Hi,

I'm trying to do a 16 by 24bit signed addition for my z80 program. Now I have this code:   ld hl,(var1)
  ld de, (var2)
  ld a, h
  rla
  sbc a,a
  add hl, de
  ld b, a
  ld a, (var1+2)
  adc a, b
  ld (var1+2), aIt seems to work in some situations, but it also sometimes gives other results.
What am I doing wrong?

btw: var1 is the 24bit var, var2 the 16bit one. I haven't fully written this routine, a part of it comes from a load 16bit to 24bit routine from calc84maniac.

many thanks in avantage.

Thanks, but it still doesn't fully work. (It gets the same results, actually.

Thanks, but it still doesn't fully work. (It gets the same results, actually.

Oh, it looks like you forgot to save the HL result back to var1. Also Xeda, I don't think that code works for a signed 16-bit value (that looks like it could work for unsigned though)

That also gives the same result...

BTW: I do save the lower 16 bits. I just forgot to copy the last line, which is :ld (tx2), hl
EDIT: double ninja'd I should try to type faster...

But if xeda's code only works for unsigned numbers, and it gave the same result as mine, does that mean if I'm just handeling the sign incorrectely? If so, then what am I doing wrong with the sign?

EDIT: I'm also not entirely sure if it's the addition that causes the problem, but as the rest of the code is almost simular to my previous triangle drawing routine (16bit and 8bit), eccept for the addition and the storing to variables.

This is what it does:

I made the two triangles 50% transparent so you can see both of them well. They should form a rectangle, but my triangle drawing routine fails on one of them.

It still gives the exact same result

It's difficoult to calculate what walues are passed to the routine by the screenshot, so as logn as i can't get wabbitemu to debug, I can't tell which arguments are passed to it.