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GB Programming: Difference between revisions
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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
To access the memory location pointed to by HL, just do... [hl]! It's the same with BC and DE.
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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
For those wondering, ''ld a, [$6511]'' leaves
==Flags==
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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
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
Initial values :
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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
{| class="wikitable"
|PUSH HL
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sbc hl, bc ; HL = HL - (BC + C flag) = $C303 - ($0300 + $00) = $C003
ld [hl], a ;
inc hl ; HL = $C004
ld b, [hl] ; B =
sub a, b ; A = A - B = $06 - $DE = $06 + (-$DE) = $06 + ($21 + $01) = $28, C flag = 0
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Notice here that doing ''sub a, b'' actually increased A's value!
inc [hl] ;
inc hl ; HL = $C005
ld [hl], b ;
ld bc, 9 ; B = $00, C = $09
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add hl, bc ; HL = HL + BC = $C005 + $0009 = $C00E
ld [hl], a ;
ld [$C00B], hl ;
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