Code for "fizzbuzz" doesn't work. Only "fizzbuzz" returned multiple times

Question

Can someone please explain why the following code does not work properly? It only returns "fizzbuzz" 100 times as the answer. Thank you.

def fizzbuzz(number)
  idx = 0
  while idx <= number
    num = number[idx]
    if num % 3 == 0 && num % 5 == 0
      puts 'fizzbuzz'
    elsif num % 5 == 0
      puts 'buzz'
    elsif num % 3 == 0
      puts 'fizz'
    else 
      puts num
    end
    idx += 1
  end
end

fizzbuzz(100)

I didn't realize that there is a subscript operator for `Fixnum`s. Very cool! http://ruby-doc.org/core-2.2.3/Fixnum.html#method-i-5B-5D — Patrick Brinich-Langlois, Feb 14 '16 at 20:05
`num = number[idx]` what it means ? it doesnot makes sense at all — Mani, Feb 14 '16 at 20:12
I using "number[idx]" as a passing variable. thats why I assigned it "num". I didn't realised I should use idx as the passing variable. — Asfand, Feb 14 '16 at 20:49
@Asfand: the question is _why_ did you use `number[idx]` and not, say, `number`? — Sergio Tulentsev, Feb 14 '16 at 21:12

score 1 · Accepted Answer · edited Feb 14 '16 at 21:33

1

You have two issues here:

num = number[idx]

The idx variable is the one that should be used for your checks not num:

def fizzbuzz(number)
  idx = 0
  while idx <= number
    if idx % 3 == 0 && idx % 5 == 0
      puts 'fizzbuzz'
    elsif idx % 5 == 0
      puts 'buzz'
    elsif idx % 3 == 0
      puts 'fizz'
    else 
      puts idx
    end
    idx += 1
  end
end

fizzbuzz(100)

The above is your solution.

The reason it is always printing 'fizzbuzz', no matter what number you pass to the fizzbuzz method, is because you are assigning number[idx] to the a local variable num where you should be using the idx counter in your conditional statements.

You need to print the numbers from 1 to the number you pass to fizzbuzz. For multiples of 3 print “Fizz” instead of the number and for the multiples of 5 print “Buzz”. For numbers which are multiples of both 3 and 5 print “FizzBuzz".

edited Feb 14 '16 at 21:33

the Tin Man

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answered Feb 14 '16 at 20:04

Cyzanfar

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"because you are number[idx] to the a local variable num. The return value number[idx] is 0" – Kimmo Lehto Feb 14 '16 at 20:23
the return value of number[idx] is actually not always 0. 100[2] for example is 1. http://ruby-doc.org/core-2.2.3/Fixnum.html#method-i-5B-5D – Kimmo Lehto Feb 14 '16 at 20:31
So is it correct to assume that it is always `0` for any `Integer` ? – Cyzanfar Feb 14 '16 at 20:33
No. For example: for 7[0]..[3] you get 0111 for 255[0]..[7] you get 1111 1111. For 65535 you get 1 sixteen times. – Kimmo Lehto Feb 14 '16 at 20:39
1

Thank you so much @Cyzanfar for your answer/help. Helped a lot. I am new to programming, just started learning. – Asfand Feb 14 '16 at 20:52
@Cyzanfar: why would you think it is always 0? That'd make quite useless operator, wouldn't it? – Sergio Tulentsev Feb 14 '16 at 21:11
Yes thanks for pointing that out, I've removed that part and updated my answer. – Cyzanfar Feb 14 '16 at 21:13

Code for "fizzbuzz" doesn't work. Only "fizzbuzz" returned multiple times

1 Answers1