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FizzBuzz
From Rosetta Code
FizzBuzz a programming puzzle. It lays out a problem which Rosetta Code users are encouraged to solve, using languages and techniques they know. Multiple approaches are not discouraged, so long as the puzzle guidelines are followed. For other Puzzles, see Category:Puzzles.
Write a program that prints the numbers from 1 to 100. But for multiples of three print "Fizz" instead of the number and for the multiples of five print "Buzz". For numbers which are multiples of both three and five print "FizzBuzz". [1]
FizzBuzz was presented as the lowest level of comprehension required to illustrate adequacy. [2]
[edit] ActionScript
The ActionScript solution works just like the JavaScript solution (they share the ECMAScript specification). The difference is that ActionScript has the trace command to write out to a console.
for (var i:int = 1; i <= 100; i++) {
if (i % 15 == 0)
trace('FizzBuzz');
else if (i % 5 == 0)
trace('Buzz');
else if (i % 3 == 0)
trace('Fizz');
else
trace(i);
}
[edit] 8086 Assembly
When assembly programs that do 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, wich 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, wich 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".
; 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"
[edit] Ada
with Ada.Text_IO; use Ada.Text_IO;
procedure Fizzbuzz is
begin
for I in 1..100 loop
if I mod 15 = 0 then
Put_Line("FizzBuzz");
elsif I mod 5 = 0 then
Put_Line("Buzz");
elsif I mod 3 = 0 then
Put_Line("Fizz");
else
Put_Line(Integer'Image(I));
end if;
end loop;
end Fizzbuzz;
[edit] ALGOL 68
main:(
FOR i TO 100 DO
printf(($gl$,
IF i %* 15 = 0 THEN
"FizzBuzz"
ELIF i %* 3 = 0 THEN
"Fizz"
ELIF i %* 5 = 0 THEN
"Buzz"
ELSE
i
FI
))
OD
)
or simply:
FOR i TO 100 DO print(((i%*15=0|"FizzBuzz"|:i%*3=0|"Fizz"|:i%*5=0|"Buzz"|i),new line)) OD
[edit] APL
⎕IO←0
(L,'Fizz' 'Buzz' 'FizzBuzz')[¯1+(L×W=0)+W←(100×~0=W)+W←⊃+/1 2×0=3 5|⊂L←1+⍳100]
[edit] AutoHotkey
Search autohotkey.com: [3]
Loop, 100
{
If (Mod(A_Index, 15) = 0)
output .= "FizzBuzz`n"
Else If (Mod(A_Index, 3) = 0)
output .= "Fizz`n"
Else If (Mod(A_Index, 5) = 0)
output .= "Buzz`n"
Else
output .= A_Index "`n"
}
FileDelete, output.txt
FileAppend, %output%, output.txt
Run, cmd /k type output.txt
A short example with cascading ternary operators and graphical output. Press Esc to close the window.
Gui, Add, Edit, r20
Gui,Show
Loop, 100
Send, % (!Mod(A_Index, 15) ? "FizzBuzz" : !Mod(A_Index, 3) ? "Fizz" : !Mod(A_Index, 5) ? "Buzz" : A_Index) "`n"
Return
Esc::
ExitApp
[edit] AWK
BEGIN {
for (NUM=1; NUM<=100; NUM++)
if (NUM % 15 == 0)
{print "FizzBuzz"}
else if (NUM % 3 == 0)
{print "Fizz"}
else if (NUM % 5 == 0)
{print "Buzz"}
else
{print NUM}
}
[edit] BASIC
Works with: QuickBasic version 4.5
[edit] If/else ladder approach
FOR A = 1 TO 100
IF A MOD 15 = 0 THEN
PRINT "FizzBuzz"
ELSE IF A MOD 3 = 0 THEN
PRINT "Fizz"
ELSE IF A MOD 5 = 0 THEN
PRINT "Buzz"
ELSE
PRINT A
END IF
NEXT A
[edit] Concatenation approach
FOR A = 1 TO 100
OUT$ = ""
IF A MOD 3 = 0 THEN
OUT$ = "Fizz"
END IF
IF A MOD 5 = 0 THEN
OUT$ = OUT$ + "Buzz"
END IF
IF OUT$ = "" THEN
OUT$ = STR$(A)
END IF
PRINT OUT$
NEXT A
See also: RapidQ
[edit] Befunge
(befunge 93)
55*4*v _ v
v <>:1-:^
|:<$ < ,*48 <
@>0"zzif">:#,_$ v
>:3%!| >0"zzub">:#,_$^
>:5%!|
v "buzz"0<>:. ^
|!%5: <
>:#,_ $> ^
[edit] C
#include<stdio.h>
int main (void)
{
int i;
for (i = 1; i <= 100; i++)
{
if (!(i % 15))
printf ("FizzBuzz\n");
else if (!(i % 3))
printf ("Fizz\n");
else if (!(i % 5))
printf ("Buzz\n");
else
printf ("%d\n", i);
}
return 0;
}
[edit] C++
#include <iostream>
using namespace std;
int main () {
int i;
for (i = 0; i <= 100; i++) {
if ((i % 15) == 0)
cout << "FizzBuzz" << endl;
else if ((i % 3) == 0)
cout << "Fizz" << endl;
else if ((i % 5) == 0)
cout << "Buzz" << endl;
else
cout << i << endl;
}
return 0;
}
Alternate version not using modulo 15:
#include <iostream>
using namespace std;
int main()
{
for (int i = 0; i <= 100; ++i)
{
bool fizz = (i % 3) == 0;
bool buzz = (i % 5) == 0;
if (fizz)
cout << "Fizz";
if (buzz)
cout << "Buzz";
if (!fizz && !buzz)
cout << i;
cout << endl;
}
}
Version computing FizzBuzz at compile time with metaprogramming:
#include <iostream>
template <int n, int m3, int m5>
struct fizzbuzz : fizzbuzz<n-1, (n-1)%3, (n-1)%5>
{
fizzbuzz()
{ std::cout << n << std::endl; }
};
template <int n>
struct fizzbuzz<n, 0, 0> : fizzbuzz<n-1, (n-1)%3, (n-1)%5>
{
fizzbuzz()
{ std::cout << "FizzBuzz" << std::endl; }
};
template <int n, int p>
struct fizzbuzz<n, 0, p> : fizzbuzz<n-1, (n-1)%3, (n-1)%5>
{
fizzbuzz()
{ std::cout << "Fizz" << std::endl; }
};
template <int n, int p>
struct fizzbuzz<n, p, 0> : fizzbuzz<n-1, (n-1)%3, (n-1)%5>
{
fizzbuzz()
{ std::cout << "Buzz" << std::endl; }
};
template <>
struct fizzbuzz<0,0,0>
{
fizzbuzz()
{ std::cout << 0 << std::endl; }
};
template <int n>
struct fb_run
{
fizzbuzz<n, n%3, n%5> fb;
};
int main()
{
fb_run<100> fb;
}
[edit] C#
using System;
namespace RosettaCode.FizzBuzz
{
class Program
{
static void Main(string[] args)
{
for (int i = 1; i <= 100; i++)
{
string output = "";
if (i % 3 == 0) output += "Fizz";
if (i % 5 == 0) output += "Buzz";
if (String.IsNullOrEmpty(output)) output = i.ToString();
Console.WriteLine(output);
}
}
}
}
[edit] Chef
This was clearly a challenge in a language without a modulus operator, a proper if statement except for checking if a variable is not exactly 0, and no way to define text except 1 character at a time on a stack.
Irish Soda Bread with Club Soda.
This is FizzBuzz
Ingredients.
1 l buttermilk
Method.
Take buttermilk from refrigerator.
Shake the buttermilk.
Put buttermilk into the 6th mixing bowl.
Serve with club soda.
Pour contents of the 1st mixing bowl into the 1st baking dish.
Clean the 1st mixing bowl.
Watch the buttermilk until shaked.
Serves 1.
Club Soda.
Gets whether to print fizz buzz fizzbuzz or number.
Ingredients.
70 g flour
105 g salt
122 ml milk
66 g sugar
117 ml vegetable oil
3 cups fizzle
5 cups seltzer
1 cup bmilk
1 cup ice
1 cup baking soda
1 g oregano
32 ml vinegar
1 g thyme
1 g sage
Method.
Put milk into the 1st mixing bowl.
Put salt into the 1st mixing bowl.
Put flour into the 1st mixing bowl.
Put vinegar into the 1st mixing bowl.
Put milk into the 1st mixing bowl.
Stir the 1st mixing bowl for 5 minutes.
Liquify contents of the 1st mixing bowl.
Put fizzle into the 3rd mixing bowl.
Combine seltzer into the 3rd mixing bowl.
Fold bmilk into the 6th mixing bowl.
Put bmilk into the 6th mixing bowl.
Put seltzer into the 6th mixing bowl.
Put bmilk into the 6th mixing bowl.
Serve with moist cake.
Fold bmilk into the 1st mixing bowl.
Fold sage into the 6th mixing bowl.
Fold sage into the 6th mixing bowl.
Put bmilk into the 4th mixing bowl.
Remove seltzer from the 4th mixing bowl.
Fold oregano into the 4th mixing bowl.
Smell the oregano.
Fold bmilk into the 6th mixing bowl.
Put bmilk into the 6th mixing bowl.
Put fizzle into the 6th mixing bowl.
Put bmilk into the 6th mixing bowl.
Serve with moist cake.
Fold bmilk into the 1st mixing bowl.
Fold sage into the 6th mixing bowl.
Fold sage into the 6th mixing bowl.
Put bmilk into the 4th mixing bowl.
Remove fizzle from the 4th mixing bowl.
Fold oregano into the 4th mixing bowl.
Crush the oregano.
Clean the 1st mixing bowl.
Fold bmilk into the 6th mixing bowl.
Put bmilk into the 1st mixing bowl.
Refrigerate.
Grind until crushed.
Refrigerate.
Shuffle until smelled.
Clean the 1st mixing bowl.
Put milk into the 1st mixing bowl.
Put vegetable oil into the 1st mixing bowl.
Put sugar into the 1st mixing bowl.
Put vinegar into the 1st mixing bowl.
Put milk into the 1st mixing bowl.
Stir the 1st mixing bowl for 5 minutes.
Liquify contents of the 1st mixing bowl.
Fold baking soda into the 3rd mixing bowl.
Fold bmilk into the 6th mixing bowl.
Put bmilk into the 6th mixing bowl.
Put baking soda into the 6th mixing bowl.
Put bmilk into the 6th mixing bowl.
Serve with moist cake.
Fold bmilk into the 1st mixing bowl.
Fold sage into the 6th mixing bowl.
Fold sage into the 6th mixing bowl.
Put bmilk into the 4th mixing bowl.
Remove baking soda from the 4th mixing bowl.
Fold oregano into the 4th mixing bowl.
Separate the oregano.
Refrigerate.
Part until separated.
Put fizzle into the 6th mixing bowl.
Serve with club soda.
Stir the 1st mixing bowl for 1 minute.
Stir the 1st mixing bowl for 7 minutes.
Stir the 1st mixing bowl for 1 minute.
Stir the 1st mixing bowl for 7 minutes.
Stir the 1st mixing bowl for 5 minutes.
Fold the oregano into the 1st mixing bowl.
Fold the oregano into the 1st mixing bowl.
Fold the oregano into the 1st mixing bowl.
Fold the oregano into the 1st mixing bowl.
Fold the oregano into the 1st mixing bowl.
Refrigerate.
Moist cake.
Mods a number
Ingredients.
1 cup chocolate
70 g wheat flour
1 cup white chocolate chips
1 cup baking powder
105 g honey
5 cups syrup
1 g vanilla
1 g rosemary
Method.
Fold chocolate into the 6th mixing bowl.
Fold syrup into the 6th mixing bowl.
Clean the 1st mixing bowl.
Put chocolate into the 5th mixing bowl.
Fold wheat flour into the 5th mixing bowl.
Put white chocolate chips into the 5th mixing bowl.
Remove white chocolate chips from the 5th mixing bowl.
Fold baking powder into the 5th mixing bowl.
Put baking powder into the 5th mixing bowl.
Fold honey into the 5th mixing bowl.
Sift the wheat flour.
Put honey into the 5th mixing bowl.
Add white chocolate chips into the 5th mixing bowl.
Fold honey into the 5th mixing bowl.
Put honey into the 5th mixing bowl.
Remove syrup from the 5th mixing bowl.
Fold vanilla into the 5th mixing bowl.
Sprinkle the vanilla.
Put white chocolate chips into the 5th mixing bowl.
Remove white chocolate chips from the 5th mixing bowl.
Fold rosemary into the 5th mixing bowl.
Set aside.
Move until sprinkled.
Recite the rosemary.
Put white chocolate chips into the 5th mixing bowl.
Remove white chocolate chips from the 5th mixing bowl.
Fold honey into the 5th mixing bowl.
Put baking powder into the 5th mixing bowl.
Add white chocolate chips into the 5th mixing bowl.
Fold baking powder into the 5th mixing bowl.
Set aside.
Repeat until recited.
Put white chocolate chips into the 5th mixing bowl.
Fold rosemary into the 5th mixing bowl.
Shuffle the wheat flour until sifted.
Put the baking powder into the 5th mixing bowl.
Combine syrup into the 5th mixing bowl.
Fold honey into the 5th mixing bowl.
Put chocolate into the 5th mixing bowl.
Remove honey from the 5th mixing bowl.
Fold chocolate into the 5th mixing bowl.
Put white chocolate chips into the 5th mixing bowl.
Fold rosemary into the 5th mixing bowl.
Siphon chocolate.
Put white chocolate chips into the 5th mixing bowl.
Remove white chocolate chips from the 5th mixing bowl.
Fold rosemary into the 5th mixing bowl.
Set aside.
Gulp until siphoned.
Quote the rosemary.
Put syrup into the 5th mixing bowl.
Fold chocolate into the 5th mixing bowl.
Set aside.
Repeat until quoted.
Put chocolate into the 1st mixing bowl.
Refrigerate.
[edit] Clojure
(map (fn [x] (cond (zero? (mod x 15)) "FizzBuzz"
(zero? (mod x 5)) "Buzz"
(zero? (mod x 3)) "Fizz"
:else x))
(range 1 101))
(def fizzbuzz (lazy-seq (map
#(cond (zero? (mod % 15)) "FizzBuzz"
(zero? (mod % 5)) "Buzz"
(zero? (mod % 3)) "Fizz"
:else %)
(iterate inc 1))))
(defn fizz-buzz
([] (fizz-buzz (range 1 101)))
([lst]
(letfn [(fizz? [n] (zero? (mod n 3)))
(buzz? [n] (zero? (mod n 5)))]
(let [f "Fizz"
b "Buzz"
items (map (fn [n]
(cond (and (fizz? n) (buzz? n)) (str f b)
(fizz? n) f
(buzz? n) b
:else n))
lst)] items))))
[edit] Common Lisp
(defun fizzbuzz ()
(loop for x from 1 to 100 do
(print (cond ((zerop (mod x 15)) "FizzBuzz")
((zerop (mod x 3)) "Fizz")
((zerop (mod x 5)) "Buzz")
(t x)))))
(defun fizzbuzz ()
(loop for x from 1 to 100 do
(format t "~&~{~A~}"
(or (append (when (zerop (mod x 3)) '("Fizz"))
(when (zerop (mod x 5)) '("Buzz")))
(list x)))))
[edit] D
module fizzbuzz;
import std.stdio;
import std.string ;
// if-else
void fizzBuzz(int n) {
for(int i = 1 ; i <= n; i++)
if (!(i%15))
writef("FizzBuzz") ;
else if (!(i%3))
writef("Fizz") ;
else if (!(i%5))
writef("Buzz") ;
else
writef(i) ;
writefln() ;
}
// use switch case
void fizzBuzz_s(int n) {
for(int i = 1 ; i <= n; i++)
switch(i%15) {
case 0:
writef("FizzBuzz") ; break ;
case 3,6,9,12:
writef("Fizz") ; break ;
case 5,10:
writef("Buzz") ; break ;
default:
writef(i) ;
}
writefln() ;
}
// recursive with concatenation
string fizzBuzz_r(int n){
string s = "" ;
if (n == 0)
return s ;
if(!(n % 5))
s = "Buzz"~ s ;
if(!(n % 3))
s = "Fizz" ~ s ;
if (s == "")
s = toString(n) ;
return fizzBuzz_r(n-1) ~ s ;
}
// alternative recursive
string fizzBuzz_r2(int n){
return (n>0) ? fizzBuzz_r2(n-1) ~
(n % 15 ? n % 5 ? n % 3 ? toString(n) :"Fizz" : "Buzz" : "FizzBuzz")
: "";
}
void main() {
fizzBuzz(100) ;
fizzBuzz_s(100) ;
writefln(fizzBuzz_r(100)) ;
writefln(fizzBuzz_r2(100)) ;
for(int i = 1 ; i <= 100;i++)
writef("%s", i % 15 ? i % 5 ? i % 3 ? toString(i) :"Fizz" : "Buzz" : "FizzBuzz") ;
}
[edit] E
for i in 1..100 {
println(switch ([i % 3, i % 5]) {
match [==0, ==0] { "FizzBuzz" }
match [==0, _ ] { "Fizz" }
match [_, ==0] { "Buzz" }
match _ { i }
})
}
[edit] Erlang
fizzbuzz() ->
F = fun(N) when N rem 15 == 0 -> "FizzBuzz";
(N) when N rem 3 == 0 -> "Fizz";
(N) when N rem 5 == 0 -> "Buzz";
(N) -> integer_to_list(N)
end,
[F(N)++"\n" || N <- lists:seq(1,100)].
[edit] Factor
USING: math kernel io math.ranges ;
IN: fizzbuzz
: fizz ( n -- str ) 3 divisor? "Fizz" "" ? ;
: buzz ( n -- str ) 5 divisor? "Buzz" "" ? ;
: fizzbuzz ( n -- str ) dup [ fizz ] [ buzz ] bi append [ number>string ] [ nip ] if-empty ;
: main ( -- ) 100 [1,b] [ fizzbuzz print ] each ;
MAIN: main
[edit] F#
#light
[1..100]
|> List.map (fun x ->
match x with
| _ when x % 15 = 0 ->"fizzbuzz"
| _ when x % 5 = 0 -> "buzz"
| _ when x % 3 = 0 -> "fizz"
| _ -> x.ToString())
|> List.iter (fun x -> printfn "%s" x)
[edit] FALSE
[\$@$@\/*=]d:
[1\$3d;!["Fizz"\%0\]?$5d;!["Buzz"\%0\]?\[$.]?"
"]f:
0[$100<][1+f;!]#%
[edit] Forth
[edit] table-driven
: fizz ( n -- ) drop ." Fizz" ;
: buzz ( n -- ) drop ." Buzz" ;
: fb ( n -- ) drop ." FizzBuzz" ;
: vector create does> ( n -- )
over 15 mod cells + @ execute ;
vector .fizzbuzz
' fb , ' . , ' . ,
' fizz , ' . , ' buzz ,
' fizz , ' . , ' . ,
' fizz , ' buzz , ' . ,
' fizz , ' . , ' . ,
[edit] or the classic approach
: .fizzbuzz ( n -- )
0 pad c!
dup 3 mod 0= if s" Fizz" pad place then
dup 5 mod 0= if s" Buzz" pad +place then
pad c@ if drop pad count type else . then ;
: zz ( n -- )
1+ 1 do i .fizzbuzz cr loop ;
100 zz
[edit] Fortran
In ANSI FORTRAN 66 or later use structured IF-THEN-ELSE (example uses some ISO Fortran 90 features):
program fizzbuzz_if
integer :: i
do i = 1, 100
if (mod(i,15) == 0) then; print *, 'FizzBuzz'
else if (mod(i,3) == 0) then; print *, 'Fizz'
else if (mod(i,5) == 0) then; print *, 'Buzz'
else; print *, i
end if
end do
end program fizzbuzz_if
In ISO Fortran 90 or later use SELECT-CASE statement:
program fizzbuzz_select
integer :: i
do i = 1, 100
select case (mod(i,15))
case 0; print *, 'FizzBuzz'
case 3,6,9,12; print *, 'Fizz'
case 5,10; print *, 'Buzz'
case default; print *, i
end select
end do
end program fizzbuzz_select
[edit] Go
package main
import "fmt"
func main() {
for i := 1; i <= 100; i++ {
switch {
case (i%3==0) && (i%5==0):
fmt.Println("FizzBuzz")
case i%3==0:
fmt.Println("Fizz")
case i%5==0:
fmt.Println("Buzz")
default:
fmt.Println(i)
}
}
}
[edit] Groovy
for (i in 1..100) {
println "${i%3?'':'Fizz'}${i%5?'':'Buzz'}" ?: i
}
[edit] Haskell
Variant directly implementing the specification:
main = mapM_ (putStrLn . fizzbuzz) [1..100]
fizzbuzz x
| x `mod` 15 == 0 = "FizzBuzz"
| x `mod` 3 == 0 = "Fizz"
| x `mod` 5 == 0 = "Buzz"
| otherwise = show x
Alternate implementation using lazy infinite lists and avoiding use of "mod":
main = mapM_ putStrLn $ take 100 fizzbuzz
fizzbuzz = zipWith (\x y -> if null y then show x else y) [1..] fbs
fbs = zipWith (++) (cycle ["","","Fizz"]) (cycle ["","","","","Buzz"])
[edit] Io
for(a,1,100,
if(a % 5 == 0) then(
"FizzBuzz" println
) elseif(a % 3 == 0) then(
"Fizz" println
) elseif(a % 5 == 0) then(
"Buzz" println
) else (
a println
)
)
[edit] Ioke
(1..100) each(x,
cond(
(x % 15) zero?, "FizzBuzz" println,
(x % 3) zero?, "Fizz" println,
(x % 5) zero?, "Buzz" println
)
)
[edit] Iptscrae
Written in Iptscrae, the scripting language for The Palace chat software.
; FizzBuzz in Iptscrae
1 a =
{
"" b =
{ "fizz" b &= } a 3 % 0 == IF
{ "buzz" b &= } a 5 % 0 == IF
{ a ITOA LOGMSG } { b LOGMSG } b STRLEN 0 == IFELSE
a ++
}
{ a 100 <= } WHILE
[edit] J
Solution 0
> }. (<'FizzBuzz') (I.0=15|n)} (<'Buzz') (I.0=5|n)} (<'Fizz') (I.0=3|n)} ":&.> n=: i.101
Solution 1
Fizz=: 'Fizz' #~ 0 = 3&|
Buzz=: 'Buzz' #~ 0 = 5&|
FizzBuzz=: ": [^:('' -: ]) Fizz,Buzz
FizzBuzz"0 >: i.100
[edit] Java
[edit] If/else ladder
public class FizzBuzz{
public static void main(String[] args){
for(int i= 1; i <= 100; i++){
if(i % 15 == 0){
System.out.println("FizzBuzz");
}else if(i % 3 == 0){
System.out.println("Fizz");
}else if(i % 5 == 0){
System.out.println("Buzz");
}else{
System.out.println(i);
}
}
}
}
[edit] Concatenation
public class FizzBuzz{
public static void main(String[] args){
for(int i= 1; i <= 100; i++){
String output = "";
if(i % 3 == 0) output += "Fizz";
if(i % 5 == 0) output += "Buzz";
if(output.equals("")) output += i;
System.out.println(output);
}
}
}
[edit] Ternary operator
public class FizzBuzz{
public static void main(String[] args){
for(int i= 1; i <= 100; i++){
System.out.println(i % 15 != 0 ? i % 5 != 0 ? i % 3 != 0 ?
i : "Fizz" : "Buzz" : "FizzBuzz");
}
}
}
[edit] Recursive
public String fizzBuzz(int n){
String s = "";
if (n == 0)
return s;
if((n % 5) == 0)
s = "Buzz" + s;
if((n % 3) == 0)
s = "Fizz" + s;
if (s.equals(""))
s = n + "";
return fizzBuzz(n-1) + s;
}
[edit] Alternative Recursive
public String fizzBuzz(int n){
return (n>0) ? fizzBuzz(n-1) +
(n % 15 != 0? n % 5 != 0? n % 3 != 0? (n+"") :"Fizz" : "Buzz" : "FizzBuzz")
: "";
}
[edit] JavaScript
Works with: NJS version 0.2.5
for (var i = 1; i <= 100; i++) {
if (i % 15 == 0) {
print("FizzBuzz");
} else if (i % 3 == 0) {
print("Fizz");
} else if (i % 5 == 0) {
print("Buzz");
} else {
print(i);
}
}
[edit] Lua
[edit] If/else Ladder
for i = 1, 100 do
if i % 15 == 0 then
print("FizzBuzz")
elseif i % 3 == 0 then
print("Fizz")
elseif i % 5 == 0 then
print("Buzz")
else
print(i)
end
end
[edit] Concatenation
for i = 1, 100 do
output = ""
if i % 3 == 0 then
output = output.."Fizz"
end
if i % 5 == 0 then
output = output.."Buzz"
end
if(output == "") then
output = output..i
end
print(output)
end
[edit] M4
define(`for',
`ifelse($#,0,``$0'',
`ifelse(eval($2<=$3),1,
`pushdef(`$1',$2)$5`'popdef(`$1')$0(`$1',eval($2+$4),$3,$4,`$5')')')')
for(`x',1,100,1,
`ifelse(eval(x%15==0),1,FizzBuzz,
`ifelse(eval(x%3==0),1,Fizz,
`ifelse(eval(x%5==0),1,Buzz,x)')')
')
[edit] Mathematica
Do[Print[Which[Mod[i, 15] == 0, "FizzBuzz", Mod[i, 5] == 0, "Buzz", Mod[i, 3] == 0, "Fizz", True, i]], {i, 100}]
[edit] MAXScript
for i in 1 to 100 do
(
case of
(
(mod i 15 == 0): (print "FizzBuzz")
(mod i 5 == 0): (print "Buzz")
(mod i 3 == 0): (print "Fizz")
default: (print i)
)
)
[edit] Metafont
for i := 1 upto 100:
message if i mod 15 = 0: "FizzBuzz" &
elseif i mod 3 = 0: "Fizz" &
elseif i mod 5 = 0: "Buzz" &
else: decimal i & fi "";
endfor
end
[edit] Modula-3
MODULE Fizzbuzz EXPORTS Main;
IMPORT IO;
BEGIN
FOR i := 1 TO 100 DO
IF i MOD 15 = 0 THEN
IO.Put("FizzBuzz\n");
ELSIF i MOD 5 = 0 THEN
IO.Put("Buzz\n");
ELSIF i MOD 3 = 0 THEN
IO.Put("Fizz\n");
ELSE
IO.PutInt(i);
IO.Put("\n");
END;
END;
END Fizzbuzz.
[edit] Oberon-2
MODULE FizzBuzz;
IMPORT Out;
VAR i: INTEGER;
BEGIN
FOR i := 1 TO 100 DO
IF i MOD 15 = 0 THEN
Out.String("FizzBuzz");
Out.Ln;
ELSIF i MOD 5 = 0 THEN
Out.String("Buzz");
Out.Ln;
ELSIF i MOD 3 = 0 THEN
Out.String("Fizz");
Out.Ln;
ELSE
Out.Int(i,0);
Out.Ln;
END;
END;
END FizzBuzz.
[edit] OCaml
let output x =
match x mod 3 = 0, x mod 5 = 0 with
true, true -> "FizzBuzz"
| true, false -> "Fizz"
| false, true -> "Buzz"
| false, false -> string_of_int x
let _ =
for i = 1 to 100 do print_endline (output i) done
[edit] Octave
for i = 1:100
if ( mod(i,15) == 0 )
disp("FizzBuzz");
elseif ( mod(i, 3) == 0 )
disp("Fizz")
elseif ( mod(i, 5) == 0 )
disp("Buzz")
endif
endfor
[edit] Oz
declare
fun {FizzBuzz X}
if X mod 15 == 0 then 'FizzBuzz'
elseif X mod 3 == 0 then 'Fizz'
elseif X mod 5 == 0 then 'Buzz'
else X
end
end
in
for I in 1..100 do
{Show {FizzBuzz I}}
end
[edit] Perl
foreach (1 .. 100) {
if (0 == $_ % 15) {
print "FizzBuzz\n";
} elsif (0 == $_ % 3) {
print "Fizz\n";
} elsif (0 == $_ % 5) {
print "Buzz\n";
} else {
print "$_\n";
};
};
More concisely:
print 'Fizz'x!($_ % 3) . 'Buzz'x!($_ % 5) || $_, "\n" for 1 .. 100;
[edit] Perl 6
Works with: Rakudo version #22 "Thousand Oaks"
Most straightforwardly:
for 1 .. 100 {
when $_ !% 3 and $_ !% 5 { say 'FizzBuzz'; }
when $_ !% 3 { say 'Fizz'; }
when $_ !% 5 { say 'Buzz'; }
default { .say; }
}
Most concisely:
say 'Fizz' x $_ !% 3 ~ 'Buzz' x $_ !% 5 || $_ for 1 .. 100;
And here's an implementation that never checks for divisibility:
for
('' xx 2, 'Fizz') xx 34 Z
('' xx 4, 'Buzz') xx 20 Z
1 .. 100
-> $f, $b, $n {
say $f ~ $b || $n;
}
[edit] PHP
[edit] if/else ladder approach
<?php
for ($i = 1; $i <= 100; $i++)
{
if (!($i % 15))
echo "FizzBuzz\n";
else if (!($i % 3))
echo "Fizz\n";
else if (!($i % 5))
echo "Buzz\n";
else
echo "$i\n";
}
>
[edit] concatenation approach
Uses PHP's concatenation operator (.=) to build the output string. The concatenation operator allows us to add data to the end of a string without overwriting the whole string. Since Buzz will always appear if our number is divisible by five, and Buzz is the second part of "FizzBuzz", we can simply append "Buzz" to our string.
In contrast to the if-else ladder, this method lets us skip the check to see if $i is divisible by both 3 and 5 (i.e. 15). However, we get the added complexity of needing to reset $str to an empty string (not necessary in some other languages), and we also need a separate if statement to check to see if our string is empty, so we know if $i was not divisible by 3 or 5.
<?php
for ( $i = 1; $i <= 100; ++$i )
{
$str = "";
if (!($i % 3 ) )
$str .= "Fizz";
if (!($i % 5 ) )
$str .= "Buzz";
if ( empty( $str ) )
$str = $i;
echo $str . "\n";
}
?>
[edit] Pike
int main(){
for(int i = 1; i <= 100; i++) {
if(i % 15 == 0) {
write("FizzBuzz\n");
} else if(i % 3 == 0) {
write("Fizz\n");
} else if(i % 5 == 0) {
write("Buzz\n");
} else {
write(i + "\n");
}
}
}
[edit] Pop11
lvars str;
for i from 1 to 100 do
if i rem 15 = 0 then
'FizzBuzz' -> str;
elseif i rem 3 = 0 then
'Fizz' -> str;
elseif i rem 5 = 0 then
'Buzz' -> str;
else
'' >< i -> str;
endif;
printf(str, '%s\n');
endfor;
[edit] PostScript
1 1 100 {
/c false def
dup 3 mod 0 eq { (Fizz) print /c true def } if
dup 5 mod 0 eq { (Buzz) print /c true def } if
c {pop}{( ) cvs print} ifelse
(\n) print
} for
or...
/fizzdict 100 dict def
fizzdict begin
/notmod{ ( ) cvs } def
/mod15 { dup 15 mod 0 eq { (FizzBuzz)def }{pop}ifelse} def
/mod3 { dup 3 mod 0 eq {(Fizz)def}{pop}ifelse} def
/mod5 { dup 5 mod 0 eq {(Buzz)def}{pop}ifelse} def
1 1 100 { mod3 } for
1 1 100 { mod5 } for
1 1 100 { mod15} for
1 1 100 { dup currentdict exch known { currentdict exch get}{notmod} ifelse print (\n) print} for
end
[edit] PowerShell
[edit] Straightforward, looping
for ($i = 1; $i -le 100; $i++) {
if ($i % 15 -eq 0) {
"FizzBuzz"
} elseif ($i % 5 -eq 0) {
"Buzz"
} elseif ($i % 3 -eq 0) {
"Fizz"
} else {
$i
}
}
[edit] Straightforward, Pipeline, Switch
1..100 | ForEach-Object {
switch ($_) {
{ $_ % 15 -eq 0 } { "FizzBuzz" }
{ $_ % 5 -eq 0 } { "Buzz" }
{ $_ % 3 -eq 0 } { "Fizz" }
default { $_ }
}
}
[edit] Concatenation
Translation of: C#
1..100 | ForEach-Object {
$s = ''
if ($_ % 3 -eq 0) { $s += "Fizz" }
if ($_ % 5 -eq 0) { $s += "Buzz" }
if (!$s) { $s = $_ }
$s
}
[edit] Prolog
Works with: SWI Prolog version 4.8.0
Maybe not the most conventional way to write this in Prolog. The fizzbuzz predicate uses a higher-order predicate and print_item uses the if-then-else construction.
fizzbuzz :-
foreach(between(1, 100, X), print_item(X)).
print_item(X) :-
( X mod 15 =:= 0
-> print('FizzBuzz')
; X mod 3 =:= 0
-> print('Fizz')
; X mod 5 =:= 0
-> print('Buzz')
; print(X)
),
nl.
[edit] PureBasic
OpenConsole()
For x = 1 To 100
If x%15 = 0
PrintN("FizzBuzz")
ElseIf x%3 = 0
PrintN("Fizz")
ElseIf x%5 = 0
PrintN("Buzz")
Else
PrintN(Str(x))
EndIf
Next
Input()
[edit] Python
for i in xrange(1, 101):
if i % 15 == 0:
print "FizzBuzz"
elif i % 3 == 0:
print "Fizz"
elif i % 5 == 0:
print "Buzz"
else:
print i
And a shorter, but less clear version, using a list comprehension and logical expressions:
for i in range(1, 101):
words = [word for n, word in ((3, 'Fizz'), (5, 'Buzz')) if not i % n]
print ''.join(words) or i
[edit] Without modulus
I came across this crazy version[4] without using the modulus operator:
messages = [None, "Fizz", "Buzz", "FizzBuzz"]
acc = 810092048
for i in xrange(1, 101):
c = acc & 3
print messages[c] if c else i
acc = acc >> 2 | c << 28
[edit] R
x <- 1:100
xx <- as.character(x)
xx[x%%3==0] <- "Fizz"
xx[x%%5==0] <- "Buzz"
xx[x%%15==0] <- "FizzBuzz"
xx
[edit] RapidQ
The BASIC solutions work with RapidQ, too.
However, here is a bit more esoteric solution using the IIF() function.
FOR i=1 TO 100
t$ = IIF(i MOD 3 = 0, "Fizz", "") + IIF(i MOD 5 = 0, "Buzz", "")
PRINT IIF(LEN(t$), t$, i)
NEXT i
[edit] Raven
100 each 1 + as n
''
n 3 mod 0 = if 'Fizz' cat
n 5 mod 0 = if 'Buzz' cat
dup empty if drop n
say
[edit] REBOL
Shortest implementation:
repeat i 100 [case/all [i // 3 = 0 [print"fizz"] i // 5 = 0 [print "buzz"] 1 [print i]]]
A long implementation that concatenates strings and includes a proper code header (title, date, etc.)
rebol [
Title: "FizzBuzz"
Author: oofoe
Date: 2009-12-10
URL: http://rosettacode.org/wiki/FizzBuzz
]
; Concatenative. Note use of 'case/all' construct to evaluate all
; conditions. I use 'copy' to allocate a new string each time through
; the loop -- otherwise 'x' would get very long...
repeat i 100 [
x: copy ""
case/all [
0 = mod i 3 [append x "Fizz"]
0 = mod i 5 [append x "Buzz"]
"" = x [x: mold i]
]
print x
]
Here are two examples by Nick Antonaccio.
repeat i 100 [
print switch/default 0 compose [
(mod i 15) ["fizzbuzz"]
(mod i 3) ["fizz"]
(mod i 5) ["buzz"]
][i]
]
; And minimized version:
repeat i 100[j:""if 0 = mod i 3[j:"fizz"]if 0 = mod i 5[j: join j"buzz"]if j =""[j: i]print j]
The following is presented as a curiosity only, not as an example of good coding practice:
m: func [i d] [0 = mod i d]
spick: func [t x y][either any [not t "" = t][y][x]]
zz: func [i] [rejoin [spick m i 3 "Fizz" "" spick m i 5 "Buzz" ""]]
repeat i 100 [print spick z: zz i z i]
[edit] Ruby
1.upto(100) do |n|
print "Fizz" if a = (n % 3).zero?
print "Buzz" if b = (n % 5).zero?
print n unless (a || b)
print "\n"
end
[edit] Scala
(1 to 100) foreach {
case x if (x % 15 == 0) => println("FizzBuzz")
case x if (x % 3 == 0) => println("Fizz")
case x if (x % 5 == 0) => println("Buzz")
case x => println(x)
}
(1 to 100) map ( x => (x % 3, x % 5) match{
case (0,0) => "FizzBuzz"
case (0,_) => "Fizz"
case (_,0) => "Buzz"
case _ => x
}) foreach println
[edit] Scheme
(do ((i 1 (+ i 1)))
((> i 100))
(display
(cond ((= 0 (modulo i 15)) "FizzBuzz")
((= 0 (modulo i 3)) "Fizz")
((= 0 (modulo i 5)) "Buzz")
(else i)))
(newline))
[edit] Seed7
$ include "seed7_05.s7i";
const proc: main is func
local
var integer: number is 0;
begin
for number range 1 to 100 do
if number rem 15 = 0 then
writeln("FizzBuzz");
elsif number rem 5 = 0 then
writeln("Buzz");
elsif number rem 3 = 0 then
writeln("Fizz");
else
writeln(number);
end if;
end for;
end func;
[edit] Slate
n@(Integer traits) fizzbuzz
[| output |
output: ((n \\ 3) isZero ifTrue: ['Fizz'] ifFalse: ['']) ; ((n \\ 5) isZero ifTrue: ['Buzz'] ifFalse: ['']).
output isEmpty ifTrue: [n printString] ifFalse: [output]
].
1 to: 100 do: [| :i | inform: i fizzbuzz]
[edit] Smalltalk
Since only GNU Smalltalk supports file-based programming, we'll be using its syntax.
Integer extend [
fizzbuzz [
| fb |
fb := '%<Fizz|>1%<Buzz|>2' % {
self \\ 3 == 0. self \\ 5 == 0 }.
^fb isEmpty ifTrue: [ self ] ifFalse: [ fb ]
]
]
1 to: 100 do: [ :i | i fizzbuzz displayNl ]
A Squeak/Pharo example using the Transcript window:
(1 to: 100) do:
[:n |
((n \\ 3)*(n \\ 5)) isZero
ifFalse: [Transcript show: n].
(n \\ 3) isZero
ifTrue: [Transcript show: 'Fizz'].
(n \\ 5) isZero
ifTrue: [Transcript show: 'Buzz'].
Transcript cr.]
The Squeak/Pharo examples below present possibilities using the powerful classes available. In this example, the dictionary can have as keys pairs of booleans and in the interaction the several boolean patterns select the string to be printed or if the pattern is not found the number itself is printed.
fizzbuzz := Dictionary with: #(true true)->'FizzBuzz'
with: #(true false)->'Fizz'
with: #(false true)->'Buzz'.
1 to: 100 do:
[ :i | Transcript show:
(fizzbuzz at: {i isDivisibleBy: 3. i isDivisibleBy: 5}
ifAbsent: [ i ]); cr]
Smalltalk does not have a case-select construct, but a similar effect can be attained using a collection and the #includes: method:
1 to: 100 do: [:n | |r|
r := n rem: 15.
Transcript show: (r isZero
ifTrue:['fizzbuzz']
ifFalse: [(#(3 6 9 12) includes: r)
ifTrue:['fizz']
ifFalse:[((#(5 10) includes: r))
ifTrue:['buzz']
ifFalse:[n]]]);
cr].
If the construction of the whole collection is done beforehand, Smalltalk provides a straightforward way of doing because collections can be heterogeneous (may contain any object):
fbz := (1 to: 100) asOrderedCollection.
3 to: 100 by: 3 do: [:i | fbz at: i put: 'Fizz'].
5 to: 100 by: 5 do: [:i | fbz at: i put: 'Buzz'].
15 to: 100 by: 15 do: [:i | fbz at: i put: 'FizzBuzz'].
fbz do: [:i | Transcript show: i; cr].
The approach building a dynamic string can be done as well:
1 to: 100 do: [:i | |fb s|
fb := {i isDivisibleBy: 3. i isDivisibleBy: 5. nil}.
fb at: 3 put: (fb first | fb second) not.
s := '<1?Fizz:><2?Buzz:><3?{1}:>' format: {i printString}.
Transcript show: (s expandMacrosWithArguments: fb); cr].
[edit] SNOBOL4
Merely posting a solution by Daniel Lyons
I = 1
LOOP FIZZBUZZ = ""
EQ(REMDR(I, 3), 0) :F(TRY_5)
FIZZBUZZ = FIZZBUZZ "FIZZ"
TRY_5 EQ(REMDR(I, 5), 0) :F(DO_NUM)
FIZZBUZZ = FIZZBUZZ "BUZZ"
DO_NUM IDENT(FIZZBUZZ, "") :F(SHOW)
FIZZBUZZ = I
SHOW OUTPUT = FIZZBUZZ
I = I + 1
LE(I, 100) :S(LOOP)
END
[edit] SNUSP
/ 'B' @=@@=@@++++#
// / 'u' @@@@@=@+++++#
// // / 'z' @=@@@@+@++++#
// // // / 'i' @@@@@@=+++++#
// // // // / 'F' @@@=@@+++++#
// // // // // / LF ++++++++++#
// // // // // // / 100 @@@=@@@=++++#
$@/>@/>@/>@/>@/>@/>@/\ 0
/ /
! /======= Fizz <<<.<.<..>>>#
/ | \
\?!#->+ @\.>?!#->+ @\.>?!#->+@/.>\ |
/ ! ! ! / |
\?!#->+ @\.>?!#->+@\ .>?!#->+@/.>\ |
/ ! \!===\! ! / |
\?!#->+ @\.>?!#->+ @\.>?!#->+@/.>\ |
/ ! ! | ! / |
\?!#->+@\ .>?!#->+ @\.>?!#->+@/.>\ |
/ \!==========!===\! ! / |
\?!#->+ @\.>?!#->+ @\.>?!#->+@/>>@\.>/
! | | |
/==========/ \========!\=== Buzz <<<<<<<.>.>..>>>#
|
\!/=dup==?\>>@\<!/back?\<<<#
\<<+>+>-/ | \>+<- /
|
/======================/
|
| /recurse\ #/?\ zero
\print=!\@\>?!\@/<@\.!\-/
| \=/ \=itoa=@@@+@+++++#
! /+ !/+ !/+ !/+ \ mod10
/<+> -\!?-\!?-\!?-\!?-\!
\?!\-?!\-?!\-?!\-?!\-?/\ div10
# +/! +/! +/! +/! +/
[edit] SQL
DROP TABLE fizzbuzz;
CREATE TABLE fizzbuzz(i int, fizz string, buzz string);
INSERT INTO fizzbuzz VALUES(1,"","");
INSERT INTO fizzbuzz SELECT i + (SELECT max(i) FROM fizzbuzz), fizz, buzz FROM fizzbuzz;
INSERT INTO fizzbuzz SELECT i + (SELECT max(i) FROM fizzbuzz), fizz, buzz FROM fizzbuzz;
INSERT INTO fizzbuzz SELECT i + (SELECT max(i) FROM fizzbuzz), fizz, buzz FROM fizzbuzz;
INSERT INTO fizzbuzz SELECT i + (SELECT max(i) FROM fizzbuzz), fizz, buzz FROM fizzbuzz;
INSERT INTO fizzbuzz SELECT i + (SELECT max(i) FROM fizzbuzz), fizz, buzz FROM fizzbuzz;
INSERT INTO fizzbuzz SELECT i + (SELECT max(i) FROM fizzbuzz), fizz, buzz FROM fizzbuzz;
INSERT INTO fizzbuzz SELECT i + (SELECT max(i) FROM fizzbuzz), fizz, buzz FROM fizzbuzz;
DROP TABLE lookup;
CREATE TABLE lookup (fizzy, buzzy, rem);
INSERT INTO lookup VALUES("fizz", "buzz", 1);
SELECT
(SELECT i FROM lookup WHERE rem = (i%3<>0)&(i%5<>0)),
(SELECT fizzy FROM lookup WHERE rem = (i%3=0)),
(SELECT buzzy FROM lookup WHERE rem = (i%5=0))
FROM fizzbuzz WHERE i <= 100;
Alternate solution by Daniel Lyons (PostgreSQL specific):
SELECT i, fizzbuzz
FROM
(SELECT i,
CASE
WHEN i % 15 = 0 THEN 'FizzBuzz'
WHEN i % 5 = 0 THEN 'Buzz'
WHEN i % 3 = 0 THEN 'Fizz'
ELSE NULL
END AS fizzbuzz
FROM generate_series(1,100) AS i) AS fb
WHERE fizzbuzz IS NOT NULL;
[edit] Standard ML
fun output x =
case (x mod 3 = 0, x mod 5 = 0) of
(true , true ) => "FizzBuzz"
| (true , false) => "Fizz"
| (false, true ) => "Buzz"
| (false, false) => Int.toString x
val () = let
fun aux i = if i <= 100 then (print (output i ^ "\n");
aux (i+1))
else ()
in
aux 1
end
[edit] Tcl
proc fizzbuzz {n {m1 3} {m2 5}} {
for {set i 1} {$i <= $n} {incr i} {
set ans ""
if {$i % $m1 == 0} {append ans Fizz}
if {$i % $m2 == 0} {append ans Buzz}
puts [expr {$ans eq "" ? $i : $ans}]
}
}
fizzbuzz 100
[edit] UNIX Shell
for n in `seq 1 100`; do
if [ $(($n % 15)) = 0 ]; then
echo FizzBuzz
elif [ $(($n % 3)) = 0 ]; then
echo Fizz
elif [ $(($n % 5)) = 0 ]; then
echo Buzz
else
echo $n
fi
done
A more portable version:
NUM=1
until (($NUM==101)) ; do
if (($NUM % 15 == 0)) ; then
echo FizzBuzz
elif (($NUM % 3 == 0)) ; then
echo Fizz
elif (($NUM % 5 == 0)) ; then
echo Buzz
else
echo "$NUM"
fi
((NUM=$NUM+1))
done
[edit] Ursala
#import std
#import nat
fizzbuzz = ^T(&&'Fizz'! not remainder\3,&&'Buzz'! not remainder\5)|| ~&h+ %nP
#show+
main = fizzbuzz*t iota 101
[edit] V
[fizzbuzz
1 [>=] [
[[15 % zero?] ['fizzbuzz' puts]
[5 % zero?] ['buzz' puts]
[3 % zero?] ['fizz' puts]
[true] [dup puts]
] when succ
] while].
|100 fizzbuzz
[edit] Second try
(a compiler for fizzbuzz)
define a command that will generate a sequence
[seq [] swap dup [zero? not] [rolldown [dup] dip cons rollup pred] while pop pop].
create a quote that will return a quote that returns a quote if its argument is an integer (A HOF)
[check [N X F : [[integer?] [[X % zero?] [N F cons] if] if]] view].
Create a quote that will make sure that the above quote is applied correctly if given (Number Function) as arguments.
[func [[N F] : [dup N F check i] ] view map].
And apply it
100 seq [
[15 [pop 'fizzbuzz' puts]]
[5 [pop 'buzz' puts]]
[3 [pop 'fizz' puts]]
[1 [puts]]] [func dup] step
[i true] map pop
the first one is much better :)
[edit] VBScript
[edit] Implementation
The EEF does eager evaluation. I'm still trying to figure out a lazy evaluation method using Eval(). One day ...
In the meantime, converting that to VB6 would mean using VB6's IIF() which would run a tad faster.
'using the IF/ELSEIF ladder
function fb( n )
if n mod 15 = 0 then
fb = "FizzBuzz"
elseif n mod 5 = 0 then
fb = "Fizz"
elseif n mod 3 = 0 then
fb = "Buzz"
else
fb = n
end if
end function
'the Mexican IF
function eef( b, p1, p2 )
if b then
eef = p1
else
eef = p2
end if
end function
'using the Mexican IF
function fb2( n )
fb2 = eef( n mod 15 = 0, "FizzBuzz", eef( n mod 5 = 0, "Fizz", eef( n mod 3 = 0, "Buzz", n ) ) )
end function
[edit] Invocation
for i = 1 to 16
wscript.stdout.write fb(i) & " "
next
wscript.echo
for i = 1 to 16
wscript.stdout.write fb2(i) & " "
next
wscript.echo
[edit] Output
1 2 Buzz 4 Fizz Buzz 7 8 Buzz Fizz 11 Buzz 13 14 FizzBuzz 16 1 2 Buzz 4 Fizz Buzz 7 8 Buzz Fizz 11 Buzz 13 14 FizzBuzz 16
[edit] Visual Basic .NET
Platform: .NET
Works with: Visual Basic .NET version 9.0+
Sub Main()
For i = 1 To 100
If i Mod 15 = 0 Then
Console.WriteLine("FizzBuzz")
ElseIf i Mod 5 = 0 Then
Console.WriteLine("Buzz")
ElseIf i Mod 3 = 0 Then
Console.WriteLine("Fizz")
Else
Console.WriteLine(i)
End If
Next
End Sub
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