FizzBuzz/Assembly
360 Assembly
<lang>FIZZBUZZ CSECT A SECTION OF CODE STARTS HERE, LABEL IT FIZZBUZZ
- HOUSE KEEPING AREA**********************
USING *,12 FOR THIS PROGRAM WE ARE GOING TO USE REGISTER 12
STM 14,12,12(13) SAVE REGISTERS 14,15, AND 0-12 IN CALLER'S SAVE AREA
LR 12,15 PUT OUR ENTRY ADDRESS(IN R15) INTO OUR BASE REGISTER
LA 15,SAVE POINT R15 AT THE *OUR* SAVE AREA (DEFINED AT THE END)
ST 15,8(13) SET FORWARD CHAIN
ST 13,4(15) SET BACKWARD CHAIN
LR 13,15 SET R13 TO THE ADDRESS OF OUT NEW SAVE AREA
- MAIN*PROGRAM****************************
LA 10,LOOP PUT THE LOOP START ADDRESS IN R10
LA 8,100 PUT THE NUMBER OF ITERATIONS IN R8
LA 5,=F'1' INITIALIZE BINARY COUNTER TO ONE
LOOP EQU * LABEL THE LOOP START
A 5,=F'1' ADD TO BINARY LOOP COUNTER
AP NUM,=PL1'1' ADD TO PACKED LOOP COUNTER
B CHK15 CHECK IF COUNTER IS % 12
LCHK15 EQU * IF NOT, COME BACK
B CHK3 CHECK IF COUNTER IS % 3
LCHK3 EQU * IF NOT, COME BACK
B CHK5 CHECK IF COUNTER IS % 4
LCHK5 EQU * IF NOT, COME BACK
MVC EOUT,EMSK PREPARE TO PKD->EBCDIC
EDMK EOUT,NUM PKD->EBCDIC
ENLOOP EQU * IF A TEST WAS POSITIVE RETURN HERE
WTO MF=(E,WTOSTART) PRINT RESULT OF LOOP
BCTR 8,10 START OVER
- HOUSE KEEPING AREA**********************
L 13,4(13) RESTORE ADDRESS TO CALLER'S SAVE AREA
LM 14,12,12(13) RESTORE REGISTERS AS ON ENTRY
XR 15,15 XOR R15 SO IT IS ALL 0 (R15 CREATES THE PROGRAM RETURN CODE)
BR 14 RETURN WHERE YOU CAME FROM
- SUBROUTINE AREA*************************
- ////////CHK3////////////////////////////////////*
CHK3 EQU * LABEL ENTRY POINT
LR 6,5 LOAD R6 WITH R5(THE BINARY LOOP INDEX)
A 6,=F'1' ADD ONE TO R6
SRDA 6,32 SHIFT RD VAL 32 BITS RIGHT(TO R7)
D 6,=F'3' DIVIDE BY 3
C 6,=F'0' IS REMAINDER 0?
BE DIV3 IF SO GOTO DIV3 ROUTINE
B LCHK3 IF NOT GO BACK TO LOOP
- ////////CHK15///////////////////////////////////*
CHK15 EQU * LABEL ENTRY POINT
LR 6,5 LOAD R6 WITH R5(THE BINARY LOOP INDEX)
A 6,=F'1' ADD ONE TO R6
SRDA 6,32 SHIFT RD VAL 32 BITS RIGHT(TO R7)
D 6,=F'15' DIVIDE BY 15
C 6,=F'0' IS REMAINDER 0?
BE DIV15 IF SO GOTO DIV15 ROUTINE
B LCHK15 IF NOT GO BACK TO LOOP
- ////////CHK5////////////////////////////////////*
CHK5 EQU * LABEL ENTRY POINT
LR 6,5 LOAD R6 WITH R5(THE BINARY LOOP INDEX)
A 6,=F'1' ADD ONE TO R6
SRDA 6,32 SHIFT RD VAL 32 BITS RIGHT(TO R7)
D 6,=F'5' DIVIDE BY 5
C 6,=F'0' IS REMAINDER 0?
BE DIV5 IF SO GOTO DIV5 ROUTINE
B LCHK5 IF NOT GO BACK TO LOOP
- ////////////////////////////////////////////////*
DIV3 EQU * LABEL ENRTY POINT
MVC EOUT,FIZZ SAY FIZZ
B ENLOOP RETURN TO LOOP
- ////////////////////////////////////////////////*
DIV5 EQU * LABEL ENTRY POINT
MVC EOUT,BUZZ SAY BUZZ
B ENLOOP RETURN TO LOOP
- ////////////////////////////////////////////////*
DIV15 EQU * LABEL ENTRY POINT
MVC EOUT,FIZZBUZ SAY FIZZBUZZ
B ENLOOP RETURN TO LOOP
- VARIABLE STORAGE************************
FIZZBUZ DC CL10'FIZZBUZZ!' CREATE A STRING IN MEMORY, LABEL THE ADDRESS FIZZBUZ FIZZ DC CL10'FIZZ!' CREATE A STRING IN MEMORY, LABEL THE ADDRESS FIZZ BUZZ DC CL10'BUZZ!' CREATE A STRING IN MEMORY, LABEL THE ADDRESS BUZZ NUM DC PL3'0' CREATE A DECIMAL IN MEMORY, MAKE IT ZERO, LABEL IT NUM TEMP DS D RESERVE A DOUBLE WORD (8 BYTES) IN MEMORY, LABEL IT TEMP EMSK DC X'402020202020' CREATE A HEX ARRAY IN MEMORY, LABEL IT EMSK WTOSTART DC Y(WTOEND-*,0) LABEL THIS WTOSTART, DEFINE A CONSTANT ADDRESS EQUAL TO
- "WTOEND" MINUS HERE(*)
EOUT DS CL10 RESERVE SPACE FOR 10 CHARACTERS, LABEL THIS EOUT WTOEND EQU * THE MEMORY ADDRESS LOCATED HERE IS LABELED WTOEND
- HOUSE KEEPING AREA**********************
SAVE DS 18F
END HELLO </lang>
6502 Assembly
The modulus operation is rather expensive on the 6502, so a simple counter solution was chosen. <lang> .lf fzbz6502.lst .cr 6502 .tf fzbz6502.obj,ap1
- ------------------------------------------------------
- FizzBuzz for the 6502 by barrym95838 2013.04.04
- Thanks to sbprojects.com for a very nice assembler!
- The target for this assembly is an Apple II with
- mixed-case output capabilities and Applesoft
- BASIC in ROM (or language card)
- Tested and verified on AppleWin 1.20.0.0
- ------------------------------------------------------
- Constant Section
FizzCt = 3 ;Fizz Counter (must be < 255) BuzzCt = 5 ;Buzz Counter (must be < 255) Lower = 1 ;Loop start value (must be 1) Upper = 100 ;Loop end value (must be < 255) CharOut = $fded ;Specific to the Apple II IntOut = $ed24 ;Specific to ROM Applesoft
- ======================================================
.or $0f00
- ------------------------------------------------------
- The main program
main ldx #Lower ;init LoopCt lda #FizzCt sta Fizz ;init FizzCt lda #BuzzCt sta Buzz ;init BuzzCt next ldy #0 ;reset string pointer (y) dec Fizz ;LoopCt mod FizzCt == 0? bne noFizz ; yes: lda #FizzCt sta Fizz ; restore FizzCt ldy #sFizz-str ; point y to "Fizz" jsr puts ; output "Fizz" noFizz dec Buzz ;LoopCt mod BuzzCt == 0? bne noBuzz ; yes: lda #BuzzCt sta Buzz ; restore BuzzCt ldy #sBuzz-str ; point y to "Buzz" jsr puts ; output "Buzz" noBuzz dey ;any output yet this cycle? bpl noInt ; no: txa ; save LoopCt pha lda #0 ; set up regs for IntOut jsr IntOut ; output itoa(LoopCt) pla tax ; restore LoopCt noInt ldy #sNL-str jsr puts ;output "\n" inx ;increment LoopCt cpx #Upper+1 ;LoopCt >= Upper+1? bcc next ; no: loop back rts ; yes: end main
- ------------------------------------------------------
- Output zero-terminated string @ (str+y)
- (Entry point is puts, not outch)
outch jsr CharOut ;output string char iny ;advance string ptr puts lda str,y ;get a string char bne outch ;output and loop if non-zero rts ;return
- ------------------------------------------------------
- String literals (in '+128' ascii, Apple II style)
str: ; string base offset sFizz .az -"Fizz" sBuzz .az -"Buzz" sNL .az -#13
- ------------------------------------------------------
- Variable Section
Fizz .da #0 Buzz .da #0
- ------------------------------------------------------
.en </lang>
68000 Assembly
This implementation uses two counters instead of divisions for the moduli. <lang 68000devpac>;
- FizzBuzz for Motorola 68000 under AmigaOs 2+ by Thorham
- Uses counters instead of divisions.
_LVOOpenLibrary equ -552 _LVOCloseLibrary equ -414 _LVOVPrintf equ -954
execBase=4
start
move.l execBase,a6
lea dosName,a1 moveq #36,d0 jsr _LVOOpenLibrary(a6) move.l d0,dosBase beq exit
move.l dosBase,a6 lea counter,a2
moveq #3,d3 ; fizz counter moveq #5,d4 ; buzz counter
moveq #1,d7
.loop
clr.l d5
- fizz
subq.l #1,d3 bne .noFizz moveq #1,d5 moveq #3,d3 move.l #fizz,d1 clr.l d2 jsr _LVOVPrintf(a6)
.noFizz
- buzz
subq.l #1,d4 bne .noBuzz moveq #1,d5 moveq #5,d4 move.l #buzz,d1 clr.l d2 jsr _LVOVPrintf(a6)
.noBuzz
- number
tst.l d5 bne .noNumber move.l d7,(a2) move.l #number,d1 move.l a2,d2 jsr _LVOVPrintf(a6)
.noNumber
move.l #newLine,d1 clr.l d2 jsr _LVOVPrintf(a6)
addq.l #1,d7 cmp.l #100,d7 ble .loop
exit
move.l execBase,a6 move.l dosBase,a1 jsr _LVOCloseLibrary(a6) rts
- variables
dosBase
dc.l 0
counter
dc.l 0
- strings
dosName
dc.b "dos.library",0
newLine
dc.b 10,0
number
dc.b "%ld",0
fizz
dc.b "Fizz",0
buzz
dc.b "Buzz",0</lang>
8080 Assembly
<lang>;; CP/M FizzBuzz in 8080 assembly
bdos: equ 5 ; CP/M calls puts: equ 9
max: equ 100 ; Amount of lines to print
org 100h mvi d,max lxi b,0305h ; Fizz and buzz counters
line: lxi h,num + 2 ; Increment the ASCII number incn: inr m mov a,m cpi '9' + 1 jnz print mvi m,'0' dcx h jmp incn
print: mvi e,0 ; Mark that we haven't output anything dcr b ; Time for fizz? cz fizzo dcr c ; Time for buzz? cz buzzo dcr e ; Output number if nothing else has been printed jz check lxi h,num call outs check: lxi h,nl ; Output a newline call outs dcr d ; More lines? jnz line ret
;; "Fizz", and reset the fizz counter fizzo: lxi h,fizz mvi b,3 inr e jmp outs
;; "Buzz", and reset the buzz counter buzzo: lxi h,buzz mvi c,5 mvi e,1
;; Output string in HL preserving registers outs: push b push d push h mvi c,puts xchg call bdos pop h pop d pop b ret
;; Strings fizz: db 'Fizz$' buzz: db 'Buzz$' num: db '000$' nl: db 13, 10, '$'</lang>
8086 Assembly
Assembly programs that output a number on the screen are programmable in two ways: calculating the number in binary to convert it next in ASCII for output,
or keeping the number in Binary Coded Decimal (BCD) notation
to speed up the output to the screen, because
no binary to decimal conversion needs to be applied.
The first approach is the most useful because the binary number
is immediately recognizable to the computer, but, in a problem
where the calculations are very few and simple and the final result
is mainly text on the screen, using binary numbers would speed up
calculations, but will greatly slow down the output.
The BCD used is based on the ASCII text encoding:
zero is the hexadecimal byte 30, and nine is the hexadecimal byte 39.
The BCD number is kept in the DX register,
the most significant digit in DH and the less significant digit in DL.
See the comments for further explaining of the program's structure,
which is meant for speed and compactness rather than modularity:
there are no subroutines reusable in another program without being edited.
This program is 102 bytes big when assembled. The program is written to be run in an IBM PC because the 8086 processor alone does not provide circuitry for any kind of direct screen output.
At least, I should point out that this program is a little bugged:
the biggest number representable with the BCD system chosen is 99,
but the last number displayed is 100, which would be written as :0
because the program does provide overflow detecting only for the units,
not for tens (39 hex + 1 is 3A, that is the colon symbol in ASCII).
However, this bug is hidden by the fact that the number 100
is a multiple of five, so the number is never displayed,
because it is replaced by the string "buzz".
<lang asm> ; Init the registers
mov dx,03030h ; For easier printing, the number is
;kept in Binary Coded Decimal, in
;the DX register.
mov ah,0Eh ; 0Eh is the IBM PC interrupt 10h
;function that does write text on
;the screen in teletype mode.
mov bl,100d ; BL is the counter (100 numbers). xor cx,cx ; CX is a counter that will be used
;for screen printing.
xor bh,bh ; BH is the counter for counting
;multiples of three.
writeloop: ; Increment the BCD number in DX. inc dl ; Increment the low digit cmp dl,3Ah ; If it does not overflow nine, jnz writeloop1 ;continue with the program, mov dl,30h ;otherwise reset it to zero and inc dh ;increment the high digit writeloop1: inc bh ; Increment the BH counter. cmp bh,03h ; If it reached three, we did
;increment the number three times
;from the last time the number was
;a multiple of three, so the number
;is now a multiple of three now,
jz writefizz ;then we need to write "fizz" on the
;screen.
cmp dl,30h ; The number isn't a multiple of jz writebuzz ;three, so we check if it's a cmp dl,35h ;multiple of five. If it is, we jz writebuzz ;need to write "buzz". The program
;checks if the last digit is zero or
;five.
mov al,dh ; If we're here, there's no need to int 10h ;write neither "fizz" nor "buzz", so mov al,dl ;the program writes the BCD number int 10h ;in DX writespace: mov al,020h ;and a white space. int 10h dec bl ; Loop if we didn't process 100 jnz writeloop ;numbers.
programend: ; When we did reach 100 numbers, cli ;the program flow falls here, where hlt ;interrupts are cleared and the jmp programend ;program is stopped.
writefizz: ; There's need to write "fizz": mov si,offset fizz ; SI points to the "fizz" string, call write ;that is written on the screen. xor bh,bh ; BH, the counter for computing the
;multiples of three, is cleared.
cmp dl,30h ; We did write "fizz", but, if the jz writebuzz ;number is a multiple of five, we cmp dl,35h ;could need to write "buzz" also: jnz writespace ;check if the number is multiple of
;five. If not, write a space and
;return to the main loop.
writebuzz: ; (The above code falls here if
;the last digit is five, otherwise
;it jumps)
mov si,offset buzz ;SI points to the "buzz" string, call write ;that is written on the screen. jmp writespace ; Write a space to return to the main
;loop.
write: ; Write subroutine: mov cl,04h ; Set CX to the lenght of the string:
;both strings are 4 bytes long.
write1: mov al,[si] ; Load the character to write in AL. inc si ; Increment the counter SI. int 10h ; Call interrupt 10h, function 0Eh to
;write the character and advance the
;text cursor (teletype mode)
loop write1 ; Decrement CX: if CX is not zero, do ret ;loop, otherwise return from
;subroutine.
fizz: ;The "fizz" string. db "fizz"
buzz: ;The "buzz" string. db "buzz"</lang>
ARM Assembly
This program runs under Linux. To make it run on another system, you would have to rewrite the print routine.
<lang>.text .global _start _start: mov r8,#100 @ 100 loops mov r9,#3 @ "Fizz" counter mov r10,#5 @ "Buzz" counter mov r11,#'0 @ Decimal representation mov r12,#'1
1: mov r6,#0 @ R6 will be set if Fizz or Buzz output subs r9,r9,#1 @ Decrement Fizz counter moveq r9,#3 @ If zero, reset to 3 moveq r6,#1 @ and set R6 ldreq r1,=fizz @ and print "Fizz" bleq print
subs r10,r10,#1 @ Decrement Buzz counter moveq r10,#5 @ If zero, reset to 5 moveq r6,#1 @ and set R6 ldreq r1,=buzz @ and print "Buzz" bleq print
tst r6,r6 @ If R6 set, don't print number bne 2f ldr r1,=num strb r12,[r1,#-1]! @ Store first digit in string cmp r11,#'0 @ If the first digit is > 0... strhib r11,[r1,#-1]! @ ...then store the first one too bl print
2: ldr r1,=nl @ Finally, print a newline bl print
add r12,r12,#1 @ Increment the low digit cmp r12,#'9 @ If that puts it higher than 9... movhi r12,#'0 @ Then reset it to 0, addhi r11,r11,#1 @ And increment the high digit instead
subs r8,r8,#1 @ Decrement loop counter bne 1b @ If not done yet, do next number mov r0,#0 @ Otherwise, exit to Linux mov r7,#1 swi #0
print: mov r2,r1 @ Print zero-terminated string on Linux 1: ldrb r0,[r2],#1 @ Get byte and advance R2 tst r0,r0 @ Zero yet? bne 1b @ If not, keep going sub r2,r2,r1 @ Calculate length mov r0,#1 @ STDOUT mov r7,#4 @ write push {lr} @ Save the link register first swi #0 pop {pc} .data .align 4 .ascii "**" num: .asciz "" fizz: .asciz "Fizz" buzz: .asciz "Buzz" nl: .asciz "\n"</lang>
Z80 Assembly
For the Amstrad CPC (should work with e.g. the built-in assembler in JavaCPC; use call &4000 to start from BASIC): <lang z80>org &4000 ; put code at memory address 0x4000 wr_char equ &bb5a ; write ASCII character in register A to screen cursor equ &bb78 ; get cursor position
push bc push de push hl
ld b,100 ; loop from 100 to 1 loop:
- check for Fizz condition
ld a,(count3) dec a jr nz,next3 push bc ld b,4 ld de,fizz printfizz: ld a,(de) call wr_char inc de djnz printfizz pop bc ld a,3 next3: ld (count3),a
- check for Buzz condition
ld a,(count5) dec a jr nz,next5 push bc ld b,4 ld de,buzz printbuzz: ld a,(de) call wr_char inc de djnz printbuzz pop bc ld a,5 next5: ld (count5),a
- test if cursor is still in first column
- (i.e., no Fizz or Buzz has been printed)
call cursor ld a,h dec a jr nz,skipnum
- print number
push bc ld b,3 ld de,count loop2: ld a,(de) call wr_char inc de djnz loop2 pop bc
skipnum:
- print carriage return/line feed
ld a,13 call wr_char ld a,10 call wr_char
- increment rightmost digit
ld hl,count+2 inc (hl) ld a,(hl)
- check if value is 10 (ASCII 58)
- if so, set to 48 (ASCII 0) and increase 10's digit
cp 58 jr nz,noinc ld a,48 ld (count+2),a ld (hl),a dec hl inc (hl)
ld a,(hl)
- check second-to-right digit, if it is 10 (0), carry over to 100's
cp 58 jr nz,noinc ld a,48 ld (count+1),a ld (hl),a dec hl inc (hl)
noinc:
djnz loop
pop hl pop de pop bc
- return to BASIC
ret
count: db "001"
count3: db 3
count5: db 5
fizz: db "Fizz"
buzz: db "Buzz"</lang>