Fizz Buzz In Many Programming Languages

view on wiki.c2.com
last modified: November 25, 2014

The FizzBuzzTest inspiring many people to code up quick and dirty solutions.

...many quick. And quite a few dirty. Many "dirty" to the point of not actually working correctly!!!

See:

http://tickletux.wordpress.com/2007/01/24/using-fizzbuzz-to-find-developers-who-grok-coding/

contains examples in

My ridiculously over-generalized solution in ScalaLanguage:

def replaceMultiples(x: Int, rs: (Int, String)*) =
        rs map {case (n, s) => Either cond (x % n == 0, s, x)}, reduceLeft ((a, b) =>
        a fold ((_ => b), (s => b fold ((_ => a), (t => Right(s + t))))))

def fizzbuzz(n: Int) =
        replaceMultiples(n, 3 -> "Fizz", 5 -> "Buzz") fold ((_ toString), identity)

1 to 100 map fizzbuzz foreach println

but at least it's DRY..."Fizz" and "Buzz" appear only once each. :)

My not so geneneralized ScalaLanguage solution that does not use the modulo operator:

object FizzBuzz extends App {

val three = (3 to 100 by 3) toSet
val five = (5 to 100 by 5) toSet

case class M(nums: Set[Int], phrase: String)

val t = List(M(three intersect five, "FizzBuzz"), M(three, "Fizz"), M(five, "Buzz"))

1 to 100 map { n => t.find(m => m.nums contains n) match { case Some(m) => m.phrase case default => n.toString }, }, foreach println

},

Obvious in VBA

Public Sub FizzBuzz()
        a = Array("", "Fizz", "", "Buzz", "")
        For n = 1 To 100
        f = a(1 + Sgn(n Mod 3)) & a(3 + Sgn(n Mod 5))
        Debug.Print IIf("" = f, n, f)
        Next n
End Sub

And R

fzbz <- function(n){c(n,"fizz","buzz","fizzbuzz")[1+(!n%%3)+2*(!n%%5)]},
for (a in 1:100) print(fzbz(a))

-- MarcThibault

Here is the one in TeX:

\newcount\x
\newcount\y
\newcount\z
\loop
        \advance\x by 1
        \advance\y by 1
        \advance\z by 1
        \ifnum\y=3 Fizz\y=0 \fi
        \ifnum\z=5 Buzz\z=0 \fi
        \ifvmode \the\x \fi\endgraf
\ifnum\x<100 \repeat
\bye

Golf:

\let~\advance\time0\day0\loop~\time1~\day1~\mit\ifnum\time=3\time0Fizz\fi\ifnum\fam=5Buzz\rm\fi\ifvmode\the\day\fi\endgraf\ifnum\day<100\repeat\bye

Readable perl version

my $i = 1;
while ($i <= 100) {
        if ($i % 3 == 0 && $i % 5 == 0) {
        print "FizzBuzz\n";
        }, elsif ($i % 3 == 0) {
        print "Fizz\n";
        }, elsif ($i % 5 == 0) {
        print "Buzz\n";
        }, else {
        print $i . "\n";
        },
        $i++
},

Lua:

for i = 1,100 do
        local n = false
        if i % 3 == 0 then
                io.write("Fizz")
                n = true
        end
        if i % 5 == 0 then
                io.write("Buzz")
                n = true
        end

        if not n then
                io.write(tostring(i))
        end

        io.write("\n")
end

ColdFusion:

<cfset index = 0 />
<cfloop index="index" from="1" to="100">
        <cfset output = "#index#" />
        <cfif index MOD 3 EQ 0 OR index MOD 5 EQ 0><cfset output = "" /></cfif>
        <cfif index MOD 3 EQ 0><cfset output &= "Fizz" /></cfif>
        <cfif index MOD 5 EQ 0><cfset output &= "Buzz" /></cfif>
        <cfoutput>#output# <br /></cfoutput>
</cfloop>

Python: for i in ("fizzbuzz" if i % 15 == 0 else ("fizz" if i % 3 == 0 else ("buzz" if i % 5 == 0 else i)) for i in range(1, 101)): print i

Lua (another): for i=1,100 do

-- Set fizzBuzz to "", "Fizz", "Buzz", or "FizzBuzz" based on mod 3 and mod 5 tests.
local fizzBuzz = (( i % 3 == 0 ) and "Fizz" or "") .. (( i % 5 == 0 ) and "Buzz" or "")
-- Print non-"" fizzBuzz or print i. 
print( (fizzBuzz ~= "" and fizzBuzz) or i )

end

C++ (with one abuse of control characters to avoid excess bools or ifs):

#include <iostream>
int main()
{
  for(int i = 1; i <= 100; ++i)
  {
    std::cout << i << '\r'; // return to beginning of line for overwriting with "Fizz" or "Buzz"
    if(i % 3 == 0) std::cout << "Fizz";
    if(i % 5 == 0) std::cout << "Buzz";
    std::cout << '\n';
  },
  return 0;
},

CategoryInManyProgrammingLanguages

Loading...