Jump to content

GB Programming: Difference between revisions

m (Text replacement - "(\bld(?:|i|l|d|h) (?:.+, ?)?)\((.+)\)" to "$1[$2]")
 
(3 intermediate revisions by 2 users not shown)
Line 123:
|DE
|HL
|([BC)]
|([DE)]
|([HL)]
|([imm16)]
|-
|A
Line 288:
|Yes
|-
|([BC)]
|Yes
|No
Line 304:
|No
|-
|([DE)]
|Yes
|No
Line 320:
|No
|-
|([HL)]
|Yes
|Yes
Line 336:
|No
|-
|([imm16)]
|Yes
|No
Line 444:
So, how does running a program works? What happens is that a special register is incremented (its value is raised by one), then the processor fetches the byte located at the address held by that register, and processes it as an opcode ; when done, everything is repeated. Instructions can be one to three opcodes (bytes) large, so this cycle may repeat for a single instruction.
 
So now, how to access memory? With parenthesesbrackets! To access memory address $CD38, you just have to use ([$CD38)]. Yay!
 
To access the memory location pointed to by HL, just do... ([hl)]! It's the same with BC and DE.
 
So, to retrieve the value at memory address $6511 into register A : ''ld a, [$6511]''
Line 454:
Remember to refer to the chart above for the legal LD combinations.
 
Obviously, ''ld [$6511], a'' will overwrite the previous value stored here. But ''ld [$6511], hl'' will store a 16-bit value, which is a word long, that is two bytes long! So, not only will ([$6511)] be overwritten, but ([$6512)] too! Always be very careful about the memory you're touching. Otherwise, stuff like the [[ZZAZZ glitch]] happen.
 
For those wondering, ''ld a, [$6511]'' leaves ([$6511)] untouched.
 
==Flags==
Line 494:
!C
|-
|INC <nowiki>{reg8 | reg16 | ([hl)]}</nowiki>
|Adds one to the operand ("increments" it)
|Affected, except for reg16
|Not affected
|-
|DEC <nowiki>{reg8 | reg16 | ([hl)]}</nowiki>
|Subtracts one to the operand ("decrements" it)
|Affected, except for reg16
|Not affected
|-
|ADD A, <nowiki>{reg8 | imm8 | ([hl)]}</nowiki>
|Adds the operand to the accumulator
|Affected
Line 514:
|Not affected
|-
|SUB <nowiki>{reg8 | imm8 | ([hl)]}</nowiki>
|Subtracts the operand from the accumulator. The syntax SUB A, <nowiki>{...}</nowiki> is also valid but less common.
|Affected
Line 558:
Because two hex digits mean one byte, $D3, as well as $61, is a byte. Since $D3 and H are leftmost in both cases, ld hl, $D361 is actually a shorter form of ld h, $D3 then ld l, $61.
 
Let's say the following instruction is ld [$2315], hl. Applying the same logic would mean H's value would be stored at ([$2315)], and L's would be at ([$2316)]. However, you just lost THE GAME; because the z80 is a "little-endian" processor, L's value (the "little-end") is stored first, at ([$2315)]. So ([$2315)] is $61, and ([$2316)] is $D3.
 
Stop here, and remember this until it becomes natural to you. Because this "little-endian"ness is very tricky for beginners. It is ''very'' important when working with memory.
 
Here is an exercise : what values will ([$C000)] to ([$C00F)] contain after this code is ran?
 
Initial values :
Line 582:
ld b, [hl]
sub a, b
inc ([hl)]
inc hl
ld [hl], b
Line 673:
where reg16 is any 16-bit register pair. AF can be used here.
 
Also meet SP, which makes all of this possible. SP is the '''hardware Stack Pointer'''. You can INC and DEC it, and you can't use it as a source in LD. Here are equivalents of ''push hl'' and ''pop hl'' (assuming we could use ([sp)], 'cause we can't :3)
{| class="wikitable"
|PUSH HL
Line 819:
ld b, [hl]
</pre>
If ([hl)] equals $63, execution jumps to placeItems.
 
Otherwise, executions continues through, increments hl twice, then jumps to "someplace"
Line 1,064:
sbc hl, bc ; HL = HL - (BC + C flag) = $C303 - ($0300 + $00) = $C003
 
ld [hl], a ; ([HL)] = ([$C003)] <- A = $06
 
inc hl ; HL = $C004
 
ld b, [hl] ; B = ([HL)] = ([$C004)] = $DE
 
sub a, b ; A = A - B = $06 - $DE = $06 + (-$DE) = $06 + ($21 + $01) = $28, C flag = 0
Line 1,074:
Notice here that doing ''sub a, b'' actually increased A's value!
 
inc ([hl)] ; ([HL)] = ([$C004)] = $DF
 
inc hl ; HL = $C005
 
ld [hl], b ; ([HL)] = B = $DE
 
ld bc, 9 ; B = $00, C = $09
Line 1,084:
add hl, bc ; HL = HL + BC = $C005 + $0009 = $C00E
 
ld [hl], a ; ([HL)] = ([$C00E)] = A = $28
 
ld [$C00B], hl ; ([$C00B)] = L = $0E, and ([$C00C)] = H = $C0
 
Initial values :
Line 1,150:
 
Heavily using [http://gbdev.gg8.se/wiki/articles/Pan_Docs the Pan Docs] for GameBoy-specific stuff.
 
The [http://marc.rawer.de/Gameboy/Docs/GBCPU_Instr.html GCISheet] is useful for understanding CPU instructions and can be combined with [https://iimarckus.org/etc/asmopcodes.txt IIMarckus's opcode to instruction page] or the copy on [[The Big HEX List]].
 
[https://tcrf.net/Help:Contents/Finding_Content/Debugger_guide/BGB Torchickens has started a tutorial for BGB emulator's debugger at The Cutting Room Floor]
 
[[Category:Arbitrary code execution]]
Cookies help us deliver our services. By using our services, you agree to our use of cookies.