet.forgetit.net:#disch): Mon Mar 22 16:22:05 2004 *** Now talking in #disch. *** Topic of #disch: Shop Smart. Shop S-Mart. | #Disch, the channel of slickness. *** Set by Atma 12771 minutes ago *** Users on #disch: windwaker101 @Yona @Disch *** End of /NAMES list. *** Mode for channel #disch is "+ntr" *** Channel #disch was created at Mon Mar 22 07:47:00 2004 baaaack yep hmm Disch... [04:03:04 PM] just like to say I love 1337 speak disch <--- hahahaha... cool do you know ASM? 6502: yes x86: no lol ok I know a little about using a debugger and that stuff and I was looking for a little lesson if possible TFG is idle right now well how much asm do you know? any? I know how to give a character infinite lives erm well ok but and how to find a byte that controls things like jumping lol basically but little to none, just using a debugger so you don't know any asm then =P lol aight lol asm is actually pretty easy, once you have the concept It's a lot like algebra cool (basic algebra... like 5 = 3 + x stuff) gotcha the NES, has 3 major 'registers' that you can use heh back registers are just like variables (like 'x' in the above example).. .all they do is hold a number hmm just a wild guess is this like STA? They are: the accumulator (A), X-Index (X), and Y-Index (Y) oh yeah i read about that in yoshi's NES doc cool yeah they're important to know ok you know about STA.. .that's one of the basic instructions *** Apophis (AyagonaBur@gamepit-28520C53.tnt1.petersburg.va.da.uu.net) has joined channel #disch STA is "Store Accumulator" ok i knew that meaning you take A and store it in memory somewhere ok another is LDA, which is "Load Accumulator", which sets A to a certain value ok LDA #$FF <-- that sets the Accumulator to $FF ok that's not to hard a concept there's also LDX/STX and LDY/STY... which do the same thing, but with X and Y instead of A loading a variable and storing a variable, right? basically yeah cool ok... now... how much binary do you know? hmm like... do you know about OR operations.. and AND operations? yes i know basic C and C++ ah great ok and logic gates another common 6502 instruction is AND AND #$3F <--- will AND the Accumulator with $3F ok which is like saying (A = A & 0x3F) in C/C++ ok I get that there's also ORA... which does an OR operation 3 letter words all the time and EOR, which does an exclusive OR operation (sometimes called XOR) ok yeah A = A | 0x3F <-- ORA A = A ^ 0x3F <-- EOR (just for examples) ok i get that the rest of the 6502 instructions are like that... they're all VERY basic cool I'm not going to go through them all... but lemme get a link for you ok There's 3 other registers you should know about besides A, X and Y ok let me guess (Link: http://www.obelisk.demon.co.uk/6502/reference.html)http://www.obelisk.demon.co.uk/6502/reference.html <--- the link you should check out sometime ok been there lol lol but i'll read it this time lol not all of it will make sense.. .at least not until you understand addressnig modes... and the other 3 registers did you want to guess? =P lol not at all I lost my instruction list, so that link help mes, too. Thanks. This is a million times better than my old list. haha... np ok the other IMPORTANT register is the Processor Status register ok which is 8-bits in size... but rather than storing a whole number, each bit is more like a flag which indicates current status of the processor This register consists of eight "flags" (a flag = something that indi- cates whether something has, or has not occurred). Bits of this register are altered depending on the result of arithmetic and logical operations. These bits are described below: right Bit No. 7 6 5 4 3 2 1 0 S V B D I Z C 'S' is usually called 'N'... to avoid confusion ok let's start with the easiest flag... Z ok Zero flag represented by 1, right? after -nearly- every instruction is executed.. the Z flag is changed depending on the result of the operation flags are always on (1) or off (0).. there's no other value ok Z is set (on) when the last operation resulted in zero ok and cleared (off) when the last operation resulted in nonzero so i get that LDA #$43 <-- Z would be cleared after this... since A is nonzero LDA #$00 <-- Z would be set, since A is zero ok simple! so they are loading stuff into the accumulator, right? that's what LDA does, yeah most instructions affect Z though... like ADC (which is for addition) ADC #$43 <-- would add $43 to the current contents of A because it wouldn't result in 0, right? yeah and if that result is zero Z would be set ok and if not it wouldn't be it -could- be zero yeah, exactly ok example... $80 + $80 = $100... but since registers are only 8-bit.. that turns out to be $00 ok so if you do: LDA #$80 ADC #$80 Z would be set ok (set after the ADC command) ok next flag is N sign flag? the 'sigN' or 'Negative' flag cool it is set... when the highest bit of the last operation was on set if the operation is negative and cleared if positive yes cool high bit on <-- means negative k so for more examples (I love examples) LDA #$23 <-- N would be clear LDA #$C5 <-- N would be set how does that result negative? the high bit is on oh gotcha C5 = 1100 0101 ^ high bit oh i get it N and Z are the 2 flags that are used most often more than a certain amount of binary digits, right? hm? that's high bit, right? having more than a certain amount of binary digits, right? the high bit is the most significant one.... when you're dealing with 8-bit registers... the 'high' would be the 8th bit h000 000l <-- h = high bit, l = low bit ok gotcha ok now: C C = "Carry" carry flag holds and carries out significant opperations and is used -only- in Artimatic operations (like addition, subtraction, and bit shifting) ok let's do ASL because it's the simplest ASL means "Arithmatic Shift Left" ok ASL A <--- would shift the accumulator left 1 bit is that a status register? oh it's an operation yeah sorry ^^ ok, I get it do you know how bitshifting works? yeah ok well after ASL... C is set if the bit that got "shifted out" was on and is cleared otherwise ok so if the accumulator was on 01 and it got shifted left it would be 02 and 01 was on ok and the C flag would be cleared ok if A was 80 and you left shift it A would be 00, and C would be SET ok since the bit that got shifted out was on it works the same way for LSR (shift right) and ROL and ROR (more bitshifting instructions) for ADC, it works a little different ADC = addition instruction yeah, you told me ADC stands for "Add to Accumulator with Carry" yeah now that means... if C is set when you do an ADC operation ok the sum will actually be 1 higher than normal ok so LDA #$20, ADC #$10 <-- could be either $30 or $31 ...depending on whether or not C was set ok I get it that's easy C is also set at the end of ADC, if the sum was greater than $FF and is cleared if the sum was less than or equal to $FF ok kinda make sense? yeah aight but what happens if the sum is greater than FF? well it wraps to $00 since A can only hold 8 bits but the C flag is set so the game can know that the addition was too great by checking the C flag ok but really... it's there for doing multi-byte addition think of it like basic adding you're back in 1st grade and you're doing math: 15 + 19 34 5 + 9 = 14 yeah ok so you write down the 4 ok and write the CARRY up in the 10's place so you know to count it for the next addition that's how it works in asm C4 then? hm? well... right Carry in place of the 10 place oh no no lol I mean in 1st grade when you did math like that on paper yeah you'd add the 1's first ok then if it was > 10, you'd put the carry up in the 10's place ooooooooooooh gotcah like say you want to add $03F2 and $0522 together in asm *gotcha that would be hard to do with only 8-bit registers.. since the highest value you can hold is $FF and both those numbers are greater than $FF so when this happens... what games do is the clear the carry flag (with the CLC instruction) after carrying the number, right? well... it's common practice to clear carry before you start adding ok so that you don't get that extra 1 added when you don't want it I would think so yeah then they add the 2 low bytes together ($F2 and $22 in this example) that gives them $14 and the C flag is set ok then they add the 2 high bytes together ($03 and $05 in this example) and since C is set it adds one? $03 + $05 = $09 yes! yeah so $0914 is the sum sweet ok SBC is the subtraction instruction and it works the same way but is weird ok SBC uses the C flag -backwards- ie... if the C flag is CLEAR, and extra 1 is subtracted and if it's SET, subtraction is normal that's what I thought this is the opposite of ADC so be careful ok so it would be subtracting without the Carry flag well say A is $08 ok and the C flag is CLAR and you do SBC #$04 take a guess at what A could be after that *would #$04 right? you'd think so ;) but no because C is clear lol you'd subtract an extra 1 if C was SET, you'd be right oh of course so it's #$03 right? yep since it's backwards you subtract one instead of adding it because C is set SBC also changes the C flag like that too ok like if subtracting causes a number to be less than zero.... C is -CLEAR- and if not, C is set ok cool? that's one of the harder flags oh cool next flag is I... which isn't all -that- important for romhacking if I is set... it disables IRQs from occuring but that don't matter this is an interrupt enable/disable flag. If it is set, interrupts are disabled. If it is cleared, interrupts are enabled. shouldn't even have mentioned it lol yeah cool lol you don't have to worry about that for ROM hacking messing with I will just mess up your ROM unless you know what you're doing ok D, B, and the 'reserverd' flag aren't really used so the only other one is 'V' which is Overflow overflow flag Overflow is kind of stupid... I haven't figured it out ADC, SBC, and CLV are the only instructions to change it when an arithmetic operation produces a result too large to be represented in a byte, V is set (well... and RTI and PLP.. but nm those) lol ok see that's misleading yea that description makes it sound more like what Carry does yeah How overflow works on ADC is like this: when Positive + Positive = Negative, V is set ok i get it or when Negative + Negative = Positive, V is set otherwise, V is clear so here's a test question ;) ok A = $62... C is clear ADC #$31 would V be set or clear? clear * Disch makes incorrect buzz noise * windwaker101 bet's he got it wrong lol $62 and $31 are oth positive *both the sum $93 <-- is negative (high bit is on ok ooooooooooooooooooooh of course positive + positive = negative ----> V set yeah ok SBC does the same thing... kinda I was comparing it to $100 instead of the high bit ok Positive - Negative = Negative <-- V set ok Negative - Positive = Positive <-- V set