16-bit FizzBuzz in x86 NASM assembly

Since this problem involves small numbers (particularly with a small loop count of 100), it's possible to ease the modulo operation setup by simply working with 16-bit and 8-bit registers:

$$\dfrac{\text{[AX] (16-bit register)}}{\text{[other 8-bit register]}} = \text{[AH] (remainder)}$$

My main concern is with the layout. Every "basic" high-level implementation I've seen has a check and print together for each case. I've found it easier to do the same thing here, but I'm not sure if that would also be too readable in assembly.

I'm also aware that it's good to minimize register-moving. Unfortunately, I still do that with every case since I'm incrementing with one register (CX) and using another (AX) for the dividend. I can stick to AX for both, but that may involve keeping a copy of the current counter value, which may just make the code a bit more complicated. I suppose it's not much of a problem here anyway.

Macros used:

• nwln - prints a newline

• PutStr - prints a defined string

• PutInt - prints a 16-bit integer value

It's not necessary to address the macros; they do work properly.

```
%include "macros.s"

.DATA

fizz_lbl: DB "Fizz", 0
buzz_lbl: DB "Buzz", 0
fizzbuzz_lbl: DB "FizzBuzz", 0

.CODE
.STARTUP

xor CX, CX ; counter

main_loop:
inc CX
cmp CX, 100

jg done

fizzbuzz_check:
mov AX, CX ; dividend = counter
mov BH, 15 ; divisor
div BH ; (counter / 15)

cmp AH, 0 ; counter divisible by 15?
je print_fizzbuzz ; if so, proceed with printing

jmp fizz_check ; if not, try checking for fizz

print_fizzbuzz:
PutStr fizzbuzz_lbl
nwln
jmp main_loop

fizz_check:
mov AX, CX ; dividend = counter
mov BH, 3 ; divisor
div BH ; (counter / 3)

cmp AH, 0 ; counter divisible by 3?
je print_fizz ; if so, proceed with printing

jmp

Since we're doing this in assembly language, it makes sense to do it much more efficiently than is typically done in high level languages. Otherwise, why bother with assembly language? So with that said, there are ways that this can be made much, much more efficient.
Avoid division

The div instruction in x86 is one of the slower instructions possible. Since we already know that we're looking for numbers divisible by 3, 5 or both, what would make far more sense is to simple keep countdown counters for both. Your initialization currently says:

xor cx, cx

It could be easily expanded to say:

xor cx, cx
    mov bx, 0503h  ; set bh = 5 counter, bl = 3 counter

Then instead of dividing, simply decrement:

inc cx
    cmp cx, 100
    jg done
;  instead of this...
;    dec bh
;    dec bl
;    cmp bx, 0
; per suggestion from @Chris Jester-Young use this:
    sub bx, 0101h
    je print_fizzbuzz
    test bl, bl
    je print_fizz
    test bh, bh
    je print_buzz
print_other:

Naturally the various print_... routines would have to reset bh, bl or both as well as printing.
Improve formatting

Generally speaking, assembly language code is not indented in the way you have your code indented. It's much more linear, with the only indentation for assembly language statements or directives.
Consider better I/O

Your output routines are not shown, but it's likely that it would be more efficient to keep the numeric output in string form, incrementing each ASCII digit and emitting the string, rather than repeatedly converting from binary register contents to a string value.

StackExchange Code Review Q#56884, answer score: 37