This topic is to catalog all your solutions to FizzBuzz, and is a companion to Why Can’t Programmers… Program?
In case any readers don’t know how to complete that test, here’s a solution in VBScript. To try it - save the code into “c:\test.vbs” and run “cscript c:\test.vbs” from a command prompt.
Dim i
For i = 1 to 100
If (i Mod 3 = 0) And (i Mod 5 = 0) Then
WScript.Echo "FizzBuzz"
ElseIf (i Mod 3 = 0) Then
WScript.Echo "Fizz"
ElseIf (i Mod 5 = 0) Then
WScript.Echo "Buzz"
Else
Wscript.Echo i
End If
Next
1 Like
I think the above solution is not elegant.
I think
If (i Mod 3 = 0) Then
WScript.Echo "Fizz"
End If
If (i Mod 5 = 0) Then
WScript.Echo "Buzz"
End If
because it covers the “both” case automatically. OK you have to care about line breaks which I don’t know in VBS (never used that crap, thank goodness).
for (int i = 1; i 101; i++)
{
if ((i % 3) == 0) Console.Write("Fizz");
if ((i % 5) == 0) Console.Write("Buzz");
Console.WriteLine();
}
bool printint;
for (int i = 1; i = 100; i++)
{
printint = true;
if (i % 3 == 0) { Console.Write("Fizz"); printint = false; }
if (i % 5 == 0) { Console.Write("Buzz"); printint = false; }
if (printint) Console.Write(i.ToString());
Console.WriteLine();
}
Heres another solution
int i = 0;
for(i = 1; i = 100; i++)
{
if(i % 15 == 0)
{
count "FizzBuzz" endl;
}
elseif(i % 5 == 0)
{
count "Buzz" endl;
}
elseif(i % 3 == 0)
{
count "Fizz" endl;
}
else
{
cout i endl;
}
}
My weapon of choice is Ruby
1.upto(100) do |i|
out = nil
out = out.to_s + 'Fizz' if i % 3 == 0
out = out.to_s + 'Buzz' if i % 5 == 0
puts out || i
end
And did you really think you could say “only 0.5% of programmers can write FizzBuzz” without people showing you how they do it? 
1 Like
Neah… Jeff just wants us to fell all fuzzy and warm… No way I’m in like the top .5% of the programmers.
Anyway… I can do the FizzBuzz thingy… hi hi hi
for i in range(1,101): print {0:"FizzBuzz", 3:"Fizz", 6:"Fizz", 9:"Fizz", 12:"Fizz" , 5:"Buzz", 10:"Buzz"}.get(i%15, i)
2 Likes
I just couldn’t resist:
for(int i=0;i100;printf(i%3==0?i%5==0?"Fizzbuzz":"FIZZ":i%5==0?"BUZZ":"%i",i++));
82 Characters
And yes, it’s as dirty as it would ever get … 
I’ve tested it with .NET 2.0 C++ and it works just fine …
public class FizzBuzz
{
public static void main(String [] args)
{
int k;
for(int i=1; i=100; i++)
{
k = 0;
if(i%3==0)
{
k = 1;
}
if(i%5==0)
{
k = k + 2;
}
switch(k)
{
case 1:
System.out.println("Fizz");
break;
case 2:
System.out.println("Buzz");
break;
case 3:
System.out.println("FizzBuzz");
break;
default:
System.out.println(i);
}
}
}
}
Forgive me, but I can’t resist…recursive fizzbuzz:
(Python)
def fizzbuzz(n):
if n:
if n % 15 == 0: return fizzbuzz(n-1) + 'fizzbuzz ';
elif n % 5 == 0: return fizzbuzz(n-1) + 'buzz ';
elif n % 3 == 0: return fizzbuzz(n-1) + 'fizz ';
else : return fizzbuzz(n-1) + ('%d ' % n)
return ''
fizzbuzz(100)
Here’s my simple C program that implements the fuzzbuzz program.
#include stdio.h
int main(void)
{
int i,j=0;
for(i=0;i=100;i++)
{
if((i%3)==0)
{
printf("Fizz");
j=1;
}
if((i%5)==0)
{
printf("Buzz");
j=1;
}
if(!j)
{
printf("%d",i);
}
printf("\n");
j=0;
}
}
well … someone had to write it in C right?
#include "stdio.h"
int main ()
{
for(int i = 1; i 101; i++)
{
if (i%3==0) printf("Fizz");
if (i%5==0) printf("Buzz");
if (i%3 i%5) printf("%d", i);
printf("\n");
}
return (0);
}
CF (SCRIPT) version, for your viewing pleasure…
for (i = 1; i LTE 100; i = i + 1) {
if ((j MOD 3) AND (j MOD 5)) {
WriteOutput(i);
} else {
if (NOT j MOD 3) {WriteOutput('FIZZ');}
if (NOT j MOD 5) {WriteOutput('BUZZ');}
}
WriteOutput(chr(13) chr(10));
}
most people are in it for the money. most people, like me, actually enjoy coding. What do I do before heading to bed? No, I dont read a book, I code. I get butteryflies in my stomach whenever my professors anounce a new C++ programming project; I start on it that same day and do not stop until complete–and perfect. I added a feature to that fizz-buzz program: it prints out the number which is divisible as well. Took me all of 30 seconds to do.
for(int loop = 1; loop = 100; loop++)
{
if ((loop % 3 == 0) (loop % 5 == 0))
{
cout loop ":\tFizzBuzz" endl ;
}else if (loop % 3 == 0)
{
cout loop ":\tFizz" endl ;
}
else if (loop % 5 == 0)
{
cout loop ":\tBuzz" endl ;
}
}
btw, my fiance gets mad when i code early in the morning; i actually had to alt-tab out of the compiler so she wouldnt see it.
And the obligatory Haskeller chimes in with:
fizzbuzz n | n `mod` 15 == 0 = "FizzBuzz"
| n `mod` 5 == 0 = "Buzz"
| n `mod` 3 == 0 = "Fizz"
fizzbuzz n = show n
main = mapM_ (putStrLn . fizzbuzz) [1..100]
That probably won’t format properly because HTML is disallowed and there’s no preview button. What’s up with that?
And now for something completely different. In Lisp. Recursive. And in backwards recursion. For no reason.
(defun fizzbuzz (n)
(when ( n 0)
(fizzbuzz (- n 1))
(format t "~a~%"
(if (= (mod n 3) 0)
(if (= (mod n 5) 0) "FizzBuzz" "Fizz")
(if (= (mod n 5) 0) "Buzz" n)))))
(progn (fizzbuzz 100) (values))
1 Like
piece of cake. took me about a minute to write it in php
then again, i started my days hacking Z80 assembly on a ZX Spectrum…
maybe this lack of “real programmers” is because of the growth of object oriented languages + visual GUIs - and move away from being at the coalface?
here’s “fizzbuzz” in php:
for ($x=1;$x=100;$x++)
{
if ($x%3==0 and $x%5==0)
{
print "$x\tfizzbuzz\n";
}
elseif ($x%3==0)
{
print "$x\tfizz\n";
}
elseif ($x%5==0)
{
print "$x\tbuzz"."\n";
}
else
{
print "$x\n";
}
}
An APL one-liner where INDEX ORIGIN is 0 :
(L,‘Fizz’ ‘Buzz’ ‘FizzBuzz’)[1+(LW=0)+W{assign}(100~0=W)+W{assign}{declose}+/1 20=3 5|{enclose}L{assign}1+{index}100]
“{index}100” creates a numeric vector from 0 to 99, “1+” brings it up to 1…100, which we {assign} to L.
“0=3 5|{enclose}L” compute the modulo of 3 and 5 for L. Enclosing vector L makes it a scalar, which can then be taken as argument of a modulo (primitives in APL requires vectors of same length on each side, or a vector and a scalar, the latter being applied to each item of said vector). This gives us a vector of two vectors (one for modulo 3, the other for modulo 5). “0=” gives us two binary vector where 1s point every solution where the number is a multiple of either 3 or 5.
“W{assign}{declose}+/1 2” multiplies these vectors by 1 and two respectively (now 1s represent modulo 3, and 2s represent modulo 5), and then we sum the two vectors. “/” is an operator which introduce the function between each item of a vector : +/1 2 3 = 1 + 2 + 3
In this case : +/(0 0 1 0 0 1 …)(0 0 0 0 2 0 …) = (0 0 1 0 0 1 …) + (0 0 0 0 2 0 …). Since the result is still a vector of vectors (with only one item), we {declose} it, which makes it a simple vector, and {assign} it to W.
“W{assign}(100~W=0)+” : we then add this vector to another where every number not 0 is replaced by 100. “~W=0” (~ means not) creates a binary vector with 1s everywhere that’s not 0, which we multiply by 100. So now we have (0 0 101 0 102 0 …). We {assign} it to W.
“1+(LW=0)+” : we use the reverse operation “=0” to create a binary vector with 1s everywhere that’s not a multiple of 3 or 5 and “L” to replace every 1 with the original number. We then add this to W, and “voil !”, every number a multiple of 3 or 5 or both is replaced by 101, 102 or 103. Since indexing starts from 0, we substract 1 from the vector “1+” (- is reserved for substraction, indicates negative numbers : 3-1 = 4).
“(L,‘Fizz’ ‘Buzz’ ‘FizzBuzz’)[ …the rest… ]” creates a vector where ‘Fizz’, ‘Buzz’ and ‘FizzBuzz’ are concatenated at the end of L, and [] indicates that we use what’s inside the brackets to index the vector to the left : ‘abc’[1 0 2] = ‘bac’
And there you go : Fizz, Buzz and FizzBuzz replace multiples of 3, 5 or both.
It is possible to create more readable code, but not as fun !
This code, which can be the body of a function, does the same thing in easy to read steps :
List{assign}1+{index}100
Fizz{assign}0=3|List
Buzz{assign}20=5|L
W{assign}Fizz+Buzz
W{assign}W+100~0=W
(L,‘Fizz’ ‘Buzz’ ‘FizzBuzz’)[1+(ListW=0)+W]
2 Likes
My simple C version:
#include stdio.h
int
main (void)
{
int i;
for (i=1; i101; i++) {
if ((i % 3 == 0) (i % 5 == 0)) {
printf ("FizzBuzz\n");
} else if (i % 3 == 0) {
printf ("Fizz\n");
} else if (i % 5 == 0) {
printf ("Buzz\n");
} else {
printf ("%d\n", i);
}
}
return 0;
}