Add with Carry Example

jonesypeter · June 2016

Hi,

I have been really struggling to understand this simple example from Mastering Machine Code and I wonder if anyone can help?

I have included the op codes, and my comments as to what I think is happening:

LD DE,3385
LD HL,7BC7
LD A,L			;A = 7B
ADD A,E			;A = 7B + 33 = AE
LD L,A			;L = AE
LD A,H			;A = C7
ADC A,D			;A = C7 + 85 = 14C so A set to 4C and carry flag set
LD H,A                  ;H = 4C
LD B,H                  ;B = 4C
LD C,L                  ;C = AE
RET

Therefore BC = 4CAE

I'm obviously getting confused because I thought with load HL etc, then the H is the last number, and L is the first.

Also the carry flag is not being set in the wrong place in my calculations, and the answer should be 44,876 (Decimal) or AF4C (Hexadecimal)!

Thanks

dbolli · June 2016

Nope.

After

LD DE,$3385
LD HL,$7BC7

D = $33
E = $85
H = $7B
L = $C7

HTH,
Derek

jonesypeter · June 2016

Thanks Derek,

I thought with LD HL etc, then the number were switched around?

In the book it says with the instruction LD HL the first byte is 21h as you would expect, the second byte is the new value of L, and the third byte is the new value of H

dbolli · June 2016

Only in the assembled code to facilitate faster loading of the registers.

Turns out the good folk at Intel named the H(i)L(o) register pair for a reason at the 8080 design stage...

:)

jonesypeter · June 2016

Ohhh, its just clicked.... You only swap the numbers round when you are typing in the Hex codes! Thank you so much. I will sleep well tonight! I owe you a beer!

dbolli · June 2016

promises, promises...

Add with Carry Example

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