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Answer 1

+6 A:

You can't do it directly. You need to copy A into a 16-bit register pair and then do the add:

LD  B, 0
LD  C, A
ADC HL, BC

Carl Norum 2010-02-11 21:44:16

Yea I think that works.. but shouldn't B be 0 and C be A, not the other way around? (wait now it doesn't work anymore since 'ld bc, a' isn't valid)

xkdkxdxc 2010-02-11 21:47:19

Yeah, quite possibly. Sorry don't remember the right order off the top of my head. Edited back.

Carl Norum 2010-02-11 21:50:35

Oh i think it should be add, not adc

xkdkxdxc 2010-02-11 21:53:03

I thought you wanted with Carry?

Carl Norum 2010-02-11 21:53:53

C holds the lower byte, B the higher byte. So, yeah, B should be the zero.

bart 2010-02-12 00:57:00

Answer 2

A:

From http://nemesis.lonestar.org/computers/tandy/software/apps/m4/qd/opcodes.html

Add Byte with Carry-In Instructions
8080 Mnemonic Z80 Mnemonic  Machine Code Operation
ADC  M        ADC A,(HL)    8E           A <- A + (HL) + Carry

Roland Bouman 2010-02-11 21:44:28

This is adding the contents of an 8 bit memory location to the 8 bit accumulator A.

Paul R 2010-02-11 21:48:22

`(HL)` means the *contents* of the byte address pointed to by HL, though... I don't *think* that's what the OP is trying to do.

bobince 2010-02-11 21:49:20

Answer 3

+1 A:

I would like to point out that the checked response (by Carl Norum) is correct, but not the best answer. The following shows the speed of the two strategies with clock cylcles. Using the right solution saves time, and won't destroy a second 16 bit register pair.

  4   ld c,a            4   add a,l
  7   ld b,0            4   ld l,a
  11  add hl,bc         4   adc a,h
                        4   sub l
                        4   ld h,a

James 2010-07-26 22:12:10

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Z80: How to add 16 and 8 bit registers?

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