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"


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


xor   CX, CX ; counter

   inc   CX
   cmp   CX, 100

   jg    done

      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

          PutStr   fizzbuzz_lbl
          jmp      main_loop

      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   buzz_check   ; if not, try checking for buzz

          PutStr   fizz_lbl
          jmp      main_loop

      mov   AX, CX ; dividend = counter
      mov   BH, 5  ; divisor
      div   BH     ; (counter / 5)

      cmp   AH, 0        ; counter divisible by 5?
      je    print_buzz   ; if so, proceed with printing

      jmp   print_other  ; if not, then can only display number

          PutStr   buzz_lbl
          jmp      main_loop

      PutInt   CX
      jmp   main_loop

  • 1
    \$\begingroup\$ Nice! +1 for the effort writing this is a low level language. May I suggest posting the result set? \$\endgroup\$
    – Phrancis
    Jul 12, 2014 at 22:59
  • 1
    \$\begingroup\$ @Phrancis: For verification of successful execution, you mean? \$\endgroup\$
    – Jamal
    Jul 12, 2014 at 23:02
  • \$\begingroup\$ Yeah basically. Though I don't doubt it did exec. \$\endgroup\$
    – Phrancis
    Jul 12, 2014 at 23:18
  • 1
    \$\begingroup\$ @Phrancis: I only hesitate because it'll take up a lot (possibly unnecessary) room. Would a link to a screenshot suffice? \$\endgroup\$
    – Jamal
    Jul 12, 2014 at 23:21
  • 1
    \$\begingroup\$ Where can we obtain macros.s? \$\endgroup\$ Jul 13, 2014 at 2:24

2 Answers 2


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

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.

  • 4
    \$\begingroup\$ Instead of dec bh; dec bl, you can just use sub bx, 101h. Then you can directly jz print_fizzbuzz without a cmp. The other comparisons can use test bl, bl and test bh, bh, to eliminate the literal 0s in the code (this makes the code smaller). \$\endgroup\$ Jul 13, 2015 at 11:10
  • 3
    \$\begingroup\$ @ChrisJester-Young All excellent suggestions. Thanks! \$\endgroup\$
    – Edward
    Jul 13, 2015 at 11:53
  • \$\begingroup\$ Interesting hack to use partial registers for this, especially in a way that avoids reading a wider register after writing one of its halves (except when you reset only one of the counters). That would cause a partial-register stall on P6-family. (Why doesn't GCC use partial registers?). Another option is to unroll by 3 and only use one down-counter, like I played around with in this FizzBuzz answer where I stored bytes into a buffer for one print. \$\endgroup\$ Dec 20, 2020 at 20:27
  • 1
    \$\begingroup\$ test reg,reg is preferred over cmp reg,0 - it's 1 byte shorter and sets FLAGS exactly equivalently in all cases (except for AF). Test whether a register is zero with CMP reg,0 vs OR reg,reg? \$\endgroup\$ Dec 21, 2020 at 23:29

Use local labels

All of your labels are global labels.

Since all these labels are trying to complete the same task and they all work together, they should all be grouped under a single global variable, and have the rest of the labels be local.

For example, you would change this label:




Also, it's just better practice.

Different conditional jump

At the end of each check for either Fizz, Buzz, or FizzBuzz, you do something like this:

  je    print_fizzbuzz   ; if so, proceed with printing

  jmp   fizz_check       ; if not, try checking for fizz


This could be shortened to:

  jne    main_loop


If the jne doesn't pass, execution will fall through to print_fizzbuzz


Right now, your code only supports Fizz, Buzz, and Fizzbuzz.

But what if you wanted to change things up a bit? Say you wanted to say "Fizz" every fourth number?

To do this, you'd be adding quite a chunk of code.

Although, there is an easier way to do this; use strucs.

Say you created this struc:

struc message

    .say: resb 10
    .num: resb 1


You could then do something create a bunch of messages easily like this:

    db "FizzBuzz", 0, 0
    db 15

    db "Buzz",0,0,0,0,0,0
    db 5

    db "Fizz",0,0,0,0,0,0
    db 3

    db 0,0,0,0,0,0,0,0,0,0; so, when iterating, can know if the end has been reached
    db 0

(The extra 0's are for filling up the 10 bytes given for the name) (Note the order: you want greatest to least)

And, you can easily

Now, in your main code, you can easily iterate through messages and, if the counter is evenly divisible by the value in the num field, then you log the say field.

Now, the code could be written like this:

xor cx, cx

    inc cx
    cmp cx, 100

    jg .done

    call search

    jmp main_loop


    mov si, messages

    mov ax, cx
    mov bh, [si + message.num]; divisor
    div bh

    cmp ah, 0; was evenly divisible
    je .print_message

    add si, message_size
    cmp byte [si], 0; the next item in `messages` is the terminator
    jne .next

    jmp .print_num

    PutStr [si + message.say]
    PutInt cx

Note: This was troublesome to test out without macros.s so if there are any issues, notify me

  • 2
    \$\begingroup\$ messages: \ .fizzbuzz: db "FizzBuzz",0 \ times 10 - ($ - .fizzbuzz) db 0 would be better for padding string fields to ten bytes long. \$\endgroup\$
    – ecm
    Dec 10, 2019 at 17:55
  • 1
    \$\begingroup\$ If you use .num == 0 as the terminator, you only have to load it once per iter (load the first iteration's divisor ahead of the first iteration, i.e. rotate and partially peel the loop to make that happen). Or with the current code, div byte [si + message.num] avoids wasting an instruction loading it into BH. Also, you can put .print_num before .print_message so the loop can fall into it. The main loop could also use a more efficient do{cx++; }while(cx<100); loop structure: Why are loops always compiled into "do...while" style (tail jump)?. \$\endgroup\$ Dec 20, 2020 at 20:39
  • 1
    \$\begingroup\$ The code in search doesn't need to be a function, it could just be the loop body, even though it's large. That could be less readable, but if you want to maximize readability, use a compiler. Usually making code more general is the last thing you want to do when implementing in asm: take advantage of as many special cases as possible. Again, otherwise use a compiler so it can take advantage of fixed constants on the fly (like 2 and 4 being powers of 2, it wouldn't use div), unless the code truly needs to handle runtime variable arrays of divisors and messages. \$\endgroup\$ Dec 20, 2020 at 20:43

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