Why isn't my FizzBuzz code processing both if statements when they both match?

Question

For those who don't know, FizzBuzz is the following problem:

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".

Every FizzBuzz solution I find is either some crazy esoteric solution made for the sake of being original, or your basic if-else chain:

for(int i = 1; i <= 100; i++) {

    if(i % 3 == 0 && i % 5 == 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);
    }
}

I am looking for a simple solution that aims to take out the "FizzBuzz" if statement. I have this in mind:

for(int i = 1; i <= 100; i++) {

    if (i % 3 == 0) 
       System.out.print("Fizz");
    if (i % 5 == 0) 
       System.out.println("Buzz")
    else
       System.out.println(i);
}

But this doesn't work. I assume it would be able to print FizzBuzz by entering both ifs, for Fizz and for Buzz, but if the number is, for example, 3, it would print Fizz3. How do I avoid this?

Because it should be `else if` instead of your second `if` statement ;) — Thomas Jungblut, May 16 '12 at 14:03
@ThomasJungblut No, because then the FizzBuzz case will necessarily not work. — Platinum Azure, May 16 '12 at 14:03
@ThomasJungblut, No he wants to chain the Fizz and Buzz together, so he is intentionally leaving out the first else if. — Igor, May 16 '12 at 14:03
Yeah, but he complains that it will print fizzX, just sayin'. — Thomas Jungblut, May 16 '12 at 14:07
After having reread your question, I think it might be worth modifying the title. — Platinum Azure, May 16 '12 at 14:09
Duplicate of [Conditional statement true in both parts of if-else-if ladder](https://stackoverflow.com/questions/26550990/conditional-statement-true-in-both-parts-of-if-else-if-ladder) — TylerH, Apr 14 '20 at 15:57

aioobe · Accepted Answer · 2012-05-16T14:12:41.133

10

What you're trying to do is

if (a)
    ...
if (b)
    ...
else // if neigther a nor b
    ...

This is simply not possible. An else can only belong to a single if. You have to go with the slightly longer variant.

To avoid doing redundant evaluations of the modulo operator, you could formulate the loop body as

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

if (fizz) 
   System.out.print("Fizz");
if (buzz)
   System.out.print("Buzz");
if (!(fizz || buzz))
   System.out.print(i);

System.out.println();

Another one would be

String result = "";

if (i % 3 == 0)   result = "Fizz";
if (i % 5 == 0)   result += "Buzz";
if (result == "") result += i;

System.out.println(result);

edited May 16 '12 at 14:12

answered May 16 '12 at 14:04

aioobe

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1

Ah, but that has an `&&` in it. ;-) – Platinum Azure May 16 '12 at 14:08
Changed according to rule of demorgan... is that cheating? :-) btw, this probably belongs at http://codegolf.stackexchange.com/ – aioobe May 16 '12 at 14:09
1

Well, I was only joking. Only the title mentions the AND operator specifically-- I think the real goal is somehow having only two if statements plus the number case, which your solution incorporates. Also, have a +1. – Platinum Azure May 16 '12 at 14:15
@PlatinumAzure, that's exactly it, I'm still thinking of a better title, haha – bpromas May 16 '12 at 14:30
Selected for the second option. Exactly what I wanted. – bpromas May 16 '12 at 16:49

score 3 · Answer 2 · edited May 23 '17 at 12:15

3

If your only goal is to avoid using &&, you could use a double negation and DeMorgan's laws:

for(int i = 1; i <= 100; i++) {

    if(!(i % 3 != 0 || i % 5 != 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);
    }
}

You can avoid && using the fact that i % 3 == 0 and i % 5 == 0 implies i % 15 == 0, as per RFC1337's answer.

Another solution is to use a switch on the remainder (mod 15, which is 5 times 3):

for(int i = 1; i <= 100; i++) {
    final int mod = i % 15;
    switch (mod) {
        case 0:
        case 3:
        case 6:
        case 9:
        case 12:
            System.out.print("Fizz");
            if (mod != 0) break;
        case 5:
        case 10:
            System.out.print("Buzz");
            break;
        default:
            System.out.print(i);
    }

    System.out.println();
}

edited May 23 '17 at 12:15

Community

1
1

answered May 16 '12 at 14:03

Platinum Azure

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just store i%15 at the top rather than calculating the modulo twice. – Randall Hunt Oct 02 '12 at 03:15
Is there a performance penalty of declaring a final variable over and over again? I seem to remember running into that years and years ago but I can't find any reference to it now. – Randall Hunt Oct 02 '12 at 18:28
I wasn't aware of anything like that. I figured compilation would take a little longer in order to ensure that the value isn't reassigned in the same scope, but I figured the compiler could also perform some optimizations based on the variable being `final`. I usually use `final` where I can. – Platinum Azure Oct 02 '12 at 19:00

score 3 · Answer 3 · answered May 16 '12 at 14:10

3

Your first if statement is all alone.

So, your code hits the first statement, which is ONLY an if statement, and then goes on to the next, which is an if/else statement.

RosettaCode has a good example without using AND operators.

   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;
   }

answered May 16 '12 at 14:10

RFC1337

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1
7

1

I think the intention was to reuse the `Fizz` when writing `FizzBuzz`. – aioobe May 16 '12 at 14:14
1

@aioobe Yeah, and I think this could be easily adapted with a little more nesting of conditionals. I was going to suggest a switch case, but [Platinum Azure](http://stackoverflow.com/a/10620230/1397668) Already provided an example for that. – RFC1337 May 16 '12 at 15:03

score 1 · Answer 4 · answered Jun 02 '16 at 14:06

This is my solution. Granted, it's a bit convoluted (as in roundabout), but I believe it suits your requirement.

int main()
{
    char fizzpass=0;

    unsigned short index=0;

    for(index=1;index<=100;index++)
    {
        if(0 == (index%3))
        {
           printf("Fizz");
           fizzpass = 1;
        }

        if(0 == (index%5))
        {
           if(1 == fizzpass)
           {                  
              fizzpass = 0;
           }

           printf("Buzz\n"); 
           continue;
        }

        if(1 == fizzpass)
        {
            fizzpass = 0;
            printf("\n");
            continue;
        }

        printf("%d\n",index);
    }

    return 0;
}

Regards.

score 0 · Answer 5 · answered Aug 15 '14 at 19:54

Just add a flag variable and use System.out.print:

package com.stackoverflow;

public class FizzBuzz {
    public static void main(String[] args) {
        for (int i = 1; i <= 100; i++) {
            boolean printed = false;
            if (i % 3 == 0) {
                printed = true;
                System.out.print("Fizz");
            }
            if (i % 5 == 0) {
                printed = true;
                System.out.print("Buzz");
            }

            if (printed) {
                System.out.println();
            } else {
                System.out.println(i);
            }
        }
    }
}

score 0 · Answer 6 · answered Apr 25 '15 at 20:48

0

This doesn't take out the if statements but does not use the && (and) operator, you could flip the binary operators.

//FizzBuzz Case
if(!(a % 3 != 0 || a % 5 != 0)){ //flips 
  result[index] = "FizzBuzz";
  index++;
  }

answered Apr 25 '15 at 20:48

TheArchon

313
5
15

score 0 · Answer 7 · answered Feb 08 '16 at 05:00

Don't use an if statement at all.

import java.util.*;
import java.lang.*;
import java.io.*;
class FizzBuzz
{
    public static void main (String[] args) throws java.lang.Exception
    {
        String[] words = {"", "Fizz", "Buzz"};
        String[] nwords = {"", ""};

        for(int i = 1; i < 101; ++i)
        {
            int fp = (i % 3 == 0) ? 1 : 0;
            int bp = ((i % 5 == 0) ? 1 : 0) * 2;
            int np = ((fp > 0 || bp > 0) ? 1: 0);

            nwords[0] = Integer.toString(i);

            System.out.print(words[fp]);
            System.out.print(words[bp]);
            System.out.println(nwords[np]);
        }
    }
}

See it on ideone.

not valid coz (i % 3 == 0) ? 1 : 0 it is an if statement in fact — maxormo, Mar 29 '17 at 23:13

score 0 · Answer 8 · edited Dec 27 '18 at 06:55

public class fizzbuzz 
{

    public static void main(String[] args) 
    {
        String result;
        for(int i=1; i<=100;i++)
        {    
            result=" ";
            if(i%3==0)
            {
                result=result+"Fizz";
            }
            if(i%5==0)
            {
                result=result+"Buzz";
            }
            if (result==" ")
            {
                result=result+i;
            }
            System.out.println(result);
        }
    }

}

This is the most efficient way I could come up with. Hope it helps! :)

score -3 · Answer 9 · answered Dec 13 '15 at 15:30

-3

Crazy albeit unrelated solution done in Python3

#!/usr/bin/python3

for i in range(1,100):
    msg = "Fizz" * bool(i%3==0)
    msg += "Buzz" * bool(i%5==0)
    if not msg:
        msg = i
print(msg)

answered Dec 13 '15 at 15:30

Harrison-SC

7
1
4

This is entirely the wrong language for the question. – TylerH Apr 14 '20 at 16:16

Why isn't my FizzBuzz code processing both if statements when they both match?

9 Answers9

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