Yeah, that's a somewhat surprising behavior for 16-bit adds, which has tripped me up more than once (and heck, I've written a Z80 emulator.)  So yeah, a 16-bit 'sbc' (with carry flag clear) followed by an 'add' really is, in essence, a 16-bit compare.  (Notice that the add will carry if and only if the sbc did.)  8080 compatibility for... the win, in this case.

Did he?  I doubt it, he's a pretty smart guy.   Note that this is only the 16-bit ADD instruction; the 8-bit ADD affects the flags as you would expect.

That code will not do what you want.  You can't load an 8-bit value from memory directly into a register, except for the A register.  TASM interprets "ld l, (foo)" as "ld l, foo"; i.e., a request to store the value "foo" in L, not to store the contents of memory location "foo" in L; the Z80 doesn't have an instruction to do the latter.  TASM is then complaining because you're trying to store a 16-bit value in an 8-bit register.

What you want to do is either "ld hl, (foo)" or "ld a, (foo) / ld l, a".  Obviously the former will destroy the contents of H, while the latter will destroy the contents of A.

7) I have to add the value of register a into register c.

   add c, aor
   ex bc, af
    add a, c
    ex bc, afdoesnt work, so I use this:
   ld b, a
Bloop:
    inc c
    djnz Bloop
    dec c
My question is if this is the most efficient way for the solution.

My question is if this is the most efficient way for the solution.

Definitely not

And seeing as you're destroying B anyway, you can use that to temporarily store A while you use Builderboy's code.

There are none. It's simply a different code formatting accross different ASM IDE/compilers.

Yep, same with the underscore in _GetKey and all those other b_calls. Each assembler assembles differently (but tiDE is going to take any, I think).