Fizz Buzz

[Catfish]

That's disgusting Bazzy, well done. What compiler do you use (GCC 4.6.1 doesn't compile it).

Here's my contribution of bad taste:
http://ideone.com/UgRe2

Edit:
http://ideone.com/d36p6
http://ideone.com/hI9Yn

[Pravesh Koirala]

I am not getting the fizzbuzz problem. Are we supposed to produce most lengthy solution??

[cire]

I'm amazed at the number of links to code on ideone.com that show incorrect output.

[Bazzy]

@Catfish
GCC 4.7.0

[chrisname]

@Pravesh,
The idea is usually to find the shortest solution, but it's much more fun to write a ridiculous one (one of mine uses a client/server architecture, for example).

[Cubbi]

Since everyone's getting crazy with languages, here's my old favorite:

: Fizz?  ( n -- b )
  3 mod 0= dup if ." Fizz" then
;

: Buzz?  ( n -- b )
  5 mod 0= dup if ." Buzz" then
;

: main ( -- )
  101 1 do
  i Fizz? i Buzz? or invert if i . then cr loop
;

main

ideone: http://ideone.com/SMdZX

[MrHutch]

Not big or clever, but he's a korn shell script that'll do it.

#!/usr/bin/ksh

typeset -i count=1

while [ $count -le 100]; do
   if [ $(($count % 15)) -eq 0 ]
   then
      print "FizzBuzz"
   elif [ $(($count % 5)) -eq 0 ]
   then
      print "Buzz"
   elif [ $(($count % 3)) -eq 0 ]
   then
      print "Fizz"
   fi
   count=$count+1
done
     

[hanst99]

I'm amazed at the number of links to code on ideone.com that show incorrect output.

Which one are you talking about? I didn't check all of them, but I've only seen correct solutions so far (although half the solutions on the last page were for FizzFuzzBuzzBizz rather than FizzBuzz).

[cire]

I missed the branching out, and only took a cursory look at the output. Shame on me.

[1yR4jE8b]

I'm still waiting for the brainfuck solution ;)

[chrisname]

I was thinking that earlier :P

[S6k9GNh0]

I started working on one but my brain imploded.

[rapidcoder]

More FizzBuzz codes:
http://rosettacode.org/wiki/FizzBuzz

BrainF* solution is there, too!

[S6k9GNh0]

I find the Befunge solution harder to understand than the BrainF*** solution.

[chrisname]

Sorry to ressurect a month-old thread, but here's a solution using list comprehensions in Haskell.

let fizzbuzz n = [ fb |
    x <- [1 .. n],
    fb <- [ f ++ b |
        f <- [if x `mod` 3 == 0 then "fizz" else ""],
        b <- [if x `mod` 5 == 0 then "buzz" else ""],
        f <- [if (f == "") && (b == "") then show x else f]
    ]
]

(paste into GHCi and then call fizzbuzz with a number).

[Cheraphy]

(define fizz-buzz
   (lambda (n set)
      (if (< n 1) set
        (cond ((= (mod n 15) 0) (fizz-buzz (- n 1) (append set 'fizzbuzz)))
              ((= (mod n 5) 0) (fizz-buzz (- n 1) (append set 'buzz)))
              ((= (mod n 3) 0) (fizz-buzz (- n 1) (append set 'fizz)))
              (else (fizz-buzz (- n 1) (append set n)))))))
(fizz-buzz 100 '())

Just started learning scheme. I feel I'm doing ok.

[chrisname]

Are you using SICP?

[Cheraphy]

I wasn't, but thank you for bringing that to my attention! Lol, I am now. O:

[BHX]

I was given this as an exercise a few years back only I was told to do Bizz, Buzz, Woof (for 3, 5, and 7):

#include <iostream>
using namespace std;

void bizzBuzz()
{
	for(int i = 1; i <= 100; i++)
	{
		bool print_number = true;
		if(i % 3 == 0)
		{
			cout << "Bizz" << ' ';
			print_number = false;
		}
		if (i % 5 == 0)
		{
			cout << "Buzz" << ' ';
			print_number = false;
		}
		if (i % 7 == 0)
		{
			cout << "Woof" << ' ';
			print_number = false;
		}
		if (print_number)
			cout << "(" << i << ")" << ' '; 
			
		// cout << endl;
	}
}

int main()
{
	bizzBuzz();
	
	return 0;
}

[hanst99]

I'm not a big fan of using monad syntax for Lists though. Oh well, while we're at it...

common lisp:

(format t "~{~a~%~}~%"
 (loop for i from 1 to 100 collecting
              (let ((fizz (= 0 (mod i 3))) (buzz (= 0 (mod i 5))))
                (concatenate 'string
                             (when fizz "Fizz")
                             (when buzz "Buzz")
                             (unless (or fizz buzz) (write-to-string i)))))) 

Or, more general:

(defmacro fizzer (begin end &rest list)                                                                                                       
           `(loop for i from ,begin to ,end collecting                                                                                                 
                 (concatenate 'string                                                                                                                  
                         ,@(loop for (num name) in list collecting                                                                                     
                               `(when (= 0 (mod i ,num)) ,name))                                                                                       
                         (unless (or ,@(loop for (num _) in list collecting                                                                            
                                            `(= 0 (mod i ,num))))                                                                                      
                           (write-to-string i))))) 

Which can be called like this:

(format t "~{~a~%~}~%" (fizzer 1 100 (3 "Fizz") (5 "Buzz")))

Please mind that I am pretty new to CL so I might have made that last part a tad bit more complicated that it needed to be.

EDIT: Fixed calls so it actually prints the results. Forgot that was part of the task.
EDIT2: There's a caveat to the macro solution - it assumes that the values passed for the fizz/buzz pairs are literals. If you use function calls with side effects there it might not work (depending on what those side effects are). I could fix this, but that would be even less readable and the macro probably wouldn't be used that way anyways.