Sometimes you need to add an 8-bit value (in A) to a 16-bit one (in either HL, BC or DE). You can extend the value to 16-bit, but it can get cumbersome as you may need to clobber registers you're using and possibly move values around (since only HL can be a destination).

Here we show how to do this without resorting to using 16-bit operations (which can avoid all those issues). Note that here "unsigned" and "signed" refers to the 8-bit value in A.

Some notes:

• In all these cases, only A gets clobbered.
• In all these cases, you can replace HL with BC or DE.
• You really should consider wrapping these into macros.

Adding A to HL is kind of tricky (you'll probably understand it better from the comments in the code snippet below). We first add the low byte. Then we add the high byte with carry and substract the low byte to get the high byte's result (which is high byte + carry).

We'll build on the code below for the other methods.

```   add   a, l    ; A = A+L
ld    l, a    ; L = A+L
adc   a, h    ; A = A+L+H+carry
sub   l       ; A = H+carry
ld    h, a    ; H = H+carry
```

Signed addition is similar, except we "extend" A to 16-bit. The "high byte" of A is always either 0 (if positive) or -1 (if negative), so we check A's sign and adjust H accordingly. After this we add the lower byte as with unsigned addition.

```   ; If A is negative, we need to subtract \$100 from HL
; (since A's "upper byte" is \$FF00)
cp    \$80
jr    c, .positive
dec   h
.positive:

; Then do addition as usual
; (to handle the "lower byte")
ld    l, a
sub   l
ld    h, a
```

Unsigned subtraction

The idea behind this is that HL - A is the same thing as HL + (-A), so we flip A's sign then perform an addition. The catch is that A's "upper byte" becomes -1, so we need to subtract 1 from HL's upper byte to account for this. Think of this as a specialized variant of signed addition.

The exception is when A is 0 (which when negated is still 0). Substracting 0 is the same as doing nothing, so we just skip the whole ordeal in that case.

The CPL instruction calculates \$FF - A, and INC to add 1 afterward makes it \$100 - A.

```   ; Flip A's sign.  If A is zero, do nothing.
cpl
inc   a
jr    z, .skip

; Subtracting \$0001 through \$00FF is like adding
; \$FFFF through \$FF01 to A.  So subtract 1 from H.
dec   h

; Now add the low byte as usual.  Two's complement
; takes care of ensuring the result is correct.
ld    l, a
sub   l
ld    h, a
.skip:
```

Signed subtraction

Signed subtraction is trivial to build on top of signed addition: flip A's sign then do the addition. The only catch is that we need to handle that -\$80 should become +\$80. and the obvious way to do that on 8080 uses the overflow flag, which isn't present on the SM83.

```   ; Values \$00-\$7F mean subtract \$0000-\$007F or add \$0000-\$FF81
; Values \$80-\$FF mean subtract \$FF80-\$FFFF or add \$0080-\$0001
; First begin to negate A (since HL - A is the same as HL + -A)
; A will be incremented later.
cpl

; Values \$00-\$7F mean add \$0001-\$0080
; Values \$80-\$FF mean add \$FF81-\$0000
; So if value at least \$80, add \$FF00
cp    \$80
jr    c, .positive
dec   h
scf
.positive:

; Now add A + 1 to HL
ld    l, a
sub   l
ld    h, a
```

With 16-bit operations

For the sake of completeness, since if the stars align right somehow (i.e. values happen to be in the right registers) these may be more useful. Also they're more compact, if space usage matters more than speed in a particular case. The additions work by sign-extending an 8-bit value to 16-bit and then adding it to HL.

Sik finds these not ideal for several reasons:

• Destination must be HL, which is more important on 8080/SM83 than on Z80 because of its use to step through an array.
• You must clobber BC or DE.
• 8080 and SM83 lack 16-bit subtraction, having only addition. (Even though Z80 adds `sbc hl`, there is no dedicated opcode for `sub hl`.)

```   ld    e, a    ; DE = A
ld    d, 0
add   hl, de  ; HL = HL+DE
```

```   ld    e, a    ; DE = A