# FizzBuzz Solution Dumping Ground

Here’s a PHP solution that actually works

``````\$x=1;

while (\$x<=100)
{
if (\$x%3==0 and \$x%5==0)
{
print "<p>\tfizzbuzz</p>\n";
}
elseif (\$x%3==0)
{
print "<p>\tfizz</p>\n";
}
elseif (\$x%5==0)
{
print "<p>\tbuzz"."</p>\n";
}
else
{
print "<p>\$x</p>\n";
}

\$x++;
}``````

I see no continue being used.

``````for (int i = 1; i <= 100; i++)
{
if (i % 3 == 0 && i % 5 == 0)
{
Console.WriteLine("Fizz-Buzz {0}", i);
continue;
}
if (i % 3 == 0 && i % 5 != 0)
{
Console.WriteLine("Fizz {0}", i);
continue;
}
if (i % 3 != 0 && i % 5 == 0)
{
Console.WriteLine("Buzz {0}", i);
continue;
}
Console.WriteLine(i);
}``````

In C I would use this beautiful oneliner:

``````for(int t=1;t<=100;t++)
( (!(t%3)&&printf("Fizz")) + ((t%5==0)&&printf("Buzz")) || printf("%d", t)), putchar('\n');
``````

Just 2 modulos needed, no continue, no caching, just pure misuse of logical expression evaluation mechanics.

1 Like

Oh man. The article was right.

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One solution in Erlang with recursion.

``````-module(fizzbuzz).
-export([print/0]).

print() ->
FizzBuzzList = eval(100, []),
lists:foreach(fun(Item)->io:format("~p~n", [Item]) end, FizzBuzzList).

eval(Count, Acc) when Count /= 0 ->
Remainders = {Count rem 3, Count rem 5},
NewAcc = case Remainders of
{0, 0} -> ["FizzBuzz"|Acc];
{0, _} -> ["Fizz"|Acc];
{_, 0} -> ["Buzz"|Acc];
{_, _} -> [Count|Acc]
end,
eval(Count-1, NewAcc);

eval(0, Acc) ->
Acc.

``````
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Old but gold, vb6:

`````` Dim i As Integer
Open "C:\output.txt" For Append As #1 ' VB6 does not have native console window support
For i=1 To 100
If i Mod 3 = 0 Then Print #1, "Fizz"
If i Mod 5 = 0 Then Print #1, "Buzz"
If i Mod 15 <> 0 Then Print #1, Cstr(i)
Next
Close #1
``````

Also remembering that VB6 does not have an official console mode support, it’s necessary to waste two lines of code, openning and closing the text file used as Output.

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My 2 cents in Perl. Not the shortest, but it is among the most compact and still readable, doesn’t cheat using a module that does the work and is not exposed, and most importantly, it works CORRECTLY (version 5.10 and above due to use of “say” instead of “print”). Yes, it is amazing how many solutions here do not follow the actual specifications (or plain don’t work):

``````while (\$i++ < 100){
my \$out;
\$out  = "Fizz" unless (\$i%3);
\$out .= "Buzz" unless (\$i%5);
say "\$out" || "\$i";
}``````

This one is simpler, more readable, and no ternary ops:

``````while (\$i++ < 100){
my \$out;
\$out  = "Fizz" unless (\$i%3);
\$out .= "Buzz" unless (\$i%5);
say "\$out" || "\$i";
}

# ... and oh, it works correctly according to the specs.  One is not supposed to print the integer if the strings "fizz", "buzz", or "fizzbuzz" are printed (many solutions in this page disregard that point.``````

Of course we couldn’t forget our both loved and hated VBA in all its glory

Normal Loop:

``````Sub FizzBuzz(n As Integer)
Dim i As Integer

For i = 1 To n
If i Mod 3 = 0 And i Mod 5 = 0 Then
Debug.Print "FizzBuzz"
ElseIf i Mod 3 = 0 Then
Debug.Print "Fizz"
ElseIf i Mod 5 = 0 Then
Debug.Print "Buzz"
Else
Debug.Print i
End If
Next i

End Sub
``````

And some recursive:

``````Sub RecursiveFizzBuzz(n As Integer)

If n > 1 Then
RecursiveFizzBuzz (n - 1)
If n Mod 3 = 0 And n Mod 5 = 0 Then
Debug.Print "FizzBuzz"
ElseIf n Mod 3 = 0 Then
Debug.Print "Fizz"
ElseIf n Mod 5 = 0 Then
Debug.Print "Buzz"
Else
Debug.Print n
End If
Else
Debug.Print n
End If

End Sub
``````

Now imagine that poor soul that made Roller Coaster 2, ALL in assembly…

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Let’s go python:

``````    def FizzBuzz(highestNumber, numbers, words):
for i in range(1, highestNumber + 1):
out = ""
for number, word in zip(numbers, words):
if i % number == 0:
out += word
print(out if out else i)

FizzBuzz(100, [3,5], ["Fizz", "Buzz"])
``````

Boom done. Reusable for different numbers and words

1 Like

Python, could be more compact but this way it is more clear what the program does and less computation only mod two times instead of four.

``````for i in range(0,101):

fizz = i % 5 == 0
buzz = i % 3 == 0

if fizz and buzz:
print("FizzBuzz")
continue
if fizz:
print("Fizz")
continue
if buzz:
print("Buzz")
continue
print(i)``````

In java, but in functional paradigm (don’t do this at work).

``````public static void main(String[] args) {
final String result = IntStream
.rangeClosed(1, 100)
.mapToObj(Scratch::fizzBuzz)
.collect(Collectors.joining(", "));

System.out.print(result);
}

static String fizzBuzz(int n) {
return buzz(n)
.andThen(fizz(n))
.apply(Function.identity())
.apply(String.valueOf(n));
}

static Function<Function<String, String>, Function<String, String>> buzz(int n) {
return test(n, 5, "buzz");
}

static Function<Function<String, String>, Function<String, String>> fizz(int n) {
return test(n, 3, "fizz");
}

static Function<Function<String, String>, Function<String, String>> test(int number, int divisor, String match) {
return f -> d -> test(number, divisor, match, f, d);
}

static String test(int number, int divisor, String match, Function<String, String> next, String defaultValue) {
return number % divisor == 0
? match + next.apply("")
: next.apply(defaultValue);
}
``````
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And since no one seems to have picked MUMPS in thirteen years of responses:

`F I=1:1:100 W !,\$S(I#15=0:"Fizzbuzz",I%3=0:"Fizz",I%5=0:"Buzz",1:I)`

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