Set by Address restrictions?

[systwo]

Hello again everyone!

I'm going through Hot Dog's tutorial on asm right now and I have encountered a slight problem. Working with variables, ld doesn't want to set a register. In the snippet below, the first line will work but the second one will not. Is there something I skipped over accidentally?

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1 2	ld a, (var1) ld b, (var1)

The error spasm outputs:

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1 2 3 4 5

Pass one... learn2.asm:7: warning: Suggest remove extra parentheses around argument Pass two... learn2.asm:7: warning: Number is too large to fit in 8 bits, truncating Done

Here is the code

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;Multiplication program #include "ti83plus.inc" .org $9d93 .db t2ByteTok, tAsmCmp B_CALL _ClrLCDFull ld a, (Mula) ;Holds the answer ld b, (Mulb) ;How many times to multiply ld c, a      ;What to add every time it multiplies, or else it will do powers BeginLoop: add a, c djnz BeginLoop ;Remember that the b is decresed before it checks ld h, 0 ld l, a B_CALL _DispHL ret Mula: .db 5 Mulb: .db 5

Thanks!


Edit:
Found the answer! For those who look at this in the future, you can only use register a to retrieve data. You need to store the memory address in hl for other registers.

[Hot_Dog]

You can also get the value in register A, and then use "ld b, a"

Wonderful job on this program so far!  Keep up the good work!

[Quigibo]

There is also another cool trick the same size and speed almost the same speed as Hot_Dog's suggestion but destroys the value of 'c' instead of 'a' (in case you need to preserve a).

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1	ld bc,(var1-1)

This works because 2 byte numbers are stored as little endian in memory.

[Runer112]

I think you mean:

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1 2	ld bc,(var1-1) ;c=(var1-1), b=(var1)

And that's the same size as Hot_Dog's suggestion, but it's 3 cycles slower.

[Quigibo]

Oops yeah, you're right, I was thinking of hl.  Fixed.

[systwo]

Wow! Thanks for the advice, I have yet to reach the multi-byte storage lesson yet, but I'm sure it will make sense to me when I get there.

[Hot_Dog]

There is also another cool trick the same size and speed almost the same speed as Hot_Dog's suggestion but destroys the value of 'c' instead of 'a' (in case you need to preserve a).

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1	ld bc,(var1-1)

This works because 2 byte numbers are stored as little endian in memory.

That's one of those tricks that totally makes sense but someone like me forgets about